Textbook Question
Recalculate the equilibrium concentrations in Problem 15.93 if the initial concentrations are 2.24 M N2 and 0.56 M O2. (This N2>O2 concentration ratio is the ratio found in air.)
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Ch.15 - Chemical Equilibrium
Chapter 15, Problem 97
In a basic aqueous solution, chloromethane undergoes a substitution reaction in which Cl- is replaced by OH-: CH3Cl(aq) + OH-(aq) ⇌ CH3OH(aq) + Cl-(aq). The equilibrium constant Kc is 1 * 10^16. Calculate the equilibrium concentrations of CH3Cl, CH3OH, OH-, and Cl- in a solution prepared by mixing equal volumes of 0.1 M CH3Cl and 0.2 M NaOH. (Hint: In defining x, assume that the reaction goes 100% to completion, and then take account of a small amount of the reverse reaction.)
Verified step by step guidance1
Step 1: Write the balanced chemical equation for the reaction: \( \text{CH}_3\text{Cl} + \text{OH}^- \rightleftharpoons \text{CH}_3\text{OH} + \text{Cl}^- \).
Step 2: Determine the initial concentrations of the reactants. Since equal volumes of 0.1 M \( \text{CH}_3\text{Cl} \) and 0.2 M \( \text{NaOH} \) are mixed, the initial concentration of \( \text{CH}_3\text{Cl} \) is 0.05 M and \( \text{OH}^- \) is 0.1 M after mixing.
Step 3: Assume the reaction goes to completion initially, meaning all \( \text{CH}_3\text{Cl} \) is converted to \( \text{CH}_3\text{OH} \). Calculate the concentrations at this point: \( [\text{CH}_3\text{OH}] = 0.05 \text{ M} \) and \( [\text{Cl}^-] = 0.05 \text{ M} \).
Step 4: Define \( x \) as the small amount of \( \text{CH}_3\text{OH} \) that reverts back to \( \text{CH}_3\text{Cl} \) and \( \text{OH}^- \). Set up the equilibrium expressions: \( [\text{CH}_3\text{Cl}] = x \), \( [\text{CH}_3\text{OH}] = 0.05 - x \), \( [\text{OH}^-] = 0.1 - x \), \( [\text{Cl}^-] = 0.05 + x \).
Step 5: Use the equilibrium constant expression \( K_c = \frac{[\text{CH}_3\text{OH}][\text{Cl}^-]}{[\text{CH}_3\text{Cl}][\text{OH}^-]} = 1 \times 10^{16} \) to solve for \( x \). Substitute the expressions from Step 4 into the equilibrium constant expression and solve for \( x \).
Related Practice
Open Question
The interconversion of l-___-lysine and l-b-lysine, for which Kc = 7.20 at 333 K, is catalyzed by the enzyme lysine 2,3-aminomutase. Lysine occurs in proteins while l-b-lysine is a precursor to certain antibiotics. At 333 K, a solution of l-___-lysine at a concentration of 3.00 * 10^-3 M is placed in contact with lysine 2,3-aminomutase. What are the equilibrium concentrations of l-___-lysine and l-b-lysine?
Textbook Question
The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25°C:
CH3CO2H1soln2 + CH3CH2OH1soln2 ∆
Acetic acid Ethanol CH3CO2CH2CH31soln2 + H2O1soln2 Kc = 3.4 (a) How many moles of ethyl acetate are present in an equi- librium mixture that contains 4.0 mol of acetic acid,
6.0 mol of ethanol, and 12.0 mol of water at 25 °C?
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Open Question
At 700 K, Kp = 0.140 for the reaction ClF₃(g) ⇌ ClF(g) + F₂(g). Calculate the equilibrium partial pressures of ClF₃, ClF, and F₂ if only ClF₃ is present initially, at a partial pressure of 1.47 atm.
Open Question
The reaction of iron(III) oxide with carbon monoxide is important in making steel. At 1200 K, Kp = 19.9 for the reaction Fe2O3(s) + 3 CO(g) ⇌ 2 Fe(l) + 3 CO2(g). What are the equilibrium partial pressures of CO and CO2 if CO is the only gas present initially, at a partial pressure of 0.978 atm?
Open Question
A 5.00-L reaction vessel is filled with 1.00 mol of H2, 1.00 mol of I2, and 2.50 mol of HI. Calculate the equilibrium concentrations of H2, I2, and HI at 500 K. The equilibrium constant Kc at 500 K for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 129.