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Ch.16 - Chemical Equilibrium
Chapter 16, Problem 82

A system at equilibrium contains I2(g) at a pressure of 0.21 atm and I(g) at a pressure of 0.23 atm. The system is then compressed to half its volume. Find the pressure of each gas when the system returns to equilibrium.

Verified step by step guidance
1
Identify the initial equilibrium condition and the equilibrium constant expression for the reaction: \( \text{I}_2(g) \rightleftharpoons 2\text{I}(g) \).
Calculate the initial equilibrium constant \( K_p \) using the initial pressures: \( K_p = \frac{(P_{\text{I}})^2}{P_{\text{I}_2}} \).
Determine the effect of compression on the system: halving the volume doubles the initial pressures of both gases.
Set up the new equilibrium expression using the doubled initial pressures and introduce a change variable \( x \) to account for the shift in equilibrium.
Solve the equilibrium expression for \( x \) to find the new equilibrium pressures of \( \text{I}_2(g) \) and \( \text{I}(g) \).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In this case, compressing the system affects the pressures of the gases, prompting a shift in equilibrium to restore balance.
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Le Chatelier's Principle

Ideal Gas Law

The Ideal Gas Law (PV=nRT) relates the pressure, volume, and temperature of an ideal gas. In this scenario, the initial pressures and the change in volume due to compression will influence the final pressures of I2 and I, allowing us to calculate the new equilibrium state.
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Ideal Gas Law Formula

Equilibrium Constant (Kp)

The equilibrium constant (Kp) for a reaction involving gases is defined in terms of the partial pressures of the reactants and products at equilibrium. Understanding Kp helps predict how the pressures of I2 and I will adjust after the system is compressed and reaches a new equilibrium.
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Related Practice
Textbook Question

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ⇌ HbCO(aq) + O2(aq) a. Use the reactions and associated equilibrium constants at body temperature given here to find the equilibrium constant for the reaction just shown. Hb(aq) + O2(aq) ⇌ HbO2(aq) Kc = 1.8 Hb(aq) + CO(aq) ⇌ HbCO(aq) Kc = 306

Textbook Question

At 650 K, the reaction MgCO3(s) ⇌ MgO(s) + CO2(g) has Kp = 0.026. A 10.0-L container at 650 K has 1.0 g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of 0.100 L. Find the mass of MgCO3 that is formed.

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Textbook Question

Consider the exothermic reaction: C2H4(g) + Cl2(g) ⇌ C2H4Cl2(g) If you were trying to maximize the amount of C2H4Cl2 produced, which tactic might you try? Assume that the reaction mixture reaches equilibrium. a. increasing the reaction volume b. removing C2H4Cl2 from the reaction mixture as it forms c. lowering the reaction temperature d. adding Cl2

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Textbook Question

Consider the endothermic reaction: C2H4(g) + I2(g) ⇌ C2H4I2(g) If you were trying to maximize the amount of C2H4I2 produced, which tactic might you try? Assume that the reaction mixture reaches equilibrium. a. decreasing the reaction volume b. removing I2 from the reaction mixture c. raising the reaction temperature d. adding C2H4 to the reaction mixture

Textbook Question

Consider the reaction: H2(g) + I2(g) ⇌ 2 HI(g) A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.621 atm and PHI = 0.101 atm. Is the second reaction at equilibrium? If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?