5 .1 M e a s u r ing e n e r g y c h a nge s Enthalpy and thermochemistry A omt is any apparatus used to measure the amount Enthalpy is an example of a state function. For a state function any of heat being exchanged change in value is independent of the pathway between the initial and with the surroundings. In an nal measurements. Other examples of state function include volume, exothermic reaction heat is temperature and pressure. generated which is transferred to the surroundings. In an For example, taking the temperature of the water in a swimming endothermic reaction heat pool early in the morning (the initial value) and then again in the is consumed. In the school afternoon (nal value), does not tell the whole story of any temperature laboratory, experiments focus uctuations that may have occurred throughout the day. The calculation on the change in temperature is a simple one. However, it does not give any indication of the heating of the reaction solvent, which and cooling which has occurred throughout the day. in most cases is water. ∆T = T T (nal) (initial) Thermochemistry is the study of heat changes that occur during exothermic reaction chemical reactions. At constant pressure, the change in enthalpy ∆H is dened as the heat transferred by a closed system during a chemical reaction. The term “change in enthalpy” or “heat of reaction” is commonly used when describing the thermodynamics of a reaction. The unit of enthalpy change ∆H is kJ. ygrene laitnetop a reactants Endothermic and exothermic reactions ∆H A chemical reaction in which heat is transferred from the system products to the surroundings is dened as an exothermic reaction with a negative ∆H. In contrast, chemical reactions that absorb heat from reaction pathway their surroundings are dened as endothermic reactions with a Figure 2 In an exothermic reaction the positive ∆H enthalpy of the products is lower than that of the reactants. The products are described The calculations involved in investigating the energetics of the reaction as being energetically more stable than the reactants. (We shall study the activation between zinc and aqueous copper(II) sulfate, CuSO are described energy of a reaction in topic 6) 4 later in the topic. Measured quantities of copper(II) sulfate solution and zinc are mixed in a calorimeter. The mixture is stirred and the change in temperature measured using a thermometer or data-logging equipment. endothermic reaction Zn(s) + CuSO (aq) → Cu(s) + ZnSO (aq) 4 4 The result of calculating ΔT shows that this single replacement reaction ygrene laitnetop products involving the displacement of the copper(II) ion by the metal zinc is an a ∆H exothermic reaction (gure 2). Ammonium nitrate, NH NO is an important component of fertilizers. 4 3 When the solid dissolves in water to form aqueous ammonium and reactants nitrate ions, the reaction requires heat to proceed. This heat is absorbed reaction pathway Figure 3 In an endothermic reaction the by the system from the surroundings, resulting in a decrease in the enthalpy of the products is greater than that of the reactants. The products are described temperature of the surroundings as recorded by a thermometer. The as being energetically less stable than the reactants apparatus containing the reaction feels cold to touch. This is an example of an endothermic reaction (gure 3). + NH NO (s) → NH (aq) + NO (aq) 3 4 3 4 141
5 ENERGE TICS AND THERMOCHEMISTRY Calculating enthalpy changes in aqueous solutions We stated earlier that the change in enthalpy ∆H or 1 K. For example, the specic heat capacity of is dened as the heat transferred by a closed system during a chemical reaction. To calculate 1 1 ∆H for a reaction we therefore need to nd the heat change. When calculating the heat change of copper is 0.385 J g K while that of ethanol is a pure substance such as water, we need to have an understanding of the physical quantity, the 1 1 specic heat capacity, c 2.44 J g K . The lower the specic heat capacity of a given substance, the higher the temperature rise achieved for the same amount of heat transferred to the sample. 1 1 Specic heat capacity is an intensive property The units for specific heat capacity are kJ kg K . that does not vary in magnitude with the size of 3 1 The specific heat capacity of water is 4.18 kJ kg the system being described. For example, a 10 cm sample of copper has the same specic heat capacity as a 1 tonne block. 1 K and this can be found in section 2 of the Data booklet Specic heat capacity is used to calculate the heat q of a system using the relationship: The specic heat capacity of a pure substance is q = mcΔT dened as the amount of heat needed to raise the temperature of 1 g of the substance by 1 °C where m is mass in kg and ∆T is the change in temperature. Wokd xmp : t tp Example 1 Example 2 When a 1.15 g sample of anhydrous lithium chloride, LiCl 180.0 J of heat is transferred to a 100.0 g sample of was added to 25.0 g of water in a coee-cup calorimeter, iron, resulting in a temperature rise from 22.0 °C to a temperature rise of 3.80 K was recorded. Calculate the 26.0 °C. Calculate the specific heat capacity of iron. enthalpy change of solution for 1 mol of lithium chloride. Solution Solution ∆T = (299 295) K = 4 K . q = mc∆T = 0.025 kg × 4.18 kJ kg 1 1 Make c the subject of the equation and solve: K × 3.80 K q __ c = m∆T = 0.397 kJ _0 _180 k_J = Conver t to energy gained for 1 mol of LiCl. 0.100 g × 4 K 0.397kJ / 1.15g LiCl × 42.394 g/mol = 14.6 kJ/mol LiCl 1 = 0.450 kJ K 1 ∆H = -q = -14.6 kJ mol co-p omt thermometer Performing reactions in a polystyrene coee cup to measure the enthalpy change is a convenient experimental procedure. The methodology introduces glass stirrer systematic errors that can be analysed and the eect of their directionality cork stopper assessed. two polystyrene cups nested stmt o are a consequence of the experimental procedure. Their together containing reactants eect on empirical data is constant and always in the same direction. With the in solution coee-cup calorimeter, the measured change in enthalpy for a reaction will Figure 4 A coee-cup calorimeter always be lower than the actual value, as heat will be transferred between the contents and the surroundings in every experiment. 142
5 .1 M e a s u r ing e n e r g y c h a nge s Investigation to nd the molar enthalpy change for a reaction Earlier we looked at the exothermic metal ● The specic heat capacity of an aqueous displacement reaction between zinc and copper(II) solution is the same as that of water. sulfate: Zn(s) + CuSO (aq) → Cu(s) + ZnSO (aq) Loss of heat from the system to the surroundings is the main source of error in this experiment and 4 4 one that is difcult to quantify. The change in temperature ∆T calculated from a graph will include The following method is used to calculate the a systematic or directional error. This loss of heat molar enthalpy change for this reaction from means that the maximum temperature recorded the equation: will be lower than the actual value, making the calculated value of q lower than the actual value. q = mc∆T The effect of errors in the procedure on the result of subsequent calculations is important in considering Experimental method to determine ΔT improvements in experimental procedures. 1 Using an electronic balance, accurately 3 3 measure the mass of 25 cm of 1.0 mol dm CuSO solution. Subtract the mass of the An accepted method of calculating the maximum 4 temperature to compensate for systematic errors in data is to look at the cooling section cylinder from the mass of the cylinder + of the curve after the reaction is complete, and extrapolate this back to the point of introduction solution following the transfer of the solution of the zinc, as shown in gure 5. A more accurate value for ΔT can then be determined. to the coffee-cup calorimeter. 2 Using a thermometer or a temperature probe and related software, record the temperature of the solution every 30 seconds for up to 3 minutes, or until a constant temperature is achieved. Calculation of molar enthalpy change 3 At 3 minutes, introduce powdered zinc (between 1.3 g and 1.4 g, previously weighed) M of opp(ii) ft oto/ 28.8 M of z/ 1.37 and commence stirring. 4 Continue to take temperature readings for up c tmpt T() T(t)/°c 39.0 to 5 minutes after the maximum temperature Table 1 Sample results has been reached. 5 Produce a temperature versus time graph to Taking the results in table 1 we can calculate the molar enthalpy change as follows: determine the change in temperature. q = mc∆T Assumptions and errors 1 1 × 39.0 K A number of assumptions are made when using = 0.0288 kg × 4.18 kJ kg K this method: = 4.69 kJ ● The heat released from the reaction is completely transferred to the water. ● The coffee cup acts as an insulator against C°/erutarepmet heat loss to the surroundings. However, the coffee cup also has a heat capacity and heat ∆T is transferred to it from the water. It would be difcult to quantify the heat capacity of a polystyrene cup, so it is assumed to be zero. ● The maximum temperature reached is an accurate representation of the heat evolved during the reaction. 0 2 4 6 8 10 time/min Figure 5 Determination of the change in temperature in calorimetry experiments 143
5 ENERGE TICS AND THERMOCHEMISTRY amount of CuSO = 1.37 g × _1_ The shape of the graph and the change in 4 1 temperature from a lower to a higher value lead 65.38 g mol to the conclusion that the reaction is exothermic: = 0.0210 mol 1 4.69 kJ ΔH = -223 kJ mol __ molar enthalpy change = 0.0210 mol 1 = 223 kJ mol TOK Tmpt The SI unit of temperature is the kelvin (K). Note that a change in temperature ΔT In theory of knowledge there calculated from experimental data in Celsius will be identical to the value of ΔT are eight specic ways of calculated in kelvin. knowing. These are: language, Throughout the world, the majority of countries use the Celsius scale for the sense perception, emotion, everyday description of temperature. As the Celsius and kelvin scales are linked, reasoning, imagination, you will often see both scales being used in an IB question. The USA uses a faith, intuition, and mixture of metric and imperial units of measurement. For example, the Fahrenheit memory. Scientists perform scale of temperature is used in the USA . experiments and process the raw data to enable us °F to draw conclusions. We 100 compare experimental and 80 theoretical values. What 60 criteria do we use when 40 making these comparisons? 20 Are our judgments subjective 0 or objective? When analysing 20 and appraising experimental limitations and making Figure 6 We use SI units in science, but a mix ture of imperial and metric systems of theoretical assumptions, measurement is used in dierent countries which of the ways of knowing are we utilizing? The tdd tp of to ΔH is determined Enthalpy change of formation 298 The change in enthalpy during a reaction can be determined using the following equation: at temperature 25 °C/298 K and pressure 100 kPa with all species in their standard state. stdd ΔH reaction = ∑(ΔH products) – ∑(ΔH reactants) ff odto are denoted by the symbol . ΔH is the standard enthalpy change of formation of a substance. f This is the energy change upon the formation of 1 mol of a substance from its constituent elements in their standard state. We can use existing enthalpy of formation data to calculate the enthalpy of std tp reaction. The value and sign of the calculated enthalpy of formation Section 12 in the Data booklet gives the standard enthalpy of informs us about the energetics of the reaction. formation for a large number of common compounds. In For example, the standard enthalpy change of formation for methane is: examinations, questions will provide any other values not 1 included in the Data booklet C(s) + 2H (g) → CH (g) ∆H = -74.9 kJ mol 144 f 2 4 It is important to note that the elements carbon and hydrogen are represented in their standard states. Equations for ∆ H must represent f the formation of 1 mol of a substance. In some cases, such as the
5 .1 M e a s u r ing e n e r g y c h a nge s formation of phenol shown below, this results in 1 mol of diatomic 2 oxygen O appearing on the reactant side: Qk qto Write equations to describe 2 the standard enthalpy change of formation for the following _1 1 compounds and state the O enthalpy value by referring to 6C(s) + 3H (g) + (g) → C H OH(s) ∆H = -165.0 kJ mol the Data booklet 2 f 2 6 5 2 Enthalpy change of combustion The standard enthalpy change of combustion ∆H is the heat c evolved upon the complete combustion of 1 mol of substance. ) propane The enthalpies of combustion found in section 13 of the Data booklet b) chloromethane are values derived under standard conditions. Butane, one of the gases classied as liqueed petroleum gas (LPG) is highly ammable: ) ethanol d) benzoic acid CH (g) + _13 O (g) → 4CO (g) + 5H O(l) 2 2 2 2 4 10 ) carbon monoxide 1 f) methylamine ∆H = -2878 kJ mol c This thermochemical equation can also be written with the 1 enthalpy of combustion value included in the equation. The negative enthalpy change indicates an exothermic reaction so the value would compod ∆H /kJ mo be included on the product side: f C H (l) +49.0 6 6 CO (g) 393.5 2 _13 1 2 CH (g) + O (g) → 4CO (g) + 5H O(l) + 2878 kJ mol 2 4 10 2 2 H O(l) 285.8 2 Working method Table 2 Standard enthalpy changes of formation. Benzene, C H is highly ammable, producing a sooty ame: 6 6 2C H (l) + 15O (g) → 12CO (g) + 6H O(l) 6 6 2 2 2 ∆H reaction = ∑(∆H products) - ∑(∆H reactants) ff = [12 × ( 393.5) + 6 × ( 285.8) 2 × (+49.0) 15 × 0] kJ =( 4722 1714.8 98.0) kJ = -6535 kJ Investigation to nd the enthalpy change of combustion 3 The enthalpy change of combustion of common 2 Accurately determine the mass of 30 cm of alcohols can be determined in the laboratory. 3 water contained in a 250 cm beaker. From a homologous series of alcohols, patterns 3 Place the beaker or metal calorimeter on a in enthalpy change of combustion values can be determined and subsequently analysed. tripod with the spirit burner beneath. 4 Using either a temperature probe or a Experimental method thermometer, determine and record the initial temperature of the water. The following procedure utilizes equipment available in a standard school laboratory. Five 5 A spirit burner is lit under the calorimeter and spirit burners are required, each containing one of the alcohols methanol, ethanol, propan-1-ol, the alcohol is burnt to heat the water. The butan-1-ol and pentan-1-ol. period over which it burns can be monitored in one of two different ways: 1 Determine the initial mass of the spirit burners a) allow each alcohol to burn until a using an electronic balance. temperature change of 30 °C is achieved 145
