Consider the reaction: HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + C2H3O2-(aq) Kc = 1.8⨉10-5 at 25°C If a solution initially contains 0.210 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 °C?
Ch.16 - Chemical Equilibrium
Chapter 16, Problem 61
Consider the reaction: Br2(g) + Cl2(g) ⇌ 2BrCl(g) with Kp = 1.11 * 10^-4 at 150 K. A reaction mixture initially contains a Br2 partial pressure of 755 torr and a Cl2 partial pressure of 735 torr at 150 K. Calculate the equilibrium partial pressure of BrCl.

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Identify the initial conditions: The initial partial pressures are 755 torr for Br2 and 735 torr for Cl2.
Set up an ICE (Initial, Change, Equilibrium) table to track the changes in partial pressures as the system reaches equilibrium. Let x be the change in pressure for Br2 and Cl2, and 2x for BrCl.
Write the expression for the equilibrium constant Kp: \( K_p = \frac{{(P_{\text{BrCl}})^2}}{{P_{\text{Br}_2} \cdot P_{\text{Cl}_2}}} \).
Substitute the equilibrium expressions from the ICE table into the Kp expression: \( K_p = \frac{{(2x)^2}}{{(755 - x)(735 - x)}} \).
Solve the equation for x, which represents the change in pressure, and use it to find the equilibrium partial pressure of BrCl, which is 2x.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Constant (Kp)
The equilibrium constant (Kp) is a numerical value that expresses the ratio of the partial pressures of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction Br2(g) + Cl2(g) ⇌ 2BrCl(g), Kp is calculated using the formula Kp = (P_BrCl^2) / (P_Br2 * P_Cl2). A small Kp value, like 1.11 * 10^-4, indicates that at equilibrium, the concentration of reactants is much greater than that of the products.
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Le Chatelier's Principle
Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to counteract that change and restore a new equilibrium. In this reaction, if the partial pressures of Br2 or Cl2 are increased, the equilibrium will shift to the right, producing more BrCl until a new equilibrium is established.
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Partial Pressure
Partial pressure is the pressure exerted by a single component of a gas mixture. In the context of the reaction, the initial partial pressures of Br2 and Cl2 are given as 755 torr and 735 torr, respectively. These values are crucial for calculating the equilibrium concentrations of all species involved in the reaction, as they directly influence the equilibrium position and the final partial pressure of BrCl.
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Related Practice
Textbook Question
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Textbook Question
Consider the reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kc = 2.99⨉10-7 at 227 °C If a reaction mixture initially contains 0.175 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 °C?
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Textbook Question
Consider the reaction: A(g) ⇌ B(g) + C(g) Find the equilibrium concentrations of A, B, and C for each value of Kc. Assume that the initial concentration of A in each case is 1.0 M and that the reaction mixture initially contains no products. Make any appropriate simplifying assumptions. b. Kc = 0.010
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Textbook Question
Consider this reaction at equilibrium: 2 BrNO(g) ⇌ 2 NO(g) + Br2(g) Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance. a. NO is added to the reaction mixture.