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 \( x \).
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Constant (Kc)
The equilibrium constant, Kc, quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. In this case, Kc = [CH3OH][Cl-] / [CH3Cl][OH-]. A large Kc value, such as 1 * 10^16, indicates that at equilibrium, the products are favored, suggesting that the reaction proceeds almost to completion.
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Substitution Reaction
A substitution reaction involves the replacement of one functional group in a molecule with another. In this scenario, chloromethane (CH3Cl) undergoes nucleophilic substitution where the chloride ion (Cl-) is replaced by the hydroxide ion (OH-), resulting in the formation of methanol (CH3OH). Understanding this mechanism is crucial for predicting the products and the direction of the reaction.
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Stoichiometry and Concentration Calculations
Stoichiometry involves the calculation of reactants and products in chemical reactions based on their molar ratios. In this problem, the initial concentrations of CH3Cl and NaOH must be considered to determine the changes in concentration as the reaction proceeds. By assuming the reaction goes to completion and then accounting for the reverse reaction, one can calculate the equilibrium concentrations of all species involved.
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Related Practice
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?