Oxford IBDP Chemistry Course Book 2014 Part 1

1 7. 1 T h e e q u I l I b r I u m l a w Solution Stdy tip Have you ever wondered H SO why no units are given for equilibrium constants? 2 4 CH CH OH(l) + CH COOH(l) ⇋ CH COOCH CH (aq) + H O(l) 3 2 3 3 2 3 2 I 1.0 1.0 0.0 0.0 C α α +α +α In general no units are included for the equilibrium constant. E 1.0 α = 0.60 1.0 α = 0.60 0.0 + α 0.0 + α The reason for this is that when ● The coefcients for α, the change in concentration, must reect the coefcients of the balanced equation. In this reaction the ratios of reactants and products are 1 :1:1:1, so the coefcients for α are the equilibrium constant is all 1. We shall develop this in the following examples. derived from thermodynamic measurements, the equilibrium constant is dened in terms of activities, instead of ● If 0.6 mol of reactants remains at equilibrium, then 1.0 α = 0.60 concentrations (or par tial and so α = 0.40. We can therefore complete the calculations above pressures in the case of a and substitute the values into the equilibrium expression. gas). For an ideal system, the 3 ● The volume of the reaction mixture is 1 dm so the concentrations ctivity of a substance is the in mol dm 3 are the same as the amounts of reactants and ratio of the concentration, in products in mol. 3 (or par tial pressure mol dm [CH COOCH CH ][H O] in Pa) to its standard value _3 _2 _3 2 K = c reference, which is taken [CH CH OH][CH COOH] 3 2 3 3 5 as 1 mol dm (or 10 Pa). (0.0 + 0.40)(0.0 + 0.40) 2 Hence if the concentration of ___ (0.40) _ = = = 0.44 (1.0 0.40)(1.0 0.40) 2 (0.60) a substance in an equilibrium The equilibrium constant has no units. was 0.25 mol dm 3 , its activity would be 0.25 (0.25 mol dm 3 / 1 mol dm 3 ), meaning that the units cancel. Therefore TOK activities have no units and the Mathematics is an integral par t of the universe. From the symmetry in nature to numerical value of the activity the presence of geometric shapes in structures and organisms and Fibonacci spirals in plants and animals, mathematics is all around us. is equivalent to the numerical Mathematics can be used to create models that explain the equilibrium systems that we investigate in chemistry. Do scientists create mathematical value of the concentration. models that mirror what occurs in an equilibrium system or is mathematics a par t of the models that we use because reality is intrinsically mathematical? Interestingly in non-ideal systems (real systems), the activities will not be precisely equal to the concentrations, and notable dierences can be involved. In this text all the examples are assumed to be based on ideal systems. In the corresponding case of pure solids and pure liquids, the activities will be equal to one. Remember also from topic 5 that the concentrations of pure solids and pure liquids are not included when writing the equilibrium constant expression, K c ▲ Figure 1 The spiral structure of the Nautilus sea shell and its relationship to the Fibonacci sequence and the golden ratio has been the subject of much debate. 391

17 EQUILIBRIUM (AHL) TOK Worked example Many problems in science Sulfur dioxide, SO is oxidized in the presence of the catalyst involve assumptions that simplify the mathematics. In 2 solving quantitative problems of equilibrium systems, vanadium(V) oxide, V O . In this reaction 2.0 mol of SO and 1.4mol assumptions are made. What is the role of intuition in 2 5 2 problem solving? of O are mixed in a 3.0 3 sealed container and the system is 2 dm allowed to come to equilibrium. At 700K a conversion rate of 15 % is achieved. Calculate the equilibrium constant for this reaction. Solution 2SO (g) + O (g) ⇋ 2SO (g) 2 2 3 I 2.0 1.4 0.0 C 2α α +2α E 1.7 1.25 0.30 ● The change in concentrations reects the coefcients of the balanced equation. For example for SO the change is 2α 2 ● The conversion rate of 15 % means that 15% of SO is converted 2 into product. Hence the equilibrium amount for SO will be: 3 2.0 × 15% = 0.30 α can then be found as follows: 0.30 = +2α α = 0.15 ● Substitute this value into the equilibrium constant expression, remembering to convert the amounts into concentrations. [SO 2 ] _3 K = c 2 [SO ] [O ] 2 2 2 0.30 _ () 2 K 3.0 = 7.5 × 10 __ = c 2 _1.7 _1.25 ( )( ) 3.0 3.0 Gis f ngy nd iii The Gibbs free energy G describes the spontaneity and temperature dependence of a reaction (sub-topic 15.2). The free energy will change as reactants are converted into products. The reaction will be spontaneous in the direction that results in a decrease in free energy (or the direction in which the free energy value becomes more negative). During this discussion we shall explore the relationship between the free energy, entropy, and position of the equilibrium. When the equilibrium constant K is determined for a given reaction, its value indicates whether products or reactants are favoured at equilibrium. The Gibbs free energy change ∆G for a given reaction is an indication of whether the forward or reverse reaction is favoured. The 392

