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Ch.15 - Chemical Equilibrium
Chapter 15, Problem 58

At 373 K, 𝐾𝑝 = 0.416 for the equilibrium 2 NOBr(𝑔) β‡Œ 2 NO(𝑔) + Br2(𝑔) If the equilibrium partial pressures of NOBr(𝑔) and Br2(𝑔) are both 0.100 atm at 373 K, what is the equilibrium partial pressure of NO(𝑔)?

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1
Identify the equilibrium expression for the reaction: \( K_p = \frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2} \).
Substitute the known values into the equilibrium expression: \( K_p = 0.416 \), \( P_{NOBr} = 0.100 \) atm, and \( P_{Br_2} = 0.100 \) atm.
Rearrange the equation to solve for \( (P_{NO})^2 \): \( (P_{NO})^2 = K_p \times (P_{NOBr})^2 / P_{Br_2} \).
Substitute the known values into the rearranged equation: \( (P_{NO})^2 = 0.416 \times (0.100)^2 / 0.100 \).
Calculate \( P_{NO} \) by taking the square root of \( (P_{NO})^2 \).

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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 ratio that expresses the relationship between the partial pressures of the products and reactants at equilibrium for a given reaction at a specific temperature. For the reaction 2 NOBr(g) β‡Œ 2 NO(g) + Br2(g), Kp is calculated using the formula Kp = (P_NO^2 * P_Br2) / (P_NOBr^2), where P represents the partial pressures of the gases involved.
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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 equilibrium reaction, the partial pressures of NOBr, NO, and Br2 are crucial for calculating the equilibrium constant and determining the concentrations of each species at equilibrium. The total pressure of the gas mixture is the sum of the partial pressures of all components.
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Stoichiometry of the Reaction

Stoichiometry refers to the quantitative relationship between the reactants and products in a chemical reaction. In this equilibrium, the stoichiometric coefficients indicate that for every 2 moles of NOBr that dissociate, 2 moles of NO and 1 mole of Br2 are produced. This relationship is essential for calculating the equilibrium partial pressure of NO based on the known partial pressures of NOBr and Br2.
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Related Practice
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At 25Β°C, the reaction CaCrO4(𝑠) β‡Œ Ca2+(π‘Žπ‘ž) + CrO42βˆ’(π‘Žπ‘ž) has an equilibrium constant 𝐾𝑐 = 7.1Γ—10βˆ’4. What are the equilibrium concentrations of Ca2+ and CrO42βˆ’ in a saturated solution of CaCrO4?

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Consider the following equilibrium, for which Δ𝐻<0

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Consider the reaction 4 NH3(𝑔) + 5 O2(𝑔) β‡Œ 4 NO(𝑔) + 6 H2O(𝑔), Δ𝐻 = βˆ’904.4 kJ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (c) decrease [O2]

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Consider the reaction 4 NH3(𝑔) + 5 O2(𝑔) β‡Œ 4 NO(𝑔) + 6 H2O(𝑔), Δ𝐻 = βˆ’904.4 kJ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (d) decrease the volume of the container in which the reaction occurs

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