Chapter 16 sample diploma questions 2022 (2)

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Chapter 16 Sample Diploma Questions 1 2 Skip Try this question after finishing Chapter 16 3 4 5
6 For the steam-hydrocarbon reforming process, the equilibrium law expression is Kc = [CO 2 (g)][H 2 (g)] 4 [CH 4 (g)][H 2 O(g)] 2 The reaction described by this equilibrium is: a. CH 4 (g) + 2 H 2 O(g) CO 2 (g) + 4 H 2 (g) b. CO 2 (g) + 4 H 2 (g) CH 4 (g) + 2 H 2 O(g) c. CH 4 (g) + H 2 O(g) CO 2 (g) + H 2 (g) d. CO 2 (g) + H 2 (g) CH 4 (g) + H 2 O(g) 7 Coal and natural gas contain trace amounts of sulfur compounds which, when burned, may lead to acid rain pollution. A reaction that is related to acid rain is as follows: 2 H 2 S(g) + 3 O 2 (g) 2 H 2 O(g) + 2 SO 2 (g) The equilibrium law expression for this reaction is: a. Kc = [H 2 O(g)] 2 +[SO2(g)] 2 b. Kc = [H 2 S(g)] 2 +[O 2 (g)] 3 [H 2 S(g)] 2 + [O 2 (g)] 3 [H 2 O(g)] 2 +[SO2(g)] 2 c. Kc = [H 2 O(g)] 2 [SO 2 (g)] 2 d. Kc = [H 2 S(g)] 2 [O 2 (g)] 3 [H 2 S(g)] 2 [O 2 (g)] 3 [H 2 O(g)] 2 [SO2(g)] 2 8 9
10 11 Some of the SO 2 (g) produced from the burning of coal and natural gas can react with NO 2 (g) in the atmosphere according to the equation. SO 2 (g) + NO 2 (g) NO(g) + SO 3 (g) ΔH = -41.9 kJ The equilibrium concentration of SO 3 (g) in the reaction could be increased by: a. Raising the temperature b. Adding a catalyst c. Removing SO 2 (g) d. Adding NO 2 (g) 12 13
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14 Solutions of carbolic acid, commonly known as phenol (HC 6 H 5(aq) ) are widely used as disinfectants. One such solution has a concentration of 6.44 x 10 -2 mol/L and a pH of 5.60. Carboxylic acid dissociates in water according to the equation HC 6 H 5 O(aq) + H 2 O(l) C 6 H 5 O-(aq) + H 3 O+(aq) The Kc expression for the reaction is a. Kc = [C 6 H 5 O-(aq)][H 3 O+(aq)] b. Kc = [HC 6 H 5 O(aq)][H 2 O(l)] [HC 6 H 5 O(aq)][H 2 O(l)] [C 6 H 5 O-(aq)][H 3 O+(aq)] c. Kc = [C 6 H 5 O-(aq)][H 3 O+(aq)] d. Kc = [HC 6 H 5 O(aq)] [HC 6 H 5 O(aq)] [C 6 H 5 O-(aq)][H 3 O+(aq)] 15 Nitrogen fixation occurs slowly in the atmosphere. The equation for this reaction is: N 2 (g) + O 2 (g) + 180.4 kJ 2 NO(g) K eq = 4.0 x 10 -31 The equilibrium, expression for this reaction is: a. Kc = 2 [NO(g)] b. Kc = [NO(g)] 2 [N 2 (g)][O 2 (g)] [N 2 (g)][O 2 (g)] c. Kc = [N 2 (g)][O 2 (g)] d. Kc = [N 2 (g)][O 2 (g)] 2[NO(g)] [NO(g)] 2 16 Use the information given in the previous questions to answer the following question. At equilibrium, if the [O 2 (g)]= [N 2 (g)], then: a. [NO(g)] = [N 2 (g)] b. [NO(g)] > [N 2 (g)] c. [NO(g)] = 2 [N 2 (g)] d. [NO(g)] < [N 2 (g)] 17 Equilibrium has been achieved when the: a. Total pressure does not change b. Rate of the forward reaction equals the rate of the reverse reaction c. Rate of the forward reaction is twice the rate of the reverse reaction d. Total energy changes 18 In an experiment at high temperature, 0.500mol/L of hydrogen bromide gas is placed into a sealed container and decomposes into hydrogen and bromine gases a) Write the equilibrium equation and law for this reaction. b) The equilibrium concentrations in this system are: [HBr(g)]=0.240 mol/L, [H 2 (g)]=[Br 2 (g)]= 0.130 mol/L. Calculate the equilibrium constant. 19 If the equilibrium constant for the formation of HI(g) from H 2 (g) and I 2 (g) is 40, determine the equilibrium constant for the decomposition of HI(g) at the same temperature.
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22 Sample written response #1 A chemist preformed three trials of an experiment involving a closed system at equilibrium at 700 K. the system can be represented by the equation H 2 (g) + I 2 (g) → 2 HI(g) ∆H = -6.70 kJ The chemist recorded the equilibrium concentrations of each gas in the table shown below. Equilibrium concentrations Trial H 2 (g) H 2 (g) HI(g) 1 0.165 0.0978 0.945 2 0.103 0.179 1.013 3 unknown unknown 1.55 1a. Provide the equilibrium constant expression for the formation of HI(g) 1b. Using the information from trials 1 and 2 to calculate an average value for the equilibrium constant , K c , for the formation of KI(g).
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1c. In trial 3, the equilibrium was established by allowing a sample of HI(g) to decompose. Calculate the equilibrium concentration of H 2 (g) and I 2 (g) for trial 3. 1d. Identify a stress that would increase the value of the equilibrium constant and explain how the stress is responsible for the increase. Sample written response #2 (June 2001) The formation of a pollutant gas, nitrogen monoxide (NO(g)), by the reaction of nitrogen with oxygen in a gasoline engine can be affected by changing the combustion temperature within the engine. The equilibrium constant for the production of one mole of NO(g) at 25 o C is 1.0 x 10 -17 Explain how an increase in temperature could affect the concentration of the pollutant gas and the equilibrium constant. Your response should also include: Relevant chemical equation(s) and values from the chemistry data booklet A description of two ways that car manufacturers could reduce the NO(g) emissions in new model vehicles. Sample Problem #4 Predict the effect of each of the following changes on this system at equilibrium using the following equation 2 NO 2 (g) N 2 O 4 (g) ∆H = -85.0 kJ 1. Add dinitrogen tetraoxide 2. Remove nitrogen dioxide 3. Increase the volume 4. Decrease the temperature 5. Add a catalyst Answers: 1. A, 2D, 3B, 4A, 5) 87.7 , 6) A 7) C 8) 43.7 9) C, 10) D, 11) D, 12) D, 13) A , 14)C, 15) B, 16) D, 17) B, 18) 𝐾𝑐 = [𝐻2(𝑔)][𝐵𝑟2(𝑔)] [𝐻𝐵𝑟] 2 , Kc=0.0704, 19) 1/40, 20) B 21)C, 22) D