5 ENERGE TICS AND THERMOCHEMISTRY Qk qto b) allow each alcohol to burn for a period of 2 minutes. Write equations to describe the standard enthalpy change of 6 After this time period, extinguish each spirit burner by replacing the combustion for the following compounds, and state the cap, re-weigh each one and record the change in mass of the alcohol. enthalpy value by referring to section 134 of the Data booklet Calculation of enthalpy of combustion aoo ∆m of oo/ ∆T/°c M of wt/ methanol 0.348 30.0 31.2 ) octane, C H 8 18 Table 3 Sample results b) chloroethane, C H Cl (hint: 2 5 a corrosive strong acid is q = mc∆T one of the products) 1 1 = 0.0312 kg × 4.18 kJ kg K × 30.0 K ) cyclohexanol, C H O 6 12 = 3.91 kJ d) methanoic acid, CH O 0.348 g __ 2 2 1 amount of CH OH = 32.05 g mol = 0.0109 mol 3 ) glucose, C H O 6 12 6 3.91 kJ __ molar energy change = 0.0109 mol 1 = 359 kJ mol As in all investigations, rst 1 determine the dependent and ΔH = –359 kJ mol independent variables and the variables that will be controlled. Assumptions ● Heat loss to the environment is negligible (in reality, it is signicant but cannot be quantied). ● All the alcohols are pure and undergo complete combustion. Dt o can be used to Obt d t ott of food record temperature changes accurately and the associated The world increase in obesity software to perform data analysis and graphing. The Obesity, eating disorders, and unhealthy diet are serious health issues facing many use of data-logging equipment cultures throughout the world, as many societies become more auent and food is demonstrates the practical readily available. Obesity is generally dened as an excessive accumulation of fat application of technology in the that can lead to health problems. The bod m dx (BMi) is found by taking a laboratory. person’s mass (in kilograms) and dividing it by the square of their height (in metres). Energetics experiments An adult with a BMI above 25 is considered overweight while one with a BMI provide a useful set of greater than 30 is obese. The World Health Organization (WHO) has been raw data and involve monitoring the eect of changes in diet on dierent nations for decades. Their experimental procedures research has found the following: that can be evaluated for random and systematic errors ● In 2013 the occurrence of obesity worldwide was more than double the level (topic 11). The identication of the systematic errors in 1980. and examination of their directionality is an essential ● In 2008 over 1.4 billion adults worldwide were overweight, with approximately aspect of the analysis of experimental results. 200 000 000 men and 300 000 000 women being classied as obese. ● 65% of the world’s population reside in countries where more people die from obesity-related causes than from being underweight. ● More than 40 000 000 children under the age of 5 years were overweight in 2010. ● Obesity is a preventable disease. 146
5 .1 M e a s u r ing e n e r g y c h a nge s In China, the rapid increase in auence and the globalization of the economy has seen an unprecedented expansion in the fast-food industry and of nutritional choices. With these has come a signicant increase in the number of children who are overweight and obese. Type 2 diabetes is normally associated with adults, but the rise in the prevalence of the disease amongst children in China and in other countries is seen as a signicant threat to the wellbeing of future generations. Food labelling and determination of energy content Governments throughout the world have a responsibility to their citizens to provide leadership, education, and guidance in health and nutrition. Linked to the globalization of economies and free-trade agreements has been the standardization of labelling of food products to include an analysis of the contents including energy content. To determine their energy content, foods were traditionally placed in a calorimeter surrounded with water and completely burnt, causing the water to rise in temperature. This temperature change was then used to calculate the energy content (sometimes referred to as “caloric value”) of the food. Today the preferred method of calculation is using the Atwater system. This system relies on average energy values for proteins, carbohydrates, fats, and alcohol being applied to foods of a known composition. The National Data Laboratory (NDL) in the USA holds information on the energy content of over 6000 foods. Figure 7 Nutritional information displayed on food packaging 147
5 energe Tics anD TherMOcheMisTry 5.2 h’ w Understandings Applications and skills ➔ The enthalpy change for a reaction that is ➔ Application of Hess’s law to calculate enthalpy carried out in a series of steps is equal to changes. the sum of the enthalpy changes for the ➔ Calculation of ∆H reactions using ∆H data. f individual steps. ➔ Determination of the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes. Nature of science ➔ Hypotheses – based on the conser vation of energy and atomic theory, scientists can test the hypothesis that if the same products are formed from the same initial reactants then the energy change should be the same regardless of the number of steps. TOK Testing hypotheses In TOK , a primary focus is on questions about knowledge Experimental evidence enables scientists to prove or disprove a which are open ended hypothesis. Based on the principles of conservation of energy and with multiple perspectives atomic theory, scientists are able to test experimentally the hypothesis and expressed without that when products are formed from the same set of reactants, the using subject-specic change in enthalpy should be identical regardless of the route taken language. Hess’s law can be and the number of chemical reactions involved. Quantitative data considered an application can be analysed and used as evidence for thishypothesis. of the law of conservation of energy. What are the Overall and net reactions challenges in applying general principles of a law If you have travelled to New York, Tokyo, London, Hong Kong, Paris, to something as specic as Beijing, Berlin, or Seoul, you will have experienced the subways that Hess’s law? criss-cross these enormous cities and transport millions of people every day. In any transport network there is more than one way to travel 148 between point A and B. For the adventurous traveller, half the fun is often working out which is the fastest route. The same idea can be true in the eld of chemistry. A chemical equation usually shows the net reaction – it is a summary of a number of different reactions, which when added together result in an overall reaction. Hess’s law is an application of the conservation of energy law: Regardless of the route by which a chemical reaction proceeds, the enthalpy change will always be the same providing the initial and nal states of the system are the same. Figure 2 shows that summing all the equations, taking into account the direction and magnitude of each, results in an overall equation. Hess’s law can then be applied to nd the enthalpy change for the reaction.
5.2 he ss’s l aW A H y H x C H z B Figure 2 Hess’s law Figure 1 You can take many alternative routes on the Paris Metro Worked example : calculating enthalphy change Using Hess’s law, and the following information, reversing the chemical equation we must change the sign of the enthalpy value: calculate the enthalpy change ΔH for the reaction: 4 _1 _1 O CO + 2H O → CH OH + 1 O 2 2 3 C + 2H + → CH OH ∆H (4) 2 2 3 4 (1) 2 2 2 _1 ∆H = +676 kJ 1 CH OH + 1 O → CO + 2H O 3 2 2 2 2 ● Because oxygen is found in all three equations, the next point of focus ∆H = -676 kJ should be hydrogen. We require 2 mol of 1 hydrogen as a reactant. Therefore, equation 3 can be used in the direction as written but C+O → CO ∆H = -394 kJ (2) with double the number of moles. This means 2 that the enthalpy value must be doubled: 2 2 H + _1 → HO ∆H = -242 kJ (3) 2 O 2 3 2 2 IB, May, 2006 2H + O → 2H O ∆H = -484 kJ 2 2 2 3 Solution ● Now we can add the three equations together, eliminating those species common to both ● Look at the overall equation 4 for the enthalpy sides and summing the enthalpy values as ● of formation of methanol. shown in table 1. From equations 1–3, the reactant carbon and rtt Podt etp the product methanol should be the main focus of your methodology. ● For carbon, we require a reaction that uses C+ → ΔH = –394 kJ 1 mol of carbon as a reactant. Carbon is a 2 reactant in equation 2, so this equation can be used as written: + O → _ ΔH = +676 kJ 1 CH OH + 1 3 2 _1 2H + O → ΔH = –484 kJ 2 3 2 ΔH = –202 kJ 2 4 2 C+O → CO ∆H = -394 kJ 2 2 2 C + 2H → CH OH 2 3 ● For methanol we need to use equation 1, _1 but we need to reverse the equation so that methanol is a product not a reactant. When + O 2 2 Table 1 149
5 ENERGE TICS AND THERMOCHEMISTRY The combination of these reactions can also C + 2H + 1 CH OH be represented diagrammatically as shown in 3 gure 3. 2 O 2 You will notice that the enthalpy cycle diagram shows the combustion of methanol equation 2 reversed, in the same way it was during the summation of equations method. 1 O O 1 O 2 2 2 2 CO + 2H O 2 2 Figure 3 Enthalpy cycle for the formation of methanol T t of With the world population exceeding 7 billion people, phosphor dust, and other hazardous substances. Guiyu in and increasing international concern over the world’s China has become a centre for e-waste disposal and the resources, recycling of materials has become mainstream rapid expansion of recycling processes in villages around in many countries. Recycling developed from a desire to Guiyu has resulted in heavy metal contamination of the use raw materials more eciently, to reduce energy use groundwater and soil, and air pollution from the burning of in the production of goods, to protect the environment plastics within the waste. from excessive pollution, and to utilize waste materials and thereby reduce landll waste disposal. However, Despite good intentions, the end result of recycling can economic and other pressures to recycle can lead be extremely negative. What is more, the eciency to potentially harmful impacts on the environment. of recycling processes in energy terms varies widely. For example, electronic or e-waste is an escalating Many countries and environmental organizations are problem throughout the world. This waste contains investigating how we can address the long-term heavy metals such as lead and cadmium, highly toxic eects of recycling programmes on the environment and communities. Figure 4 Technology waste recycling in Guiyu, China 150
5.2 he ss’s l aW Worked example Determine the enthalpy of formation of ethane, In summary: CH using the enthalpy of combustion data in 2 6 ■ section 13 of the Data booklet The equation for the combustion of carbon is doubled. ■ The equation for the combustion of hydrogen is tripled. Solution ● Write a balanced chemical equation for the ■ The equation for the combustion of ethane is reversed. formation of 1 mol of ethane: ∆H f 2C(s) + 3H (g) C H (g) 2 2 6 2C(s) + 3H (g) → C H (g) 2 2 6 ● Write equations for the combustion of carbon, hydrogen, and ethane and determine their enthalpy values from the Data booklet: ∆H = –84 kJ c 1....C(s) + O (g) → CO (g) Figure 5 shows how the enthalpy of formation can be found using an enthalpy cycle diagram. 2 2 ∆H = -393.5 kJ c ∆H f 2....H (g) + _1 O (g) → H O(l) 2C(s) + 3H H (g) 2 2 2 2 2 2 6 ∆H = -286 kJ c 3....C H (g) + _7 (g)→ 2CO (g) + 3H O(l) O 2 2 2 6 2 2 3 7 O O ∆H = -1561 kJ 2 2 c 2 2 ● Multiply or reverse the sign of the enthalpy 2O change for each equation accordingly. 2 3(-286) + 1561 2(-394) ● Now combine the equations to form the net enthalpy of formation equation for ethane: 1....(×2): 2C(s) + → (g) 2 2 2CO (g) + 3H O(l) 2 2 ∆H = -787 kJ c Figure 5 Alternative method: enthalpy cycle to nd the __ enthalpy of formation of ethane 2....(×3): 3H (g) + → 2 2 2 ∆H = -858 kJ c 3....(reversed): 2CO (g) + → 2 2 2 _ (g) + 2 ∆H = +1561 kJ c std tp You may nd the summation of equations method easier when working out the direction of the equations and the mole coecients. During examinations you may be asked to use the summation of equations method and/or construct an enthalpy cycle. Often candidates make simple arithmetical errors when calculating the enthalpy of reaction. It is advisable to clearly show your full working rather than simply recording the nal answer. This gives the examiner the oppor tunity to assign par t marks where applicable. 151
5 energe Tics anD TherMOcheMisTry 5.3 Bod tp Understandings Applications and skills ➔ Bond forming releases energy and bond ➔ Calculation of the enthalpy changes from breaking requires energy. known bond enthalpy values and comparison ➔ Average bond enthalpy is the energy needed of these with experimentally measured values. to break 1 mol of a bond in a gaseous molecule ➔ Sketching and evaluation of potential energy averaged over similar compounds. proles in determining whether reactants or products are more stable and if the reaction is exothermic or endothermic. ➔ Discussion of the bond strength in ozone relative to oxygen in its impor tance to the atmosphere. Nature of science ➔ Models and theories – measured energy changes can be explained based on the model of bonds broken and bonds formed. Since these explanations are based on a model, agreement with empirical data depends on the sophistication of the model and data obtained can be used to modify theories where appropriate. Modelling energy changes changes in reactions can be understood using models of bond breaking and bond making. The Scientic models are developed to explain degree of agreement between these models and certain processes that cannot be observed the empirical data obtained in the laboratory is directly. Based on a the o r e ti ca l und e r s t a n di n g dependent on the validity of the model and the of the processes, such models can produce accuracy of the data. evidence in support of the theories, or can inform modications to the theories. Energy Bond enthalpy The breaking of the hydrogen molecule into individual hydrogen atoms requires energy. The bond enthalpy of a bond (the H H bond in this example) is dened as the energy required to break 1 mol of bonds in gaseous covalent molecules under standard conditions. Bond breaking is an endothermic process and has a positive enthalpy value, for example: 1 H (g) → 2H(g) ∆H = +436 kJ mol 2 Bond enthalpy is also referred to as bond dissociation enthalpy , and selected values are provided in section 11 of the Data booklet and in table1. These are average values and are therefore only an 152
