**Equilibrium Constant Calculation****Problem 20:**Given the equilibrium constant, \( K_c \), for the decomposition of COBr\(_2\) is 0.190.\[ \text{COBr}_2(g) \rightleftharpoons \text{CO}(g) + \text{Br}_2(g) \quad \text{with} \quad K_1 = 0.19 \]Determine \( K_2 \) for the reaction:\[ 2\text{CO}(g) + 2\text{Br}_2(g) \rightleftharpoons 2\text{COBr}_2(g) \]**Multiple Choices:**a. 0.0361 b. 2.63 c. 5.62 d. 27.7 e. 10.5To solve for \( K_2 \), it is essential to understand the relationship between the equilibrium constants of these reactions. The given reaction is the reverse of the balanced decomposition reaction, but multiplied by 2.\[ K_1 = 0.19 \]For the reversed reaction, \( K_{reverse} = \frac{1}{K_1} \)Since the reaction is doubled, we raise \( K_{reverse} \) to the power of 2:\[ K_2 = \left(\frac{1}{K_1}\right)^2 \]\[ K_2 = \left(\frac{1}{0.19}\right)^2 = \left(5.263\right)^2 = 27.7 \]Therefore, the correct answer is: d. 27.7

20). The equilibrium constant, Ke, for the decomposition of COBR2 is 0.190. COBr2(g) =* CO(g) + Br2(g) K, = 0.19 What is K2 for the following reaction? 2C0(g) + 2B12(g) * 2COBr2(g) a. 0.0361 b. 2.63 c. 5.62 d. 27.7 е. 10.5

20). The equilibrium constant, Ke, for the decomposition of COBR2 is 0.190. COBr2(g) =* CO(g) + Br2(g) K, = 0.19 What is K2 for the following reaction? 2C0(g) + 2B12(g) * 2COBr2(g) a. 0.0361 b. 2.63 c. 5.62 d. 27.7 е. 10.5

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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