Ch.15 - Chemical Equilibrium
Chapter 15, Problem 77
Which of the following reactions yield appreciable equilib- rium concentrations of both reactants and products? (a) 2 Cu1s2 + O21g2 ∆ 2 CuO1s2; Kc = 4 * 1045
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Textbook Question
The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp = 7.2 * 107 at 1073 K. At the same temperature, what is Kp for each of the following reactions? (a) As21g2 + 3 H21g2 ∆ 2 AsH31g2
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Textbook Question
The vapor pressure of water at 25 °C is 0.0313 atm. Cal- culate the values of Kp and Kc at 25 °C for the equilibrium H2O1l2 ∆ H2O1g2.
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Textbook Question
For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, also write the equilibrium constant expression for Kp. (a) Fe2O31s2 + 3 CO1g2 ∆ 2 Fe1l2 + 3 CO21g2
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Textbook Question
The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilib- rium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3?
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Textbook Question
Calculate the equilibrium concentrations at 25 °C for the reaction in Problem 15.84 if the initial concentrations are
3N2O44 = 0.0200 M and 3NO24 = 0.0300 M.
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Textbook Question
The industrial solvent ethyl acetate is produced by the reac-tion of acetic acid with ethanol:
CH3CO2H1soln2 + CH3CH2OH1soln2 ∆
CH3CO2CH2CH31soln2 + H2O1soln2
Ethyl acetate (b) A solution prepared by mixing 1.00 mol of acetic acid and 1.00 mol of ethanol contains 0.65 mol of ethyl ace- tate at equilibrium. Calculate the value of Kc. Explain why you can calculate K without knowing the volume of the solution.
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