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Ch.15 - Chemical Equilibrium
Chapter 15, Problem 77a

Consider the hypothetical reaction A(g) ⇌ 2 B(g). A flask is charged with 0.75 atm of pure A, after which it is allowed to reach equilibrium at 0°C. At equilibrium, the partial pressure of A is 0.36 atm. (a) What is the total pressure in the flask at equilibrium?

Verified step by step guidance
1
Start by writing the balanced chemical equation for the reaction: A(g) ⇌ 2 B(g). This indicates that one mole of A produces two moles of B.
Determine the change in pressure for A. Initially, the pressure of A is 0.75 atm, and at equilibrium, it is 0.36 atm. The change in pressure for A is therefore 0.75 atm - 0.36 atm = 0.39 atm.
Use the stoichiometry of the reaction to find the change in pressure for B. Since 1 mole of A produces 2 moles of B, the change in pressure for B is 2 times the change in pressure for A, which is 2 × 0.39 atm = 0.78 atm.
Calculate the equilibrium pressure of B. Since B is formed from A, its equilibrium pressure is equal to the change in pressure, which is 0.78 atm.
Add the equilibrium pressures of A and B to find the total pressure in the flask at equilibrium. Total pressure = pressure of A + pressure of B = 0.36 atm + 0.78 atm.

Key Concepts

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

Equilibrium Constant

The equilibrium constant (K) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction A(g) ⇌ 2 B(g), K can be calculated using the equilibrium partial pressures of A and B. Understanding K helps predict the direction of the reaction and the extent to which reactants are converted to products.
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Equilibrium Constant K

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. In the context of the reaction A(g) ⇌ 2 B(g), the total pressure at equilibrium is the sum of the partial pressures of A and B. Knowing the partial pressures allows for the calculation of total pressure and helps in understanding gas behavior in reactions.
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Partial Pressure Calculation

Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, and temperature. The ideal gas law (PV=nRT) is particularly relevant here, as it relates the total pressure, volume, and temperature of the gas mixture. Understanding these laws is essential for calculating changes in pressure and volume during chemical reactions involving gases.
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Related Practice
Textbook Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2 NOBr(𝑔) ⇌ 2 NO(𝑔) + Br2(𝑔) An equilibrium mixture in a 5.00-L vessel at 100°C contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. (b) What is the total pressure exerted by the mixture of gases?

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Textbook Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2 NOBr(g) ⇌ 2 NO(g) + Br2(g) An equilibrium mixture in a 5.00-L vessel at 100°C contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. (c) What was the mass of the original sample of NOBr?

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Open Question
Consider the hypothetical reaction A1g2 Δ 2 B1g2. A flask is charged with 0.75 atm of pure A, after which it is allowed to reach equilibrium at 0 _x001F_C. At equilibrium, the partial pressure of A is 0.36 atm. (a) What is the total pressure in the flask at equilibrium?
Textbook Question

Consider the hypothetical reaction A(g) ⇌ 2 B(g). A flask is charged with 0.75 atm of pure A, after which it is allowed to reach equilibrium at 0°C. At equilibrium, the partial pressure of A is 0.36 atm. (c) What could we do to maximize the yield of B?

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Open Question
As shown in Table 15.2, the equilibrium constant for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) is Kp = 4.34 * 10^-3 at 300 _x001F_C. Pure NH3 is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are 1.05 g NH3 in the equilibrium mixture. (a) What are the masses of N2 and H2 in the equilibrium mixture? (b) What is the total pressure in the vessel?
Textbook Question

As shown in Table 15.2, the equilibrium constant for the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) is Kp = 4.34 × 10-3 at 300°C. Pure NH3 is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are 1.05 g NH3 in the equilibrium mixture. (b) What was the initial mass of ammonia placed in the vessel?

1061
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