5 .3 BOnD enThalP y approximation. They are derived from experimental data involving the breaking of the same bond found in a wide variety of compounds. Bod av bod 1 For example, the C H bond enthalpy will vary through the alkane HH OO tp/kJ mo series as the chemical environment of the individual bonds changes. O=O 436 OH 144 If a molecule of methane underwent a series of steps in which one CH 498 CC 463 hydrogen atom was removed at a time, the bond dissociation enthalpy C=C 414 C≡C 346 would be different each time, as the chemical environment changes CO 614 C=O 839 upon the removal of successive hydrogen atoms. Additionally, the bond C=N 358 C≡N 804 enthalpy for molecules in the gaseous state does not take account of the NN 615 N=N 890 intermolecular forces that exist. These limitations are not considered to N≡N 158 Cl Cl 470 be signicant and the average bond enthalpy is an accepted value to use Br Br 945 242 for enthalpy of reaction calculations. II 193 151 Bond length Consider the molecules hydrogen uoride H F, hydrogen chloride H Cl, hydrogen bromide H Br, and hydrogen iodide H I. In topic 3 we discussed how as you move down group 17 the atomic radius increases with increasing atomic number, Z. The consequence of this is that bond length increases, and bond strength decreases, in the hydrogen halides as you move down group 17 (table 2). Bod HF H Cl H Br HI Bod t/pm 92 128 141 160 Table 2 Bond lengths of the hydrogen halides Bond strength Table 1 Average bond enthalpies at 298 K The bond enthalpy reects the strength of the covalent bond. As we move from single to double to triple bonds the number of electrons in the bond increases resulting in an increase in electrostatic forces and a shortening of the bond length. This trend can be seen in the carbon– carbon bond of the homologous series of alkanes, alkenes, and alkynes. Bond polarity The polarity of a bond can be described by the difference in electronegativity of the bonded atoms (tables 3 and 4). atom etot vt Bod ∆(etot vt) Bod tp/ 1 kJ mo H 2.2 HF 1.8 567 F 4.0 H Cl 1.0 431 Cl 3.2 Br 3.0 H Br 0.8 366 δ F Table 3 Electronegativity values Table 4 Bond polarity and bond enthalpy at 298 K δ+ H Fluorine has the highest electronegativity value of any element. The polarity of the H F bond results in a partial charge on each atom. The bond is said to have ionic character. The partial charges Figure 1 The polar H F bond attract one another, increasing the strength of the bond (gure 1). 153
5 ENERGE TICS AND THERMOCHEMISTRY Worked example: using bond enthalpies to nd the enthalpy change of reaction Using the data from section 11 of the Data booklet nd the enthalpy change for the electrophilic addition of hydrogen bromide to ethene to form bromoethane. CH (g) + HBr (g) → C H Br (g) 2 4 2 5 ∆H = ∑(BE bonds broken) - ∑(BE bonds formed) ∆H = [4BE + BE + BE ] [5BE + BE + BE ] C–H C=C H–Br C–H C–C C–Br ∆H = [(4 × 414) + 614 + 366] [(5 × 414) + 346 + 285] 1 ∆H = 2636 2701 = -65 kJ mol The bond enthalpy calculated will vary signicantly from the calculation using ∆H reaction = ∑∆H (products) - ∑∆H (reactants) f f ( 105.7kJ) because bond enthalpies are average values. Additionally, when liquids are involved in the reaction, bond enthalpy calculations do not take into account the intermolecular forces within the liquids. std tp A frequent error made by candidates is to confuse the dierent equations for the calculation of a change in enthalpy. For bond enthalpy, think in terms of bond breaking and formation: ∆H = ∑(BE bonds broken) - ∑(BE bonds formed) For enthalpy of formation, think in terms of products and reactants: ∆H reaction = ∑(∆H products) - ∑(∆H reactants) f f Bond enthalpy values and enthalpies of combustion Gasoline or petrol is produced by the fractional distillation of petroleum and used to power various modes of transport. The automotive industry has witnessed signicant changes in its markets as the demand for automobiles in developed economies is being overtaken by that of the developing economies. The enthalpy of combustion of octane, C H can be calculated using 8 18 bond enthalpy values from section 11 of the Data booklet: _25 CH (g) + O (g) → 8CO (g) + 9H O(g) 2 8 18 2 2 2 ∆H = ∑(BE bonds broken) - ∑(BE bonds formed) c = (18BE + 7BE + _25 ) (16BE + 18BE ) C–H C–C BE C=O O=O O–H 2 = (18(414) + 7(346) + _25 (16(804) + 18(463)) (498)) 2 1 = 16 099 21 198 = -5099 kJ mol 154
5 .3 BOnD enThalP y In comparison, the experimentally determined value for the enthalpy change of combustion ∆H for octane (section 13 of the Data booklet) is Qk qto c 1 5470 kJ mol 1 In the chlor-alkali industry The reason for the difference is that when calculating the enthalpy chlorine gas is combined change using bond enthalpy values, it is assumed that the reaction takes place in the gaseous state, with no intermolecular forces involved. with hydrogen gas However, the experimentally derived enthalpy of combustion involves octane and water in their standard states, namely liquid. Additionally, to produce hydrogen as mentioned above, all bond dissociation enthalpy values are averaged across a wide range of related compounds so they represent only an chloride: approximation of the true value. Cl (g) + H (g) → 2HCl(g) 2 2 If the enthalpy change for his reaction is 185 kJ, calculate the bond enthalpy for the H–Cl bond. Ozone 2 Methane and chlorine react Ozone, O is both created and destroyed in the stratospheric layer of 3 to produce chloromethane Earth’s a tmosphere . Ul tr a v io l e t ( UV ) r a ys fr o m t h e s un a r e a bso r be d and hydrogen chloride. by oxygen, O , splitting the molecule into single oxygen atoms. ) Write the balanced 2 These oxygen atoms can then combine with oxygen molecules to chemical equation for form ozone: this reaction. UV b) Using bond enthalpy O (g) → O (g) + O (g) 2 values from the Data O (g) + O (g) → O (g) 3 2 booklet, determine Ozone is very effective at absorbing harmful long- and short-wavelength the enthalpy change UV radiation. This absorption breaks down the ozone molecule to reform molecular oxygen and a single oxygen atom. Without the presence for this reaction. of ozone in the stratosphere, life on Earth would change forever, as harmful UV radiation would damage cells in both plants and animals. ) Deduce whether this reaction is exothermic or endothermic. The bond dissociation enthalpy of an oxygen molecule is 498 kJ mol 1 . In d) Which are more comparison, the energy required to break an oxygen–oxygen bond within energetically stable, 1 an ozone molecule is 364 kJ mol . The consequence of this is that an the reactants or the ozone molecule is decomposed by UV rays more readily than an oxygen products? molecule. The ozone photolysis reaction described above is an endothermic reaction with the required energy coming from the UV radiation. The potential energy prole for this reaction is shown ingure 2. Examination of this energy prole reveals that the oxygen molecule and oxygen atom have a greater combined energy than the reactant ozone molecule. The products of this reaction are said to be less stable, as they exist at a higher energy. ygrene laitnetop O +O 2 products a ∆H O 3 reactants reaction pathway Figure 2 Endothermic energy prole for ozone photolysis 155
5 ENERGE TICS AND THERMOCHEMISTRY Dpto of t ozo Since the early 1980s scientists have been monitoring ozone depletion globally, par ticularly the giant holes in the ozone layer which have appeared above the Arctic and Antarctic polar icecaps (gure 3). Figure 3 ERS-2 satellite map of Antarctic ozone hole in 2010 Chlorouorocarbons (CFCs) are a type of hydrocarbon containing carbon, hydrogen, and halogen atoms. The compounds themselves are considered to be non-toxic and have a low level of both ammability and reactivity. First used in industry in the late 1920s, a CFC called Freon gas became an industry standard refrigerant in domestic refrigerators throughout the world. Over the following decades the use of CFCs in commercial air conditioning and the automobile industry, as an aerosol propellant, and as a solvent saw the demand for this class of compounds increase rapidly. Scientific research in 1973 discovered that while CFCs remained harmless close to the Ear th’s su rf a ce, w h en ex po s e d to U V ra d ia tio n i n the s trato sp her e these compounds under went chemical reactions resulting in the release of chlorine. The chlorine released then had a catalytic effect on the destruction of ozone in this layer of the atmosphere. As the chlorine is not directly consumed in the reaction, small amounts of CFCs were found to be responsible for the destruction of large quantities of ozone. Cl(g) + O (g) → ClO (g) + O (g) 3 2 ClO (g) + O (g) → Cl(g) + 2O (g) 3 2 Science has made many advances that have improved people’s daily lives and extended life expectancy and quality. As a consequence the world’s population is increasing. The use of CFCs has had a massive economic (positive) and environmental (negative) impact on the world. Whether the blame for the environmental consequences lies with multinational companies who utilize technology or the scientists who invent it is the focus of many discussions. 156
Que sTiOns Questions 1 Consider the specic heat capacity of the 5 Use data from section 12 of the Data booklet to following metals (table 4). calculate the enthalpy change of reaction for each of these chemical reactions: 1 1 Mt sp t pt/J k K a) C H (g) + H (g) → C H (g) 2 4 2 2 6 Cu 385 Ag 234 b) CH (g) + H O(g) → CO(g) + 3H (g) 4 2 2 Au 130 Pt 134 6 a) Dene the term standard enthalpy change of Table 4 formation, ΔH . [2] f Which metal will show the greatest temperature increase if 50 J of heat is supplied b) (i) Use the information in table 5 to to a 0.001 kg sample of each metal at the same initial temperature? calculate the enthalpy change for the complete combustion of but-1-ene according to the following equation. A. Cu C Au C H (g) + 6O (g) → 4CO (g) + 4H O(g) 4 8 2 2 2 B. Ag D. Pt [1] compod C H (g) CO (g) H O(g) 2 2 IB, May 2007 4 8 1 ∆H /kJ mo +1 394 42 f [3] 2 When 40 joules of heat are added to a sample of Table 5 solid H O at 16.0 °C the temperature increases to 2 8.0 °C. What is the mass of the solid H O sample? (ii) Deduce, giving a reason, whether 2 thereactants or the products are morestable. 1 1 Specic heat capacity of H O(s) = 2.0 J g K 2 [2] A. 2.5 g C. 10 g (iii) Predict, giving a reason, how the B. 5.0 g D. 160 g [1] enthalpy change for the complete IB, Nov 2007 combustion of but-2-ene would compare with that of but-1-ene based on average bond enthalpies. [1] 3 The temperature of a 2.0 g sample of aluminium IB, May 2007 increases from 25 °C to 30 °C. How many joules of heat energy were added? (Specic heat 1 1 capacity of aluminium = 0.90 J g K .) 7 The ∆H values for the formation of two oxides of nitrogen are given below. A. 0.36 C. 9.0 B. 2.3 D. 11 [1] _1 (g) + O (g) → NO (g) ∆H 1 N 2 = –57 kJ mol 2 2 2 IB, May 2003 1 = +9 kJ mol N (g) + 2O (g) → N O (g) ∆H 2 2 2 4 Use these values to calculate ∆H for the following reaction (in kJ): 4 What is the energy change (in kJ) when the temperature of 20 g of water increases by 10 °C? 2NO (g) → N O (g) 2 2 4 A. 20 × 10 × 4.18 a) 105 c) +66 B. 20 × 283 × 4.18 b) –48 d) +123 C. _20 × 10_× 4.18 IB, November 2007 1000 D. 20 × 283 × 4.18 [1] __ 1000 IB, November 2003 157
5 ENERGE TICS AND THERMOCHEMISTRY 8 The standard enthalpy change of three Reaction II combustion reactions is given below in kJ. SO (g) + _1 (g) ⇋ SO (g) ∆H = –92 kJ 2 O 3 2 2 2C H (g) + 7O (g) → 4CO (g) + 6H O(l) 2 6 2 2 2 ∆H = –3120 a) State the name of the term ∆H . State, with a reason, whether reaction I would be 2H (g) + O (g) → 2H O(l) ∆H = –572 2 2 2 accompanied by a decrease or increase in C H (g) + 3O (g) → 2CO (g) + 2H O(l) temperature. [3] 2 4 2 2 2 ∆H = –1411 b) At room temperature sulfur trioxide, SO , is 3 Based on the above information, calculate a solid. Deduce, with a reason, whether the the standard change in enthalpy, ∆H , for the following reaction. ∆H value would be more negative or less negative if SO (s) instead of SO (g) were 3 3 formed in reaction II. [2] C H (g) → C H (g) + H (g) [4] 2 6 2 4 2 c) Deduce the ∆H value of this reaction: [1] IB, November 2009 _1 S(s) + 1 O (g) → SO (g) 3 2 2 IB, November 2005 9 Approximate values of the average bond 1 enthalpies, in kJ mol , of three substances are shown in table 6. HH 430 12 But-1-ene gas burns in oxygen to produce carbon dioxide and water vapour according to the following equation. FF 155 CH + 6O → 4CO + 4H O 2 4 8 2 2 HF 565 a) Use the data in table 7 to calculate the value Table 6 of ∆H for the combustion of but-1-ene. [3] What is the enthalpy change, in kJ, for this Bod CC C=C CH O=O C=O OH reaction? av 348 612 412 496 743 463 2HF → H +F bod 2 2 A. +545 B. +20 C. 20 D. 545 tp/ 1 kJ mo IB, May 2006 Table 7 b) State and explain whether the reaction 10 The reaction between ethene and hydrogen gas above is endothermic or exothermic. [1] is exothermic. IB, May 2006 a) Write an equation for this reaction. [1] b) Deduce the relative stabilities and 13 Given the following data: energies of the reactants and products. [2] 1 C(s) + 2F (g) → CF (g); ∆H = –680 kJ mol 1 c) Explain, by referring to the bonds in 2 4 the molecules, why the reaction is 1 F (g) → 2F(g); ∆H = +158 kJ mol 2 2 exothermic. [2] 1 C(s) → C(g); ∆H = +715 kJ mol 3 IB, November 2007 1 calculate the average bond enthalpy (in kJ mol ) for the C F bond. 11 Two reactions occurring in the manufacture IB, November 2003 of sulfuric acid are shown below: Reaction I S(s) + O (g) ⇋ SO (g) ∆H = –297 kJ 2 2 158
