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Chapter 13: Problem 32

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

Short Answer

Expert verified
The value of the equilibrium constant K for the given reaction at 1100 K is \(4.37 \times 10^{-4}\).

Step by step solution

01

Write down the given information

We are given the following information: - The balanced chemical equation: \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) - The temperature, T = 1100 K - The Kp value at 1100 K = 0.25
02

Calculate the relationship between Kp and K

For any given reaction, Kp and K are related by the following formula, \[K_p = K(RT)^{\Delta n}\] where - \(K_p\) is the equilibrium constant in terms of pressure (given), - K is the equilibrium constant that we need to find, - R is the gas constant, \(R = 8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\), - T is the temperature in Kelvin, - \(\Delta n = \text{sum of coefficients of products - sum of coefficients of reactants}\).
03

Calculate Δn for the given reaction

The balanced chemical equation is \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\). From this equation, we can calculate \(\Delta n\): \[Δn = \text{sum of coefficients of products} - \text{sum of coefficients of reactants}\] \[Δn = (2) - (2 + 1) = -1\]
04

Use given values to find K

Since we know Kp, T, and Δn, we can use these values and the relationship equation from Step 2 to find K. \[0.25 = K(8.314 \cdot 1100)^{-1}\] First, calculate the value inside parentheses: \(8.314 \cdot 1100 = 9145.4\) Insert this value into the equation: \[0.25 = K(9145.4)^{-1}\] To find the value of K, raise both sides to the power of -1: \[K = \frac{1}{0.25(9145.4)}\] Now, calculate the value of K: \[K = \frac{1}{2286.35} = 4.37 \times 10^{-4}\]
05

Write the final answer

The value of the equilibrium constant K for the given reaction at 1100 K is \(4.37 \times 10^{-4}\).

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Most popular questions from this chapter

At a particular temperature, a 3.0-L flask contains \(2.4\) moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\), and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.
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