DHS Y5 Revision Lecture Chem Equilibria (With suggested answer)

2024 Y5 CHEMISTRY REVISION LECTURE Term 4 Week 1 Discussion Questions 2 (a) Nitrogen monoxide in the air can be converted to nitric acid, which results in acid rain. Both nitrogen monoxide and nitrogen dioxide also cause ozone layer depletion. One way of forming nitrogen monoxide is through the dissociation of nitrogen dioxide. 2NO2(g) ⇌ 2NO(g) + O2(g) ∆H = +114.2 kJ mol-1 At 494 °C, the value of Kp for the above reaction is 36.9 kPa. When nitrogen dioxide is placed in a sealed container at 494 °C, equilibrium is reached when 40% of the original nitrogen dioxide has dissociated. (i) Given that the standard enthalpy change of formation of NO(g) is +90.3 kJ mol-1, calculate the standard enthalpy change of formation of NO2(g). [1] (ii) Write an expression for the equilibrium constant, Kp, for the reaction. [1] (iii) Calculate • the initial partial pressure of nitrogen dioxide and • the total pressure at equilibrium. [3] (iv) State and explain the impact on the equilibrium yield of NO(g) and value of Kp if the process was now conducted at 400 oC. [2] 1 (a) A sample of chlorodifluoromethane, CHClF2, undergoes pyrolysis in a closed vessel of volume 18.5 dm3 according to the following equilibrium. 2CHClF2(g) ⇌ C2F4(g) + 2HCl(g) ∆H = +128 kJ mol−1 At the reaction temperature, the degree of dissociation was found to be 75%. (i) Write the expression for the equilibrium constant, Kc, for the reaction. [1] (ii) Calculate the initial amount of CHClF2 in the sample given that the magnitude of Kc at the reaction temperature was 0.780. [2] (iii) Explain how the Kc of the reaction will change when the reaction is carried out at a lower temperature. [2]

(v) The volume of the sealed container was instantaneously doubled at constant temperature, and the system allowed to reach equilibrium. Explain the effect this will have on the partial pressures of the individual gases. [3] (vi) Hence, predict how the value of Kp will change. [1] N2018/P2/2 MCQ 1 A gas P decomposes to two other gases, Q and R, according to the equation: 2P(g) ⇌ 3Q(g) + R(g) The graph below represents the decomposition of 1.0 mole of P in the presence of a catalyst at various temperatures. Which of the following statements about the above system are correct? 1 The equilibrium constant, Kp, for the decomposition reaction increases with increasing temperature. 2 At T1 K, the mole fraction of P is greater than that of R at equilibrium. 3 The decomposition reaction is spontaneous at all temperatures. Amount of P present/mol tim T1 K T2 K

2 The graph below shows how the fraction of a substance, X represented by one of the following compounds in the equilibrium mixture shown below varies with temperature at pressures of Y Pa and Z Pa. 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) ∆H = −1267 kJ mol−1 Identify X and the correct relative magnitudes of Y and Z. X Pressure A N2 Z > Y B O2 Y > Z C H2O Y > Z D NH3 Z > Y

2025 Y5 CHEMISTRY REVISION LECTURE 1 (a) A sample of chlorodifluoromethane, CHClF2, undergoes pyrolysis in a closed vessel of volume 18.5 dm3 according to the following equilibrium. 2CHClF2(g) ⇌ C2F4(g) + 2HCl(g) ∆H = +128 kJ mol−1 At the reaction temperature, the degree of dissociation was found to be 75%. (i) Write the expression for the equilibrium constant, Kc, for the reaction. [1] [1] Kc = [𝑪𝑪𝟐𝟐𝑭𝑭𝟒𝟒][𝑯𝑯𝑪𝑪𝑯𝑯]𝟐𝟐 [𝑪𝑪𝑯𝑯𝑪𝑪𝑯𝑯𝑭𝑭𝟐𝟐]𝟐𝟐 (ii) Calculate the initial amount of CHClF2 in the sample given that the magnitude of Kc at the reaction temperature was 0.780. [2] Let initial amount of CHClF2 be x 2CHClF2(g) ⇌ C2F4(g) + 2HCl(g) Initial amt / mol x 0 0 Change amt / mol –0.75x +0.375x +0.75x Equil amt / mol 0.25x 0.375x 0.75x Kc = [𝑪𝑪𝟐𝟐𝑭𝑭𝟒𝟒][𝑯𝑯𝑪𝑪𝑯𝑯]𝟐𝟐 [𝑪𝑪𝑯𝑯𝑪𝑪𝑯𝑯𝑭𝑭𝟐𝟐]𝟐𝟐 = �𝟎𝟎.𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏.𝟑𝟑 ��𝟎𝟎.𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏.𝟑𝟑 � 𝟐𝟐 �𝟎𝟎.𝟐𝟐𝟑𝟑𝟑𝟑 𝟏𝟏𝟏𝟏.𝟑𝟑 � 𝟐𝟐 = 0.780 [1] Working to find the 3 equilibrium amounts in ratio. [1] x = 4.28 mol

2 (a) Nitrogen monoxide in the air can be converted to nitric acid, which results in acid rain. Both nitrogen monoxide and nitrogen dioxide also cause ozone layer depletion. One way of forming nitrogen monoxide is through the dissociation of nitrogen dioxide. 2NO2(g) ⇌ 2NO(g) + O2(g) ∆H = +114.2 kJ mol-1 At 494 °C, the value of Kp for the above reaction is 36.9 kPa. When nitrogen dioxide is placed in a sealed container at 494 °C, equilibrium is reached when 40% of the original nitrogen dioxide has dissociated. (i) Given that the standard enthalpy change of formation of NO(g) is +90.3 kJ mol-1, calculate the standard enthalpy change of formation of NO2(g). [1] ∆H = ∆Hf (products) – ∆Hf (reactants) +114.2 = 2(90.3) + 0 – (2 × ∆Hf NO2) ∆Hf NO2 = +33.2 kJ mol-1 [1] (ii) Write an expression for the equilibrium constant, Kp, for the reaction. [1] 𝐾𝐾𝑝𝑝 = 𝑃𝑃𝑁𝑁𝑁𝑁2𝑃𝑃𝑁𝑁2 𝑃𝑃𝑁𝑁𝑁𝑁2 2 [1] (iii) Explain how the Kc of the reaction will change when the reaction is carried out at a lower temperature. [2] When temperature decreases, by LCP , the position of equilibrium shifts left towards the exothermic reaction to produce heat. [1] Concentration of the products C2F4 and HCl decreases and concentration of the reactants CHC lF2 increases. Kc will decrease. [1]

(iii) Calculate • the initial partial pressure of nitrogen dioxide and • the total pressure at equilibrium. [3] Let the initial pressure of NO2 be x kPa. 2 NO2 (g) 2 NO (g) + O2 (g) Initial pressure (kPa) x 0 0 Change in Pressure (kPa) -0.40x +0.40x +0.20x Equilibrium pressure (kPa) 0.60x 0.40x 0.20x [1] Kp = (0.40𝑥𝑥)2(0.20𝑥𝑥) (0.60𝑥𝑥)2 = 36.9 x = 415 kPa (or 415125 Pa) Hence, initial pressure of NO2 = 415 kPa [1] Equilibrium pressure = 0.60x + 0.40x + 0.20x Or (249075 + 166050 + 83025) = 1.2 x = 1.2 x 415 = 498 kPa [1] (iv) State and explain the impact on the equilibrium yield of NO(g) and value of Kp if the process was now conduct