Que sTiOns 14 Methanol is made in large quantities as it c) The Data booklet value for the enthalpy of is used in the production of polymers and in fuels. combustion of methanol is 1 726 kJ mol The enthalpy of combustion of methanol can be determined theoretically or experimentally. Suggest why this value differs from the values calculated in parts a) and b) (i) Part a) [1] (ii) Part b) [1] _1 CH OH(l) + 1 O (g) → CO (g) + 2H O(g) 3 2 2 2 2 IB, May 2011 a) Using the information from section 11 of the Data booklet, determine the 15 One important property of a rocket fuel mixture theoretical enthalpy of combustion is the large volume of gaseous products formed of methanol. [3] which provide thrust. Hydrazine, N H , is 2 4 b) The enthalpy of combustion of methanol often used as a rocket fuel. The combustion of can also be determined experimentally in hydrazine is represented by the equation below. a school labora tor y. A b ur ne r c ontai ni n g N H (g) + O (g) → N (g) + 2H O(g) methanol was weighed and used to heat 2 4 2 2 2 water in a test tube as illustrated in 1 = –585 kJ mol ∆H c gure 4. a) Hydrazine reacts with uorine to produce nitrogen and hydrogen uoride, all in the thermometer gaseous state. State an equation for the reaction. [2] b) Draw the Lewis structures for hydrazine and nitrogen. [2] test tube with water c) Use the average bond enthalpies given in shield stand burner section 11 of the Data booklet to determine Figure 4 the enthalpy change for the reaction in part a) above. [3] d) Based on your answers to parts a) and c), suggest whether a mixture of hydrazine and uorine is a better rocket fuel than a mixture of hydrazine and oxygen. [2] IB, May 2010 The data shown in table 8 were collected. Initial mass of burner and methanol/g 80.557 16 Two students were asked to use information from the Data booklet to calculate a value for the Final mass of burner and methanol/g 80.034 enthalpy of hydrogenation of ethene to form ethane. Mass of water in test tube/g 20.000 Initial temperature of water/°C 21.5 C H (g) + H (g) → C H (g) 2 4 2 2 6 Final temperature of water/°C 26.4 John used the average bond enthalpies from section 11. Marit used the values of enthalpies Table 8 of combustion from section 12. i) Calculate the amount, in mol, of a) Calculate the value for the enthalpy of methanol burned. [2] hydrogenation of ethene obtained using theaverage bond enthalpies given in ii) Calculate the heat absorbed, in kJ, section 11. [2] by the water. [3] b) Marit arranged the values she found in iii) Determine the enthalpy change, 1 section 12 into an energy cycle (gure 5). in kJ mol , for the combustion of 1 mole of methanol. [2] 159
5 ENERGE TICS AND THERMOCHEMISTRY ∆H (hydrogenation) C H (g) + H (g) C H (g) 2 2 4 2 6 1 1 2 lom Jk 1141 2 lom Jk 682 - O3 2 l O o 1 m J k 0 6 5 1 - 2CO (g) + 3H O(l) 2 2 Figure 5 Calculate the value for the enthalpy [1] of hydrogenation of ethene from the energycycle. c) Suggest one reason why John’s answer isslightly less accurate than Marit’s answer. [1] d) John then decided to determine the enthalpy of hydrogenation of cyclohexene to produce cyclohexane. CH (l) + H (g) → CH (l) 2 6 10 6 12 (i) Use the average bond enthalpies to deduce a value for the enthalpy of hydrogenation of cyclohexene. [1] (ii) The percentage difference between thesetwo methods (average bond enthalpies and enthalpies of combustion) is greater for cyclohexenethan it was for ethene. John’s hypothesis was that it would bethe same. Determine why the use ofaverage bond enthalpies is less accurate for the cyclohexene equationshown above, than it was forethene. Deduce what extra information is needed to provide a moreaccurate answer. [2] IB, May 2009 160
6 CHEMICAL KINETICS Introduction In this topic we will discuss a very important aspect of chemical reactions, namely chemical kinetics. We begin the chapter by exploring the rate of a chemical reaction. We will then describe the kinetic-molecular theory of gases and collision theory and outline the importance of the principle of Occam’s razor in science as a guide to developing theories. We will also learn that the greater the probability that molecules collide with sufcient energy and proper orientation the higher the rate of reaction. In addition we consider the effect of a catalyst on rate. Throughout the topic the analysis of graphical and numerical data obtained from rate experiments will form a key part of our treatment of this branch of physical chemistry. 6.1 C oo o o co Understandings Applications and skills ➔ Species react as a result of collisions of ➔ Description of the kinetic theory in terms of sufficient energy and proper orientation. the movement of par ticles whose average ➔ The rate of reaction is expressed as the change kinetic energy is propor tional to temperature in concentration of a par ticular reactant/ in kelvin. product per unit time. ➔ Analysis of graphical and numerical data from ➔ Concentration changes in a reaction can be rate experiments. followed indirectly by monitoring changes in ➔ Explanation of the effects of temperature, mass, volume, and colour. pressure/concentration, a nd p ar ti cle s i z e on ➔ Activation energy (E ) is the minimum energy a rate of reaction. that colliding molecules need in order to have ➔ Construction of Maxwell-Boltzmann successful collisions leading to a reaction. energy distribution curves to account for ➔ By decreasing E , a catalyst increases the rate a the probability of successful collisions and of a chemical reaction, without itself being factors affecting these, including the effect of permanently chemically changed. a catalyst . ➔ Investigation of rates of reaction experimentally and evaluation of the results. ➔ Sketching and explanation of energy profiles with and without catalysts. Nature of science ➔ The principle of Occam’s razor is used as a guide to developing a theory - although we cannot directly see reactions taking place at the molecular level, we can theorize based on the current atomic models. Collision theory is a good example of this principle. 161
6 CHEMIC AL K INE TICS Studying reaction rates When a chemical reaction takes place, four questions are of interest to chemists: 1 What occurs during the reaction? 2 What is the extent of the chemical reaction? 3 Does the reaction happen rapidly or slowly? 4 What energy transfer is involved in the reaction and could a reaction potentially occur given sufcient time? Answering these four questions involves the following: 1 This information can be obtained from the balanced chemical equation which identies the reactants, products (and their states), and stoichiometry (sub-topic 1.1). 2 This is discussed in terms of chemical equilibrium sub-topic 7.1). 3 This is the study of the rate of chemical reactions, a branch of chemistry called chemical kinetics, which is the focus of this topic. 4 This is the eld of thermodynamics (topic 5), which is the study of energy or heat ow in a chemical reaction. Thermodynamics tells us nothing about how quickly or slowly a given reaction takes place. One of the main considerations when examining a chemical reaction is whether it will take place fast enough for it to be useful. There is not much point in carrying out a reaction if it takes 150 years for the product to be formed! Many chemical reactions occur very quickly, such as the rapid ination of airbags in cars, while others such as rusting take place over a period of years. Rate of reaction The rate of reaction is dened as the change in concentration of reactants or products per unit time. 3 1 3 1 , mol dm min , etc. Units: mol dm s Experimental measurements of reaction rates Δc, the change in concentration, can be measured by monitoring a property that will change when the reactants are converted into products. Examples include: 1 change in pH (for acid base reactions) 2 change in conductivity (for reactions involving electrolytes) 3 change in mass or volume (for reactions involving solids or gases) 4 change in colour (for reactions involving transition metals or other coloured compounds). In order to determine the rate of reaction, at time t, a graph of concentration (or the property associated with concentration) versus time is plotted. The rate of reaction is then determined from the slope or gradient of the tangential line at time t. 162
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n Let us consider the following reaction between calcium carbonate (limestone), CaCO and hydrochloric acid, HCl: 3 CaCO (s) + 2HCl(aq) → CaCl (aq) + CO (g) + H O(l) 3 2 2 2 In this reaction, gaseous carbon dioxide is one of the products. If the volume, V,of carbon dioxide gas produced is recorded over time t, the rate of reaction can be determined. Figure 1 shows the experimental set-up for this reaction: delivery tube clamp measuring cylinder carbon dioxide ask trough hydrochloric acid water calcium carbonate Figure 1 Experimental set-up for measuring the rate at which carbon dioxide is produced in the reaction between calcium carbonate and hydrochloric acid Table 1 shows the data recorded during this experiment. t/ 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 0.0 19.0 30.0 37.5 45.0 50.0 52.0 53.0 53.0 53.0 53.0 3 V(CO )/cm 2 Table 1 Volume, V, of carbon dioxide evolved at time t in the reaction between calcium carbonate and hydrochloric acid 60 Figure 2 shows a plot of V against t 50 The rate of reaction can be expressed in three ways: mc/)2OC(V 40 3 ● average rate 30 ● instantaneous rate 20 ● initial rate. 10 0 20 40 60 80 100 0 Average rate t/s The average rate is a measure of the change in concentration of Figure 2 Plot of volume of carbon dioxide gas reactant or product in a given time interval, t Mathematically, average rate can be expressed as: evolved V(CO ) against time t in the reaction 2 between calcium carbonate and hydrochloric acid _Δc average rate = Δt where: Δc = change in concentration of reactant or product Δt = time interval over which the change in concentration was measured For gases, the average reaction rate can be also expressed as the change in volume: _ΔV average rate = Δt where ΔV is the change in volume of the gas produced or consumed during the reaction. 163
6 CHEMIC AL K INE TICS Hence in this example, as the reaction is complete at 70.0 s, then: 53.0 1 3 1 _ cm average rate = = 7.57× 10 s 70.0 Instantaneous rate The instantaneous rate is given by: lim Δc dc _ _ = Δt → 0 dt Δt Initial rate When t = t , the instantaneous rate = the initial rate. 0 In order to deduce the initial rate of reaction at t = 0 s, a tangent is drawn to the curve at 0 s, and the slope or gradient of the tangential line gives the initial rate (gure 3). 60 (x , y ) = (20, 50) 50 40 2 2 30 (x , y ) = (50, 50) 3 3 2 (x , y ) = (20, 39) mc/) OC(V 4 4 3 20 10 (x , y ) = (0, 0) 1 1 0 0 20 40 60 80 100 120 t/s Figure 3 Plot of volume of carbon dioxide evolved versus time showing tangents used to measure both initial rate at t = 0 s and instantaneous rate at t = 50 s (x , y ) = (0.0, 0.0) 1 1 (x , y ) = (20.0, 50.0) 2 2 Δy _50.0 0.0 3 1 _ cm Initial rate = = = 2.50 s Δx 20.0 0.0 So at the start of the reaction, carbon dioxide gas is being generated at a 3 rate of 2.50 cm per second. In order to deduce the instantaneous rate of reaction, at t = 50.0 s, a tangent is drawn to the curve at point t, and the slope or gradient of the tangential line gives the instantaneous rate at this time (gure 3). (x , y ) = (50.0,50.0) 3 3 (x , y ) = (20.0,39.0) 4 4 Δy _50.0 _39.0 1 3 1 _ instantaneous rate = = = 3.67 × 10 cm s Δx 50.0 20.0 164
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n Rate equation A rate equation is a mathematical differential expression, which shows rate expressed in terms of concentration. For example, in the reaction A + B → C + D, a rate equation can be written as follows: rate = d[A] = d[B] = d[C] = d[D] _ _ _ _ - - + + dt dt dt dt In the general case where: xA + yB → qC + pD where x, y, q, and p are the stoichiometric coefcients: _1 d[A] _1 d[B] _1 d[C] _1 d[D] - _ _ + _ _ · rate = = - · = = + · x y q p dt dt dt dt Kinetic–molecular theory of gases The ideal gas equation (sub-topic 1.3) is given by in kelvin. The movement of atoms or the expression: molecule s is due to the ir the r m a l e ne r g y. At a given instant in time, some of the pV = nRT particles will be travelling at greater ve locitie s tha n the o the r pa rt ic l e s . H owe ve r, The equation describes how gases behave, but as the temperature increases the average kine tic e ne rg y wi l l b e hig he r. A t a g i ve n does not explain why gases act as they do. In temperature, the particles of all gases will have the sa me ave r a g e ki ne tic e ne r gy. order to understand the physical properties of gases, a model was devised in 1857 by Rudolf Clausius called the kinetic molecular theory of gases, more simply termed the kinetic Hence, the kinetic molecular theory can offer theory of gases. The theory can be summarized an explanation of temperature and pressure at asfollows: the molecular level. Starting with the kinetic 1 Gases consist of a large number of particles, molecular theory postulates, the ideal gas which are moving at high velocities in equation can be derived. randomdirections. 2 The size of a gaseous particle is negligible. In this model there is the assumption that, even though gaseous particles have mass, their volumes are considered negligible. This can be justied since, at normal pressure, the space between atoms or molecules is very large compared to the size of the atom or molecule. 3 Collisions between one gaseous particle and another are completely elastic. When particles collide, there is no net loss in energy energy is simply transferred. A Figure 4 In the kinetic molecular theory of gases, a sample useful analogy here is what happens when of gas is modelled as consisting of a collection of par ticles two snooker balls collide. The total kinetic moving at high velocities in random directions. The sizes of energy is equivalent before and after the the par ticles are negligible. The par ticles collide with each collision of the snooker balls. other and with the walls of the container containing the gaseous sample. All the collisions are c. The pressure 4 The average kinetic energy of the particles exer ted by the gas results from collisions of its par ticles is proportional to the absolute temperature with the walls of the container 165