1 7. 1 T h e e q u I l I b r I u m l a w relationship between Gibbs free energy and the equilibrium constant K is summarized in table 1. eiii constnt Dsciption Gis f ngy cng K=1 at equilibrium, neither ∆G = 0 K>1 reactants nor products favoured products favoured ∆G < 0 (negative value) K<1 reactants favoured ∆G > 0 (positive value) ▲ Table 1 The relationship between the equilibrium constant and the Gibbs free energy change At a given temperature, a negative ∆G value for a reaction indicates that Stdy tip the reaction is spontaneous and the equilibrium concentrations of the The expression ∆G = -RT ln K products are larger than the equilibrium concentrations of the reactants. is provided in section 1 of the The equilibrium constant is greater than 1. The more negative the value Data booklet of ∆G, the more the forward reaction is favoured and the larger the value of K. The gas constant (R) has the value and units of The quantitative relationship between standard Gibbs free energy 8.31 J K 1 1 change, temperature, and the equilibrium constant is described in the mol . This is equation: provided in section 2 of the Data booklet. The standard ∆G = -RT ln K Gibbs free energy, ΔG°, has units of kJ mol 1 . When using By rearranging this equation it is possible to calculate the equilibrium this expression, as shown in constant, and hence deduce the position of equilibrium for the reaction. the worked example below, ln K = _∆G - either R has to be changed to RT 0.00831 kJ K 1 1 mol or ΔG° The standard Gibbs free energy change can be calculated using the conver ted to J mol 1 . methods described in sub-topic 15.2. Worked example Calculate the equilibrium constant at 300 K for Rearranging the equation to solve for K, the oxidation of iron: ∆G = -RT ln K 3 2Fe(s) + O (g) ⇋ Fe O (s) 2 2 2 3 ∆G _ ln K = - RT 1 ∆H = -824.2 kJ mol 3 1 _743.1_× 10 J_mol ln K = - = 298 1 1 1 1 mol mol ∆S = -270.5 J K 8.31 J K × 300 K 298 129 K= e = 2.6 × 10 Solution The very large value of K demonstrates that the oxidation of iron at room temperature is highly First nd ∆G : favoured. Reactions of this nature are said to be irreversible. ∆G = ∆H - T∆S 1 = -824.2 - (300 × -0.2705) kJ mol 1 = -743.1 kJ mol 393

17 EQUILIBRIUM (AHL) Questions 1 Consider the following equilibrium reaction. A. 0.64 Cl (g) + SO (g) ⇋ SO Cl (g) ∆H = -84.5 kJ B. 1.3 2 2 2 2 3 In a 1.00 dm closed container, at 375 °C, C. 2.6 8.60× 10 3 and 8.60 × 10 3 mol of SO mol 2 D. 64 [1] of Cl were introduced. At equilibrium, 7.65 × 2 10 4 of SO Cl was formed. IB, May 2010 mol 2 2 a) Deduce the equilibrium constant expressionK for the reaction. [1] c 4. a) The production of ammonia is an important b) Determine the value of the equilibrium industrial process. constant K . [3] c N (g) + 3H (g) ⇋ 2NH (g) 2 2 3 c) If the temperature of the reaction is i) Using the average bond enthalpy changedto 300 °C, predict, stating a reason values in Table 10 of the Data Booklet, in each case, whether the equilibrium determine the standard enthalpy concentration of SO Cl and the value of 2 2 change for this reaction. [3] K will increase ordecrease. [3] c d) If the volume of the container is changed ii) The standard entropy values, S, at 298 3 K for N (g), H (g) and NH (g) are 193, to1.50 dm , predict, stating a reason 2 2 3 in eachcase, how this will affect the 131 and 192 JK 1 1 mol respectively. equilibrium concentration of SO Cl and Calculate ∆S for the reaction and with 2 2 thevalue of K . [3] reference to the equation above, explain c the sign of ∆S . [4] e) Suggest, stating a reason, how the additionof a catalyst at constant iii) Calculate ∆G for the reaction pressure and temperature will affect the at 298 K. [1] equilibriumconcentrationof SO Cl . [2] 2 2 IB, November 2009 iv) Describe and explain the effect of increasing temperature on the spontaneity of the reaction. [2] 2 When a mixture of 0.100 mol NO, 0.051 mol b) The reaction used in the production of H and 0.100 mol HO were placed in a 1.0 3 ammonia is an equilibrium reaction. 2 2 dm ask at 300 K, the following equilibrium was Outline the characteristics of a system at established. equilibrium. [2] 2NO(g) + 2H (g) ⇋ N (g) + H O(g) c) Deduce the equilibrium constant expression, 2 2 2 K , for the production of ammonia. [1] c At equilibrium, the concentration of NO was 3 found to be 0.062 mol dm . Determine the d) i) 0.20 mol of N (g) and 0.20 mol of H (g) 2 2 equilibrium constant, K , of the reaction at this were allowed to reach equilibrium c 3 temperature. in a 1 dm closed container. At equilibrium the concentration of NH (g) 3 IB, May 2009 was 0.060 mol dm 3 . Determine the 3. 0.50 mol of I (g) and 0.50 mol of Br (g) equilibrium concentrations of N (g) 2 2 2 and H (g) and calculate the value 2 [3] are placed in a closed ask. The following of K . c equilibrium is established. ii) Predict and explain how increasing the temperature will affect the I (g) + Br (g) ⇋ IBr(g) 2 2 value of K . [2] c The equilibrium mixture contains 0.80 mol of IB, May 2010 IBr(g). What is the value of K ? c 394