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Question 6 of 13 The Kp for the reaction A (g) = 2 B (g) is 0.0730. What is Kp for the reaction 2 A (g) = 4B (g)?

Question 6 of 13 The Kp for the reaction A (g) = 2 B (g) is 0.0730. What is Kp for the reaction 2 A (g) = 4B (g)?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The given image presents a question related to chemical equilibrium, specifically focusing on the equilibrium constant for pressure, \( K_p \). ### Question: The \( K_p \) for the reaction \( A \; (g) \rightleftharpoons 2 \; B \; (g) \) is 0.0730. What is \( K_p \) for the reaction \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \)? ### Explanation: To solve this problem, we need to understand how the equilibrium constant changes with the stoichiometry of the reaction. If the coefficients of a balanced equation are multiplied by a factor \( n \), the equilibrium constant \( K_p \) for the new equation is raised to the power of \( n \). In this case, the original reaction is doubled: - Original reaction: \( A \; (g) \rightleftharpoons 2 \; B \; (g) \) - New reaction: \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \) Since the coefficients are doubled, the equilibrium constant for the new reaction becomes: \[ K_p' = (K_p)^2 = (0.0730)^2 \] ### Additional Information: The image also shows a simple on-screen calculator interface, below which numbers and basic arithmetic operators are present, perhaps to assist in solving the question. By using the calculator: \[ K_p' = 0.0730 \times 0.0730 = 0.005329 \] Hence, \( K_p \) for the reaction \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \) is approximately 0.005329.
Transcribed Image Text:The given image presents a question related to chemical equilibrium, specifically focusing on the equilibrium constant for pressure, \( K_p \). ### Question: The \( K_p \) for the reaction \( A \; (g) \rightleftharpoons 2 \; B \; (g) \) is 0.0730. What is \( K_p \) for the reaction \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \)? ### Explanation: To solve this problem, we need to understand how the equilibrium constant changes with the stoichiometry of the reaction. If the coefficients of a balanced equation are multiplied by a factor \( n \), the equilibrium constant \( K_p \) for the new equation is raised to the power of \( n \). In this case, the original reaction is doubled: - Original reaction: \( A \; (g) \rightleftharpoons 2 \; B \; (g) \) - New reaction: \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \) Since the coefficients are doubled, the equilibrium constant for the new reaction becomes: \[ K_p' = (K_p)^2 = (0.0730)^2 \] ### Additional Information: The image also shows a simple on-screen calculator interface, below which numbers and basic arithmetic operators are present, perhaps to assist in solving the question. By using the calculator: \[ K_p' = 0.0730 \times 0.0730 = 0.005329 \] Hence, \( K_p \) for the reaction \( 2 \; A \; (g) \rightleftharpoons 4 \; B \; (g) \) is approximately 0.005329.
Expert Solution
Step 1: Writing the given information,

A(g) ⇌ 2B(g)    Kp = 0.0730

2A(g) ⇌ 4B(g)  Kp = ?

Note: 

Equilibrium constant when coefficient are changed:

If a balanced chemical equation coefficient are multiplied by factor n, then  the original equilibrium constant (K) is raised to the nth power to obtain the new equilibrium constant value.


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