6 CHEMIC AL K INE TICS Collision theory Occam’s razor is a principle attributed to the of quantum mechanics. The principle of Occam’s fourteenth-century English Franciscan friar, theologian, and logician William of Ockham, razor is used as a guide in the development of though many references suggest that the principle was known much earlier than this. The principle a theory although we cannot directly see states that: reactions taking place at the molecular level, we “Entities should not be multiplied unnecessarily.” can formulate theories based on current atomic Scientists have formulated the principle like this: models. Collision theory is a good example of this “When two competing theories exist that principle. explain observed facts, both giving essentially the same predictions, then the simpler of the For a chemical reaction to occur between two two theories is the optimum one that should reacting particles, a number of conditions must be used until more evidence transpires to befullled: proveotherwise.” 1 The two particles must collide with each other, The British theoretical physicist Stephen Hawking of the University of Cambridge and author of that is, there must be physical contact. ABrief History of Time said: 2 The colliding particles must have the correct “We could still imagine that there is a set of laws that determines events completely mutual orientations. for some supernatural being, who could observe the present state of the universe 3 The reacting particles must have sufcient without disturbing it. However, such models of the universe are not of much interest to kinetic energy to initiate the reaction. us mortals. It seems better to employ the principle known as Occam’s razor and cut out These three conditions form the basis of the all the features of the theory that cannot be observed.” collision theory. This theory is a model that The principle is often encountered widely in helps us understand why rates of chemical chemistry. For example, it has been used as a justication of the idea of uncertainty in the eld reactions depend on temperature. The model is based on the kinetic molecular theory. For most chemical reactions, only a small fraction of collisions lead to a reaction taking place. In most collisions, the reacting particles bounce off one another and remain unchanged, resulting in no new product formation. In the gaseous reaction between hydrogen, H , and iodine, I , to form 2 2 hydrogen iodide, HI, the reaction proceeds very slowly at room temperature, since only a small number of collisions results in a reaction: H (g) + I (g) ⇋ 2HI(g) 2 2 do o cvo Let us consider each of these three conditions in the box above separately: g, E 1 The rst, that physical contact is necessary for a reaction to occur, is The activation energy is the relatively obvious to appreciate. minimum energy that colliding par ticles need for a reaction 2 Figure 5 illustrates the second condition. In (a) the orientation of to occur. the colliding particles is not favourable, so no reaction will occur. In (b) the particles have a suitable orientation, so the collision can be effective and result in a chemical reaction, provided the reacting particles have sufcient kinetic energy . 166
(a) ineective collision 6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n + No reaction occurs, since (b) eective collision orientation is not favourable. + +2 Orientation is favourable, so may result x in reaction if there is sucient kinetic energy. Figure 5 The possibility of a collision leading to a reaction depends on the a orientation of the par ticles y A useful analogy in understanding activation energy is to imagine somebody trying to push a large rock over a hill (gure 6). available energy release Initially, there must be a minimum input of energy in order for the person to shift the rock over the hill. Once over the hill, the rock can fall to point z z Figure 6 Analogy of activation energy, E a Catalysts The analogy just described can be conveyed in a potential energy prole (gure 7). (a) (b) transition state transition state a ygrene laitnetop ygrene laitnetop a reactants products ∆H ∆H reactants products reaction coordinate reaction coordinate Figure 7 Potential energy prole for (a) an exothermic reaction with ΔH<0; (b) an endothermic reaction with ΔH>0 The arrangement of atoms at the crest of the energy prole is termed the do o c transition state or activated complex A catalyst is a substance Catalysts may be either homogeneous or heterogeneous. that increases the rate of a chemical reaction, but is not Homogeneous catalysts consumed in the reaction A homogeneous catalyst is in the same physical phase or state as the itself. A catalyst provides reactants. The destruction of gaseous ozone, O , by chlorine atoms is an an alternative pathway for the reaction and lowers the 3 activation energy, E example of homogeneous catalysis, since chlorine atoms (which act as the catalyst) have the same state as the gaseous reactants. a In the stratosphere (upper atmosphere), ozone in the ozone layer absorbs over 95% of the UV radiation reaching Earth from the sun, protecting us from this harmful radiation. 167
6 CHEMIC AL K INE TICS hν O (g) → O (g) + O (g) 3 2 O (g) + O (g) → O (g) + heat 2 3 Thus, there is a net energy conversion from UV to heat energy. With the progressive depletion of the ozone layer (gure 8), more UV radiation can reach the Earth’s surface. This can be associated with an increased risk of skin cancers (melanomas) and cataracts. Uu rouc: Ozo ho Wc - images, data and information for both the southern and nor thern hemispheres can be found on this website hosted by NASA - Goddard Space Flight Centre: http://ozonewatch.gsfc.nasa.gov Figure 8 The largest ozone hole to date seen here shown in purple was recorded on September 24, 2006 for the Antarctic Hemisphere Chlorine atoms are produced in the reaction of a chlorouorocarbon (CFC) with UV light. CFCs were previously used in air conditioning units, refrigerators, and aerosols. Freon, CF Cl (g), is one example of a CFC. In the 2 2 presence of UV light, the weaker C Cl bond is broken (the C F bond strength is greater), and radicals are produced. The chlorine radicals then attack ozone. hν CF Cl (g) → CF Cl (g) + Cl (g) 2 2 2 rst step: Cl (g) + O (g) → ClO (g) + O (g) 3 2 second step: ClO·(g) + O (g) → Cl·(g) + 2O (g) 3 2 overall reaction: 2O (g) → 3O (g) 3 2 Chlorine acts as a catalyst and is regenerated in the second step. We can represent the potential energy prole for this reaction scheme as shown in gure 9. transition states ygrene laitnetop Cl•(g) + 0 (g) (uncatalysed) 3 a a + O•(g) (catalysed) ∆H Cl•(g) + 20 (g) (catalysed) 2 (catalysed) reaction coordinate Figure 9 Potential energy prole for the catalytic destruction of ozone, showing both the catalysed and uncatalysed pathways 168
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n As can be seen from gure 9, there are actually two transition states for the catalysed pathway. This is specic for this particular reaction and, in general, the more typical representation of a potential energy prole showing the catalysed and uncatalysed pathways shows just one transition state for the catalysed pathway. This is shown in gure 10. Although ozone in the stratosphere acts as a protective shield from the harmful effects of high-energy UV-a and UV-b radiation, high concentrations of ozone in the troposphere (the lower atmosphere) can lead to respiratory problems such as asthma and emphysema. Ozone may be formed in the reaction of gaseous nitrogen oxides, NO and NO , 2 from car exhaust gases with VOCs (volatile organic compounds). Ozone in the troposphere also acts as a greenhouse gas transition states at crests ygrene laitnetop (uncatalysed) a reactants (catalysed) a ∆H products reaction coordinate Figure 10 Typical potential energy prole showing catalysed and uncatalysed pathways Heterogeneous catalysts A heterogeneous catalyst is in a different phase or state from the ac v Can you name ve countries reactants. Typically, the heterogeneous catalyst is in the solid phase around the world that still continue to use leaded petrol? and the reactants are in the liquid or gaseous states. An example of a 169 heterogeneous catalyst is the catalytic converter used in the exhaust system of a car. In a car engine pollutants such as carbon monoxide, CO, nitrogen monoxide, NO, and unburned hydrocarbons, C H , are x y produced. The carbon monoxide comes from the incomplete combustion of hydrocarbon fuels. Nitrogen monoxide is produced from the reaction of atmospheric nitrogen and oxygen at high temperature in the engine of the car. The catalytic converter converts these substances into less harmful substances, namely carbon dioxide, CO , water, H O, and nitrogen, N : 2 2 2 catalyst 2NO(g) N (g) + O (g) 2 2 catalyst 2NO (g) N (g) + 2O (g) 2 2 2 catalyst 2CO(g) + O (g) 2CO (g) 2 2 catalyst CH CH CH (g) + 5O (g) 3CO (g) + 4H O(g) 3 2 3 2 2 2 propane fuel Examples of catalysts used in catalytic converters are platinum, palladium, and rhodium, and transition metal oxides such as vanadium(V) oxide, V O , copper(II) oxide, CuO, and chromium(III) 2 5 oxide, Cr O . Leaded petrol (leaded gasoline) is not used in modern cars – 2 3 if a car tted with a catalytic converter used leaded petrol, the lead would poison the catalyst.
6 CHEMIC AL K INE TICS Maxwell-Boltzmann energy distribution and temperatureygrene citenik htiw selcit rap fo noitcarf Rates of reaction in the gas phase can be interpreted at the molecular level using different approaches: 1 the collision theory 2 Maxwell Boltzmann energy distribution curve 3 temperature effects on kinetic energies. area A for catalysed a We have already explored the collision theory reaction model. We shall now look at approaches 2 and 3. area B kinetic energy Maxwell–Boltzmann energy distribution Figure 11 Maxwell Boltzmann energy distribution cur ve cur ve showing the activation energy for a catalysed reaction. Area A shows the fraction of particles in the sample that do not The kinetic-molecular theory says that particles have sucient energy to react. Area B shows the fraction of gas move randomly in different directions at of particles in the sample that have the minimum energy high velocities. However, these velocities differ required to initiate a reaction with the use of a catalyst and, since the particles are constantly colliding with other particles and with the sides of the container, the velocity of a single gaseous particle ygrene citenik htiw selcit rap fo noitcarf changes constantly. It is not realistic to discuss the velocity of an individual gaseous particle in this scenario. What needs to be considered is the distribution of velocities of the particles . The distribution of velocities is described by the for catalysed Maxwell Boltzmann energy distribution a reaction area A curve. This is a plot of the fraction of particles E for uncatalysed a reaction with a given kinetic energy (that is the probability of that value of kinetic energy occurring) versus area B area C kinetic energy. As can be seen from gure 11, kinetic energy the representation is asymmetric. The area under Figure 12 Maxwell Boltzmann energy distribution cur ve the curve represents the total number of gaseous showing the activation energy for an uncatalysed reaction. particles in the sample. At a certain temperature, Area A shows the fraction of par ticles in the sample that do the majority of gaseous particles will have a not have sucient energy to react. Area B + area C shows kinetic energy near the mean value. At a given the fraction of par ticles that have the minimum energy time, some of the gaseous particles will have required in order to initiate a reaction using a catalyst. Area C shows the fraction of par ticles in the sample that have the either high or low velocities, but the majority will minimum energy required to initiate a reaction without the have velocities and hence kinetic energies close to use of a catalyst the mean kinetic energy. 170
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n Temperature eects on kinetic energies ygrene citenik htiw selcit rap fo noitcarf T 1 With increasing temperature T, the proportion of particles that have sufcient kinetic energy T to overcome the activation energy barrier will 2 increase. As T increases, the mean velocity of the particles increases and so there will be an increase in the proportion of particles having a greater kinetic energy. As a result, the Maxwell Boltzmann energy distribution curve becomes atter and broader at a higher temperature. Therefore, with increasing temperature, the kinetic energy frequency of collisions will increase. There will be more successful collisions, since there are Figure 13 Maxwell Boltzmann energy distribution cur ves now more particles which have sufcient kinetic energy to overcome the activation energy barrier. for two temperatures, T > T . Notice that, at the higher This results in an increase in the rate of reaction. Typically with an increase in temperature of 2 1 10°C, the reaction rate will double. temperature, the energy distribution is broader and the mean kinetic energy is greater. The proportion of particles that have sucient thermal energy to overcome the activation energy barrier has increased. The area under both curves is the same as this signies the total number of gaseous particles in the sample tOK Temperature can be considered as a measure of the average amount of kinetic su p energy of par ticles, which is the energy due to motion. As the temperature In a potential energy prole, the y-axis is the potential increases, so does the kinetic energy of the par ticles. The lowest temperature energy and the x-axis is the reaction coordinate, which that can be attained theoretically is bou zo, which is 273.15 °C. This is represents the progress of the reaction. In a Maxwell taken as the zero point on the Kelvin scale. At 0 K , all thermal energy has been Boltzmann energy distribution curve, the y-axis is the removed from a substance and the motion of par ticles has eectively ceased. fraction of par ticles with a cer tain kinetic energy and the However, if you consider Heisenberg’s uncer tainty principle, all molecular x-axis is kinetic energy. Don’t mix up the two axis labels in movement may not cease entirely at 0 K . Consider why this may be the case. these two representations. The Kelvin scale gives a natural measure of the kinetic energy of a gas, and is independent of physical proper ties, whereas the ar ticial Celsius scale is based on the physical proper ties of water. The Celsius scale is dened about an arbitrary zero point and hence negative °C values occur. Are physical proper ties such as temperature invented or discovered? Could Anders Celsius, the Swedish astronomer credited with the Celsius scale, have chosen the melting and boiling points of another substance other than water to devise the ar ticial Celsius scale? 171
6 CHEMIC AL K INE TICS Factors that aect the rate of a chemical reaction There are four factors that can increase the rate of 4 Decreasing the par ticle size of a chemical reaction: reactants in the solid phase 1 increasing the temperature at which the In a heterogeneous reaction involving a gas (or reaction is conducted a liquid) and a solid, the rate of reaction will increase if the surface area of the solid is increased 2 addition of a catalyst by breaking the solid up into smaller pieces. The reason for this is that the reaction takes place only 3 increasing the concentration of the reactants on the surface of the solid reactant. So, for example, in the case of a liquid reacting with a solid, only the 4 decreasing the particle size of reactants in the surface particles of the solid will have direct contact with the liquid reactant. If a nely divided solid solid phase. or powder is used, there will be an increase in the surface area and there will be a greater number of Let us examine each of these individually. solid particles available for reaction. As a result, the rate of reaction will increase. For example, nely 1 Increasing the temperature at which divided grain can combust explosively. This has led to a number of major explosions in conned spaces the reaction is conducted in grain factories. We have just discussed this factor. A good ac v example of the effect of temperature on rate Find out some examples from dierent locations involves the refrigeration of milk. At room around the world where this has happened. temperature, milk can turn sour over a period of time due to bacterial reactions. This process is slowed down if the temperature is brought to just above 0 °C in a refrigerator. 2 Addition of a catalyst Measuring the rate of a chemical We have also already discussed this in detail. 3 Increasing the concentration of the reaction reactants We shall look at a number of techniques that can be used to monitor the rate of a chemical reaction If in a xed volume the concentration of as stated previously: reactant species increases, there will be a corresponding increase in the frequency of 1 Change in pH (for acid-base collisions. Hence there will be an increase in the number of successful collisions, and therefore reactions) the rate of reaction will increase. An example of this involves the destruction of statues made In a reaction where either hydronium cations, of limestone (calcium carbonate, CaCO ). If the H + (or simply + or hydroxide ions, OH , are 3 O H) concentration of the pollutant sulfur dioxide, SO , in the atmosphere increases, the rate of 3 2 present as either the reactant or product species, destruction of the limestone statues increases. the change in pH can be monitored using a pH probe and meter 2 Change in conductivity (for reactions 2CaCO (s) + 2SO (g) + O (g) → 2CaSO (s) 3 2 2 4 involving electrolytes) + 2CO (g) 2 Consider the following reaction: The solubility of calcium sulfate is greater than IO (aq) + 5I (aq) + + → 3I (aq) + 3H O(l) that of calcium carbonate, leading to erosion of 3 6H (aq) 2 2 the limestone. In this reaction of iodate ions with iodide ions in an acidic medium there is a net decrease in the concentration of ions from a total of 12 on 172
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n the reactant side to zero on the product side. absorbance This decrease in the concentration of ions can be monitored using a conductivity probe and light photoelectric meter. If the net number of ions decreases source during the reaction, then the total electrical cell conductivity will also decrease and vice versa. lter coloured 3 Change in mass or volume (for reactions involving gases) meter We have seen an example of this physical method solution earlier. Figure 14 Schematic diagram of a colorimeter that records 4 Change in colour (for reactions absorbance. According to Beer ’s law the absorbance A is directly involving transition metals or other proportional to the concentration c, that is A ∝ c or A = cεl, coloured compounds) where l is the path length and ε is the extinction coecient. Colorimetry (gure 14) is used to monitor In colorimetry, light of a certain wavelength chemical reactions thathave a coloured is passed through the coloured solution being reactant or a coloured product. The change in monitored. The colorimeter or spectrophotometer colour intensity corresponding to the change then measures the intensity of the transmitted light. in concentration of the coloured reactant or The absorbance is indicative of the amount of light product can be monitored using a colorimeter. absorbed by the reaction mixture. A calibration One example of this involves the ethanedioate curve can be plotted of absorbance versus manganate(VII) reaction: concentration of the standard coloured solution (gure 15). 2 + ecnabrosba (aq) + 16H (aq) 2MnO (aq) + 5C O 4 2 4 purple 2+ → 2Mn (aq) + 10CO (g) + 8H O(l) 2 2 pale pink 3 concentration/mol dm Figure 15 Sketch of a typical calibration cur ve Worked examples: rates of reaction 1 b) M(CO ) = 12.01 + 2 × 16.00 = 44.01 g mol 2 Example 1 _0.25 3 In the chemical reaction of calcium carbonate mol with hydrochloric acid, 0.25 g of carbon dioxide n(CO )= = 5.7 × 10 was generated in 60.0 s. 2 44.01 3 _5.7 × 10 5 1 Average rate = = 9.5 × 10 mol s 60.0 a) Deduce the balanced chemical equation forthe reaction, including state symbols. su p Be careful with gc gu in a question like b) -1 this. Since division is involved the answer should be Calculate the average rate, in mol s expressed with the smallest number of signicant gures from the experimental data, which in this Solution case will be two. a) CaCO (s) + 2HCl(aq) → CaCl (aq) + CO (g) + 3 2 2 H O(l) 2 173
6 CHEMIC AL K INE TICS Example 2 Then as the reaction proceeds with time, the gradient decreases as the amount of carbon Figure 16 shows how the volume of carbon dioxide generated slows as the concentration dioxide formed varies with time, when a of acid decreases. hydrochloric acid solution is added to excess calcium carbonate in a ask. The plot eventually attens as hydrochloric acid is consumed. ii) 2 ) OC( V 2 ) OC( V 0 time 0 Figure 16 time i) Explain the shape of the curve. [3] Figure 17 ii) Copy the graph and sketch the curve Decreasing the concentration decreases the rate of reaction. In gure 17 the curve (blue) youwould obtain if double the volume is less steep, but the same maximum volume of gas will be evolved over a longer time. Since of hydrochloric acid solution of half the acid concentration is halved, collisions will be less frequent and hence the rate of reaction the concentration, as in the example will be slower. Since there is the same amount of hydrochloric acid, the same volume of above,isused instead, with all other variables carbon dioxide will be produced. kept constant from the original.Explain why the shape of thecurve is different. [4] iii) Outline one other way in which the rateof this reaction can be studied in a schoollaboratory. Sketch a graph to illustrate iii) There are a number of possible answers here. One method is to measure the rate at which how the selected variable wouldchange with the mass decreases as the gas is given off. This would simply involve a plot of mass of ask time. [2] + contents versus time or a plot of mass loss versus time (gure 18). iv) Dene the term activation energy and state one reason why the reaction between calcium carbonate and hydrochloric acid takes place at a reasonably fast rate at room temperature. [2] ssol ssam stnetnoc + ksa fo ssam IB, May 2010 Solution i) Rate = gradient of graph. time time Figure 18 At the beginning of the reaction, carbon dioxide is produced quickest, as the concentration of hydrochloric acid is greatest. 174
6 .1 C O lli s iOn t h e Or y a n d r at e s Of r e a C t i O n Other possibilities might include monitoring b) Draw a graph of total volume of oxygen the pH or the pressure. versus time. iv) Activation energy is the minimum energy the colliding particles need to have in order for 3 1 a reaction to occur. The reaction between , calcium carbonate and hydrochloric acid c) Calculate the average rate, in cm min takes place at a reasonably fast rate at room temperature because the activation energy correct to one decimal place. barrier is quite low. 3 Example 3 d) Deduce, in s, how long it took for 40 cm of The data in table 2 were recorded for the decomposition of hydrogen peroxide using a oxygen to be collected. manganese(IV) oxide catalyst. The total volume of oxygen gas collected was measured at 3 1 differenttimes. e) Determine the initial rate, in cm min f) Determine the instantaneous rate, in 3 1 , at t = 4 min. cm min g) Explain whether the catalyst used is homogeneous or heterogeneous. Solution 1 a) H O (aq) → H O(l) + O (g) 2 2 2 2 2 b) 70 tm/m to voum o ox g g 60 3 0 negyxo fo emulov latot 50 1 coc/cm 2 0 40 3 4 18 30 5 32 6 42 20 7 50 8 56 10 9 61 10 64 0 2 4 6 8 10 12 Table 2 64 0 time/min 64 64 Figure 19 Plot of total volume of oxygen given o versus time 3 3 1 64 cm cm _ c) Average rate = = 9.1 min 7 min d) 2.8 min = 2.8 × 60 s = 168 s e) (x , y ) = (2,50); (x , y ) = (0,0); 1 1 2 2 _0 50 3 1 cm Initial rate = = 25 min a) Deduce the balanced chemical equationfor 0 2 the reaction, including statesymbols. 175
6 CHEMIC AL K INE TICS f) (x , y ) = (6, 64); (x , y ) = (2, 36); g) Manganese(IV) oxide is solid, so is in 3 3 4 4 a different phase to aqueous hydrogen Instantaneous rate at t = 4 min: peroxide. Manganese(IV) oxide is acting as a _36 64 3 1 7cm = min 2 6 heterogeneous catalyst in this reaction. su p 2 Both the x- and y-axes must be labelled and units There are a number of points that you have to be careful about in graphical questions: included. 1 Graphs should have a title which involves a plot of y 3 When finding the slope of a tangential line, try to versus x (not the other way round!). choose two points as far apar t as possible. 176
QUe stiOns Questions 1 Consider the reaction between gaseous iodine 4 Equal masses of powdered calcium and gaseous hydrogen. carbonatewere added to separate solutions of hydrochloric acid. The calcium carbonate I (g) + H (g) ⇋ 2HI(g) ∆H = -9 kJ 2 2 was in excess. The volume of carbon Why do some collisions between iodine and dioxideproduced was measured at regular hydrogen not result in the formation of the product? intervals. Which curves in gure 20 best represent the evolution of carbon dioxide against time forthe acid solutions shown in A. The I and H molecules do not have 2 2 table 3? sufcient energy. B. The system is in equilibrium. C. The temperature of the system is too high. 2 mc/)g( OC fo emulov D. The activation energy for this reaction is I 3 II very low. [1] III IB, May 2011 IV 3 3 nitric acid 2 At 25 °C, 200 cm of 1.0 mol dm time/s is added to 5.0 g of magnesium powder. If the experiment is repeated using the same mass ▲ Figure 20 of magnesium powder, which conditions will result in the same initial reaction rate? 3 3 3 25 cm o 50 cm o 25 cm o 3 3 3 2 mo m 1 mo m 1mo m Voum Coco o tmpu / °C hC hC hC o hnO / 3 3 hnO /mo m A. I III IV 3 3 B. cm C. D. I IV III A. 200 2.0 25 B. C. I II III D. 200 1.0 50 II I III 100 2.0 25 ▲ Table 3 100 1.0 25 [1] [1] IB, May 2009 IB, May 2011 5 Hydrochloric acid is reacted with large pieces 3 Which of the following is an appropriate unit of calcium carbonate; the reaction is then for rate of reaction? repeated using calcium carbonate powder. How does this change affect the activation energy A. s and the collision frequency? B. min 3 acvo g Coo quc C. cm s increases stays constant increases 3 1 increases min increases stays constant D. mol dm stays constant stays constant A. B. C. D. [1] IB, November 2009 177
6 CHEMIC AL K INE TICS 6 Which factors can affect the rate of a chemical 9 a) A solution of hydrogen peroxide, H O , is 2 2 reaction? added to a solution of sodium iodide, NaI, acidied with hydrochloric acid, HCl. The I. The concentration of the reactants. II. The temperature at which the reaction yellow colour of the iodine, I , can be used 2 to determine the rate of reaction. takes place. H O (aq) + 2NaI(aq) + 2HCl(aq) 2 2 III. The physical state of the reactants. → 2NaCl(aq) + I (aq) + 2H O(l) 2 2 A. I and II only The experiment is repeated with some B. I and III only changes to the reaction conditions. For each C. II and III only of the changes that follow, predict, stating a D. I, II, and III reason, its effect on the rate of reaction. i) The concentration of H O is increased 2 2 at constant temperature. [2] 7 In an acid-catalysed hydrolysis reaction of ethyl ethanoate, the concentration of the ester ii) The solution of NaI is prepared from 3 3 to 0.35 mol dm changes from 1.50 mol dm a ne powder instead of from large in 3.5 min, at a given temperature, T . Which crystals. [2] 1 b) Explain why the rate of a reaction increases of the following statements are correct? when the temperature of the system I. The average rate of the reaction is 3 1 increases. [3] min 0.33mol dm IB, November 2009 II. If the reaction is carried out at a higher temperature, T , the reaction rate will 2 begreater. 10 Models can prove vital in chemistry. Discuss III. The products of the reaction will be the principles of the kinetic molecular ethanoic acid and ethanol. theory and the collision theory. A. I and II only B. I and III only C. II and III only 11 Design an appropriate experiment to measure the rate of reaction of a hydrolysis reaction D. I, II, and III (saponication) of the ester methyl ethanoate in an alkaline medium. 8 Factors that affect the rate of a chemical reaction include particle size, concentration of reactants, and the temperature of the reaction. i) Dene the term rate of a chemical 12 Design an appropriate experiment to measure the rate of reaction using the “clock reaction” reaction. [1] technique involving the reaction of magnesium with dilute hydrochloric acid solution. ii) List the three characteristic properties of reactant particles that affect the rate of reaction as described by the collision theory. [3] IB, May 2011 178
7 EQUILIBRIUM Introduction An understanding of reactions that are in equilibrium, examine the equilibrium constant K , equilibrium and how to control the position of c the equilibrium is of fundamental importance to science and society. The Haber process, used examine the information it conveys and discuss for the large-scale manufacture of ammonia, is an equilibrium system that has had a profound the effects of changing experimental conditions on impact on the history of the World. In this chapter we discuss how reactions can be in a state of the value of K applying Le Châtelier’s principle. c We will also introduce the term reaction quotient, Q, which is a measure of the relative amounts of products and reactants present in a reaction that is not in a state of equilibrium. 7.1 E Understandings Applications and skills ➔ A state of equilibrium is reached in a closed ➔ The characteristics of chemical and physical system when the rates of the forward and systems in a state of equilibrium. reverse reactions are equal. ➔ Deduction of the equilibrium constant ➔ The equilibrium law describes how the equilibrium constant (K ) can be determined for expression (K ) from an equation for a c c a par ticular chemical equation. homogeneous reaction. ➔ Determination of the relationship between ➔ The magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium dierent equilibrium constants (K ) for the c same reaction at the same temperature. and its temperature dependence. ➔ Application of Le Châtelier ’s principle to ➔ The reaction quotient (Q) measures the relative predict the qualitative eects of changes of amount of products and reactants present temperature, pressure, and concentration on during a reaction at a par ticular point in time. the position of equilibrium and on the value of Q is the equilibrium constant expression with the equilibrium constant. non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure, and temperature. ➔ A catalyst has no eect on the position of equilibrium or the equilibrium constant. Nature of science ➔ Obtaining evidence for scientic theories – ➔ Common language across dierent disciplines – isotopic labelling and its use in dening the term dynamic equilibrium is used in other equilibrium. contexts, but not necessarily with the chemistry denition in mind. 179
7 E Q U ILIBR I U M Equilibrium reactions in chemistry Many important chemical reactions are reversible and exist in a state of equilibrium, with both the forward and reverse reactions occurring simultaneously with products and reactants constantly being interconverted. An understanding of chemical equilibrium enables chemists to control reaction conditions and maximize the yield of the desired product being synthesized involving equilibrium reactions. Industry relies heavily on understanding and controlling equilibrium reactions. Systems that are in equilibrium are common in everyday life. For example, imagine you have some hot liquid in a storage container sealed with a lid. Water molecules in the liquid that have sufcient energy will leave the liquid phase and enter the gaseous phase in the process of evaporation. At the same time, some gaseous water molecules will lose energy through collision and enter the liquid phase in the process of condensation. As evaporation continues, the atmosphere in the closed system becomes saturated with water vapour and the chance of collisions between particles resulting in condensation increases. Eventually the rate of evaporation will equal the rate of condensation and there will be no further change in the amount of liquid present and the amount of gas present in the closed system. The system is then at equilibrium. The forward and reverse reactions are occurring at equal rates. The concentration of liquid and gas does not change, but molecules are constantly being interconverted between the two phases. This is described as a dynamic equilibrium H O(l) ⇋ H O(g) 2 2 liquid phase gaseous phase ▲ Figure 2 In a dynamic equilibrium the for ward and reverse reactions occur at equal rates Heterogeneous equilibrium and solubility ▲ Figure 1 Bromine in a closed A saturated solution in a closed system will establish a dynamic system will establish a dynamic equilibrium if there is excess solid present. If you mix solid sodium equilibrium between the liquid chloride, Na Cl a nd w a te r, in it ia l l y the s ol id wi l l be g i n t o di ss ol ve. and gaseous phases. Eventually The concentration of ions present in the aqueous solution will the colour stops changing as the increase. Some aqueous ions will recombine and precipitate out of equilibrium concentration of vapour solution. At this stage the rate of dissolution is faster than the rate of is reached precipitation. 180
7. 1 E q u i l i b r i u m Eventually, when the solution becomes saturated, the rate of dissolving TOK will equal the rate of precipitation. A dynamic equilibrium has been established. Scientists use specialized terminology to precipitation facilitate unambiguous communication of meaning + (aq) and understanding. They NaCl(s) ⇋ Na (aq) + Cl recognize the need for a common language. For dissolution example, a closed system is one in which no matter initial equilibrium is exchanged with the surroundings. Cl (aq) + The language of science and Na (aq) its internationally agreed terminology and symbols + aid communication and understanding. However, + Na (aq) when meaning is inexible does language also play a Na (aq) Cl (aq) role in forming boundaries to our experience of new Cl (aq) Cl (aq) knowledge? + Na (aq) ▲ Figure 3 A saturated solution is in dynamic equilibrium Chemical systems Stdy tp For non-reversible reactions, When describing chemical reactions in an open system, the terms the use of a single arrow → in a “reactants” and “products” give the impression that reactions proceed to chemical equation represents completion, producing new substances. In reality many reactions exist in the complete conversion of a state of equilibrium. The relative rates of forward and reverse reactions reactants to products. Reversible depend on the physical conditions of temperature and pressure, on reactions and systems in chemical conditions (a change in concentration of reactants and equilibrium are designated by products), and on the presence of a catalyst. the use of a double-headed arrow: ⇋ Nitrogen(IV) oxide or nitrogen dioxide, NO is a toxic brown gas that 181 2 exists in equilibrium with colourless dinitrogen tetroxide, N O : 2 4 N O (g) ⇋ 2NO (g) 2 4 2 When a sample of colourless frozen N O is placed into a container 2 4 which is then sealed, initially N O is the only substance present and no 2 4 colour is observed. At room temperature N O begins to decompose and a colour change 2 4 to brown signals the presence of NO . Your understanding of kinetic 2 theory will tell you that initially the rate of decomposition of N O will 2 4 be greatest, when the concentration [N O ] is at a maximum. As time 2 4 passes, sufcient NO molecules will be present in the closed system for 2 successful collisions to reform N O . 2 4 The concentration of N O decreases progressively while the 2 4 concentration of NO increases. The system approaches a dynamic 2 equilibrium as the rates of the forward and reverse reactions become equal.
7 E Q U ILIBR I U M At equilibrium: NO 2 ● The forward and reverse reactions are occurring at equal rates. noitartnecnoc ● There is no change in the concentrations of reactants and products. NO 2 4 ● There is no change in macroscopic properties such as colour and density. ● The equilibrium can be approached from either the forward or reverse direction. equilibrium achieved 0 time ● The equilibrium is dynamic. The forward and reverse reactions continue but no overall change in the concentration of reactants and ▲ Figure 4 Change in concentration versus products occurs. time for the reaction N O (g) ⇋ 2NO (g) 2 2 4 approaching and achieving equilibrium ● Any changes in the reaction conditions, such as temperature, pressure, or concentration of reactants or products, can affect the rate of for ward equilibrium, demonstrating its dynamic nature. reaction etar qck eston Which statement is always true for a chemical reaction that has reached equilibrium achieved equilibrium? (rates are equal) rate of reverse reaction a) The yield of product(s) is greater than 50% 0 ) The rate of the forward reaction is greater than the rate of the reverse time reaction. ▲ Figure 5 At equilibrium, the rates of the for ward c) The amounts of reactants and products do not change. and reverse reactions remain constant d) Both forward and reverse reactions have stopped. [1] ib, Nov 20 05 Animated computer The equilibrium law simulations are available in which the user can decide When a reaction system has established equilibrium, the forward and how to change the conditions reverse reactions are occurring at equal rates and there is no change of a reaction to illustrate in the concentration of reactants and products. To make best use of the concept of dynamic the reaction we need to understand the position of the equilibrium , equilibrium. that is, whether reactants or products are favoured, and how we can manipulate this to maximize the yield of products and the protability TOK of industry. Scientists have a common The law of chemical equilibrium states that at a given temperature terminology and a common reasoning process, which the ratio of the concentration of products (raised to the power of their involves using deductive logic and induction molar coefcients) to the concentration of reactants (raised to the power through analogies and generalizations. They share of their molar coefcients) is a constant. This constant is called the mathematics, the language of science, as a powerful equilibrium constant, K . The subscript ‘c’ indicates that concentration tool. Indeed, some scientic c explanations exist only in mathematical form. values for products and reactants are being used. For example: O (g) + 4HCl(g) ⇋ 2H O(g) + 2Cl (g) 2 2 2 2 2 [H O] [ Cl ] 2 2 __ K = c 4 [O [ HCl ] ] 2 The value of the equilibrium constant is specic for each chemical reaction at a given temperature. Values of K have no units. c 182
7. 1 E q u i l i b r i u m The magnitude of K tells you about the position of the equilibrium. c If K is a very large number, K >> 1, this indicates that at a given cc temperature, products are favoured over reactants. The larger the value of K , the greater the proportion of products that exists compared with c reactants at equilibrium. Conversely, a very small value of K (K << 1 ) cc indicates that the reaction is unfavourable at this given temperature. In a hoogeneos e all the reactants and products Writing equilibrium constant expressions are present in one phase. The most common example is When constructing the equilibrium constant expression for a reversible reactions that occur homogeneous reaction, the following need to be considered: in the gaseous phase. 1 For an aqueous reaction the concentration of the solvent water In a heteogeneos e the reactants or does not appear in the equilibrium constant expression, as its products exist in more than one phase, such as gaseous concentration does not change during the reaction. and solid, liquid and solid. 2 If the reaction takes place in a non-aqueous solution (such as the The equilibrium constant expressions derived here are esterication discussed in sub-topic 10.2), water must be included in for homogeneous equilibria. the K expression as any other reactant or product. c Table 1 gives some examples of equilibrium constant expressions for different reactions. Checa eaton E constant expesson 2+ [ Fe(SCN) ] 3+ 2+ ___ Fe (aq) + SCN (aq) ⇋ Fe(SCN) (aq) K= c 3+ 1 [ Fe ][ SCN ] [ C H COOCH ] 3 7 3 __ CH OH(aq) + C H COOH(aq) ⇋ C H COOCH (aq) + H O(l) K= c 3 3 7 3 7 3 2 [ CH OH ][ C H COOH ] 3 3 7 2 NH [] __3 N (g) + 3H (g) ⇋ 2NH (g) K= 3 c 2 2 3 [ N ][ H ] 2 2 ▲ Table 1 Equilibrium constant expressions for some chemical reactions TOK qck eston Scientists approach their understanding of the universe Deduce the equilibrium constant expression for each of from varying perspectives. Observations made using our senses examine phenomena at the macroscopic the following homogeneous equilibrium reactions. level. The models and theories developed by the scientic community focus on our understanding of the 1 3 microscopic world. Which of the ways of knowing (WOK) enable us to make the transition from the macroscopic to a) N (g) + H (g) ⇋ NH (g) the microscopic? 2 2 2 2 3 macoscopc properties of substances can be identied using our senses, and can be directly determined by ) ClNO (g) + NO(g) ⇋ NO (g) + ClNO(g) measurements. Examples include colour, texture, and density. mcoscopc properties exist at an atomic level 2 2 and can be determined only indirectly. c) 4NH (g) + 5O (g) ⇋ 4NO(g) + 6H O(g) 3 2 2 183
7 E Q U ILIBR I U M Worked example: deducing equilibrium constant expressions For the homogeneous equilibrium: _0.300 ×_0.300 = 2 (3.00) H (g) + I (g) ⇋ 2HI(g) 2 2 3 2 = 1.00 × 10 the equilibrium concentrations (in mol dm ) are Note: as follows: K = _1 c(1) K [ H (g) ] = 0.300, [ I (g) ] = 0.300, [HI(g)] = 3.00 c(2) 2 2 Deduce the equilibrium constant expression, K , The value of K also depends on how the chemical c c and determine the value of K for the forward and equilibrium reaction is balanced: c reverse reactions. _1 _1 2 2 H (g) + I (g) ⇋ HI(g) 2 2 Solution [HI] _ K = c(3) 1 1 For the forward reaction: 2 2 [H ] [I ] 2 2 2 _3.0_0 [HI] _ K = K = c(1) c(3) 1 1 [ H ][ I ] 2 2 2 2 ( 0.300 ) ( 0.300 ) 2 ( 3.00 ) __ = = 10.0 0.300 × 0.300 ___ 2 K = K = 1.00 × 10 c(3) √ c(1) For the reverse reaction: 2HI(g) ⇋ H (g)+ I (g) This last example shows that dividing the equation by 2 throughout gives a new equilibrium constant 2 2 which is equal to the square root of the original equilibrium constant. [ H ][ I ] 2 2 _ K = c(2) 2 [HI] qck estons 1 The equilibrium constant for the reaction between 2 The equilibrium constant for the equilibrium between hydrogen and chlorine gas to produce hydrogen N O and NO is 7.7 × 10 4 at 273 K: 2 4 2 33 chloride gas is 2.40 × 10 at 298 K . N O (g) ⇋ 2NO (g) 2 4 2 H (g) + Cl (g) ⇋ 2HCl(g) 2 2 Calculate the equilibrium constant for 1 Calculate the equilibrium constant for the reverse N O (g) ⇋ NO (g) at 273 K . reaction, namely the decomposition of HCl. 2 2 4 2 Combining equilibrium constants Table 2 summarizes the ways that the equilibrium constant expressions and equilibrium constants can be combined. Change n eacton eaton E constant expesson E constant _1 K (reverse) = (for ward) reverse the reaction inverse of the expression _1 c K K c c or K c __ halve the coecients square root of the expression √K double the coecients square the expression c 2 K c sum equations product of the expressions K= K ×K × … c c1 c2 ▲ Table 2 The equilibrium constant K for the same reaction at the same temperature can be c expressed in a number of ways 184
7. 1 E q u i l i b r i u m The eect of changing experimental conditions on the equilibrium constant When equilibrium is established, the position of the equilibrium remains constant provided that the temperature and pressure do not change. A change in experimental conditions can affect the equilibrium position. However, the value of K remains constant unless the temperature c changes (table 3). Change n condton E poston K concentration of c changes in response to a change in product or reactant [reactants] or [products] no change pressure in a reaction with gaseous reactants or no change temperature products, the pressure can aect the equilibrium position changes, catalyst unless usually changes: the direction of change depends on whether the reaction is ΔH = 0 exothermic or endothermic no change no change ▲ Table 3 The eect of changing conditions on the equilibrium position and the value of K c Le Châtelier ’s principle Le Châtelier’s principle is a useful tool for predicting the effect that changing conditions will have on the equilibrium position: If a change is made to a system that is in equilibrium, the balance between the forward and reverse reactions will shift to offset this change and return the system to equilibrium. At a given temperature, table 3 shows that changing the concentration ▲ Figure 6 The chromate–dichromate equilibrium. of reactants or products does not result in a change in the value of equilibrium constant K . The equilibrium position of the reaction will c change in response to the change in concentration so as to return K c to its original value. For example, gure 6 illustrates the chromate– dichromate equilibrium: 2 (aq) is orange while Aqueous dichromate Cr O 2 7 2 2 + 2 2H (aq) aqueous chromate CrO Cr O (aq) + H O(l) ⇋ 2CrO (aq) + (aq) is yellow. The 2 4 2 7 4 orange dichromate yellow chromate colour of the solution gives an indication of the position of equilibrium If the concentration of a reactant is increased, Le Châtelier’s principle tells us that the forward reaction will be favoured to counteract this change. Decreasing the concentration of a product will also favour the forward reaction. The addition of hydroxide ions, OH , to the reaction + + reacts with H mixture results in a reduction in H concentration as OH to form water. The equilibrium mixture becomes a paler colour as more orange dichromate reacts to form yellow chromate. The reverse reaction is favoured if the concentration of a product is increased or the concentration of a reactant is decreased. The addition + of H ions in the form of concentrated hydrochloric acid, HCl results in the reverse reaction being favoured and a deeper orange colour being observed. In both cases, the value of equilibrium constant K remains c unchanged as long as the temperature remains the same. 185
7 E Q U ILIBR I U M Le Châtelier’s principle allows industrial chemists to manipulate reaction conditions to maximize the amount of the desired product formed in an equilibrium reaction. The Haber process for the manufacture of ammonia uses the following reaction: N (g) + 3H (g) ⇋ 2NH (g) 2 2 3 The yield is increased by using high concentrations of reactant gases nitrogen and hydrogen, and removing the product ammonia from the equilibrium mixture. Cause and eect Le Châtelier’s principle is a useful tool in helping to predict the qualitative effects of changes in “Cause and effect” or “causation” refers to the temperature, pressure, and concentration on the situation where a second event (the effect) is a position of an equilibrium reaction and the value direct consequence of the rst event (the cause). of the equilibrium constant. However, the principle Scientists understand this concept and often does not provide an explanation for the effects of develop a hypothesis that suggests a relationship these changes on the equilibrium reaction: it does or causation between factors that may be not demonstrate a cause-and-effect relationship true or false. They then test the hypothesis by between changing these conditions and the experimentation and collection of empirical resultant change in the equilibrium system. data, which is used to either support, modify, or disprove the hypothesis. The Haber process a process that could manufacture ammonia on a large scale. Ammonia manufacture is now Over the past century, advances in the science one of the most widespread industrial processes of agriculture have resulted in over one-third in the world. As well as its major application of the Earth’s land being cultivated with crops in fertilizers, ammonia is also used in the and the population of the planet increasing to manufacture of plastics, bres, explosives, and over 7 billion people. Agriculture is a major pharmaceuticals. employment sector in developing countries and food security remains of great concern to With the onse t of the r s t w or l d w a r, H a be r many, including international bodies such as the World Trade Organization (WTO), the Food and worked with the German military to advance Agriculture Organization (FAO), and the World Health Organization (WHO). the production of explosives. Ammonia reacts with nitric acid to form ammonium nitrate, Global demand for food increases exponentially HNO (aq) + NH (g) → NH NO (aq) which is every decade and we will not be able to meet 3 3 demands for food and water without scientic 4 3 endeavours in the elds of biotechnology and fertilizers. More than 50 % of current food used to this da y a s a f e r ti l i z e r. Am m o ni u m production relies on the use of fertilizers. nitrate will also undergoexplosive decomposition when detonated: NH NO (s) → N O(g) + 2H O(g) 4 3 2 2 Fritz Haber (1868–1934) was a German chemist Haber is sometimes referred to as the “father of chemical warfare”, advancing the research into who was awarded the Nobel Prize in Chemistry and utilization of many poisonous gases during the rst world war. He was not alone in making in 1918 for his work on the synthesis of signicant advances in science with military applications during this time. Gustav Hertz ammonia, NH from its elements. A shortage and James Franck were physicists who devised 3 of natural fertilizers at the beginning of the twentieth century prompted Haber to research 186
7. 1 E q u i l i b r i u m investigations to support Niels Bohr’s model of Hertz, Franck, and Hahn were all members the atom and lay the groundwork for quantum of Haber’s research team who worked on mechanics (for more on quantum mechanics, developing chemical weapons. Their stories see the Physics course book, topic 12). They were serve to illustrate the Nature of Science (NOS). awarded the Nobel Prize in Physics in 1925, while Their research signicantly advanced scientic Otto Hahn received the Nobel Prize in Physics understanding while raising questions about the in 1944 for his discovery of the ssion of heavy ethical considerations of the results of some of atomic nuclei. Later in life he continued to receive their work, the concept of intellectual property an impressive range of awards for his scientic and how society goes on to utilize scientic achievements and was nominated on several discoveries. occasions for the Nobel Peace Prize. He is regarded as the founder of the atomic age. The eect of pressure on reactions in the gas phase Stdy tp When considering the eect We have seen how Le Châtelier’s principle can be used to predict the of changes in pressure on a effect of a change in concentration of a product or reactant on the reaction, you must refer to the position of equilibrium. A system that involves substances in the gaseous moles of gaseous reactants or rather than the aqueous phase will be affected by changes in applied products in the reaction. For pressure: example: 4HCl(g) + O (g) ⇋ 2H O(g) + 2Cl (g) C(s) + H O(g) ⇌ H (g) + CO(g) 2 2 2 2 2 In this reaction there are 5 moles of gas on the reactant side and 4 moles In this heterogeneous reaction, on the product side. A change in pressure applied to the system will the solid carbon is not included result in a shift in the equilibrium position. If the pressure is increased, when considering the eect of Le Châtelier’s principle says that the equilibrium will shift to reverse this changes in pressure. change. The forward reaction becomes favoured to reduce the pressure of the system. In the same way, a decrease in pressure will result in the reverse reaction being favoured. In the Haber process: N (g) + 3H (g) ⇋ 2NH (g) 2 2 3 there are 4 moles of gas on the reactant side and only 2 moles of gas on the product side. A high pressure will favour the forward reaction, to decrease the pressure of the system. Such a change in the equilibrium position will not affect the value of equilibrium constant, K , if the temperature remains constant. c Temperature and the equilibrium constant An understanding of the thermodynamics of a reaction is required when considering the effect of changing the temperature on the equilibrium constant. For example: N (g) + 3H (g) ⇋ 2NH (g) ΔH = –92 kJ 2 2 3 In this reaction energy can be considered a product and is released to the surroundings. 187
7 E Q U ILIBR I U M TOK exothermic reaction The work of Fritz Haber had mixed outcomes for a society. While his method of ammonia production had a ygrene laitnetop reactants signicant and long-lasting eect on increasing food ∆H production for humanity, his work with the German products military in the manufacture of explosives and the use of reaction progress chlorine in chemical warfare ▲ Figure 7 The potential energy prole of an exothermic reaction brought into question the role of scientists in The reverse reaction is endothermic – it requires energy from the society. The outcomes surroundings. of scientic endeavours often have signicant At equilibrium the forward and reverse reactions occur at equal rates ethical implications. and there is no net change in energy. Should scientists be held responsible for the way in For the exothermic reaction, an increase in temperature will shift the which society utilizes their discoveries? equilibrium in the direction that will consume the extra energy. In accordance with Le Châtelier’s principle, the equilibrium position will move to the left, favouring the reactants, to minimize the effect of the change; the concentrations of nitrogen and hydrogen will increase. This results in a decrease in the equilibrium constant K . Conversely, c a decrease in temperature for the exothermic reaction will shift the equilibrium to the right, favouring the forward reaction and increasing the concentration of ammonia, NH . This results in an increase in K . 3 c Table 4 summarizes the effects on the equilibrium system of changing temperature. Type of Change n E poston E eacton tepeate constant K exothec moves to the left, increase favouring reactants c endothec decreases decrease moves to the right, favouring products increases increase moves to the right, favouring products increases decrease moves to the left, favouring decreases reactants ▲ Table 4 The eects on the equilibrium system of a change in temperature 188
7. 1 E q u i l i b r i u m The eect of a catalyst on equilibrium reactions The addition of a catalyst provides an alternative pathway for a reaction, lowering the activation energy. In a reaction that goes to completion, a catalyst means that a greater proportion of reactants have sufcient energy to overcome the activation energy barrier and become products. In a reversible reaction, the lowered activation energy has an equal effect on both the forward and reverse reactions. The rates of the forward and reverse reactions increase by an equal amount. The position of the equilibrium will not change and there is no effect on the equilibrium constant K c qck eston PCl (g) ⇋ PCl (g) + Cl (g) ∆H = +92.5 kJ 5 3 2 Predict and explain any shift in the equilibrium position when: a) the temperature of the system is decreased ) additional chlorine gas is injected into the system c) the pressure applied to the system is increased d) a catalyst is added. Reaction quotient If a system has not reached equilibrium, the ratio of concentration of product to reactants will not equal K . This ratio is called the reaction c quotient Q and this helps you to determine the progress of the reaction as it moves toward equilibrium and the direction of the reaction that is favoured to establish equilibrium (table 5). Q>K The concentration of products is greater than at equilibrium and the c reverse reaction is favoured until equilibrium is reached. Q<K The concentration of reactants is greater than at equilibrium and the c forward reaction is favoured until equilibrium is reached. Q=K The system is at equilibrium and the forward and reverse reactions c occur at equal rates. ▲ Table 5 The relationship between the reaction quotient Q and the equilibrium constant K c 189
7 E Q U ILIBR I U M Questions 1 The equation for a reversible reaction used in industry to convert methane to hydrogen is Poston of Vae of e e constant shown below. shifts towards the decreases CH (g) + H O(g) ⇋ CO(g) + 3H (g) A. reactants B. shifts towards the increases 4 2 2 C. D. reactants decreases Δ = +210 kJ shifts towards the increases Which statement is always correct about products thisreaction when equilibrium has been shifts towards the reached? products A. The concentrations of methane and carbon monoxide are equal. IB, November 2003 B. The rate of the forward reaction is greater than the rate of the reverse reaction. C. The amount of hydrogen is three times the 4 The equation for one reversible reaction amount of methane. involving oxides of nitrogen is shown below: D. The value of ΔH for the reverse reaction is N O (g) ⇋ 2NO (g) ΔH = +58 kJ 210 kJ. [1] 2 4 2 IB, May 2006 Experimental data for this reaction can be represented on the following graph (gure 8). 2 Sulfur dioxide and oxygen react to form sulfur 1.0 0.8 product trioxide according to the equilibrium: 3 2SO (g) + O (g) ⇋ 2SO (g) md lom/noitartnecnoc 2 2 3 How are the amount of SO and the value 0.6 0.4 2 0.2 0 reactant 0 of the equilibrium constant for the reaction ▲ Figure 8 affected by an increase in pressure? A. The amount of SO and the value of the 3 equilibrium constant both increase. 2 4 6 8 10 B. The amount of SO and the value of the 3 time/min equilibrium constant both decrease. C. The amount of SO increases but the 3 valueof the equilibrium constant a) Write an expression for the equilibrium decreases. D. The amount of SO increases but the constant, K , for the reaction. Explain c 3 the signicance of the horizontal parts of value of the equilibrium constant does the lines on the graph. State what can be notchange. [1] deduced about the magnitude of K for the c IB, November 2007 reaction, giving a reason. [4] b) Use Le Châtelier’s principle to predict and explain the effect of increasing the 3 What will happen to the position of equilibrium temperature on the position of equilibrium. [2] and the value of the equilibrium constant when c) Use Le Châtelier’s principle to predict and the temperature is increased in the following explain the effect of increasing the pressure reaction? [1] on the position of equilibrium. [2] Br (g) + Cl (g) ⇋ 2BrCl(g) ΔH = +14 kJ d) State and explain the effects of a catalyst 2 2 onthe forward and reverse reactions, on the position of equilibrium, and on the value of K . [6] c IB, November 2005 190
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