### Determining the Equilibrium Constant \( K_c \) for a Modified Reaction#### Given Reaction:\[ 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \qquad K_c = 0.0160 \]#### Question:What is the value of \( K_c \) for the reaction below?\[ 4H_2(g) + 2S_2(g) \rightleftharpoons 4H_2S(g) \]#### Solution:When manipulating equilibrium expressions, it's essential to understand how changes in coefficients and direction of the reaction affect the equilibrium constant. Here, the original reaction is:\[ 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \quad K_c = 0.0160 \]To obtain the new reaction:\[ 4H_2(g) + 2S_2(g) \rightleftharpoons 4H_2S(g) \]we notice that it is the reverse reaction of the original one and the coefficients are doubled.For the reverse reaction:\[ K'_c = \frac{1}{K_c} \]Doubling the coefficients of all substances in the reaction changes the equilibrium constant as follows:\[ K''_c = \left( K'_c \right)^2 \]#### Calculation:1. First, reverse the original reaction:\[ K'_c = \frac{1}{0.0160} \]2. Calculate \( K'_c \):\[ K'_c = 62.5 \]3. Since the coefficients are doubled in the new reaction:\[ K''_c = \left( 62.5 \right)^2 \]\[ K''_c = 3906.25 \]Thus, the equilibrium constant \( K_c \) for the new reaction is:\[ K''_c = 3906.25 \]#### Answer:\[ K''_c = \boxed{3906.25} \]For the user to input this value, there is an input box labeled "Kc of new reaction" and an "Enter" button next to it to submit their answer.

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Answered: 2H₂S(g) 2H₂(g) + S₂(g) Kc = 0.0160 What…

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2H₂S(g) 2H₂(g) + S₂(g) Kc = 0.0160 What is the value of Kc for the reaction below? 4H₂(g) + 2S₂(g) = 4H₂S(g) K = [?]

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

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 117CP: A mixture of N2, H2, and NH3 is at equilibrium [according to the equationN2(g)+3H2(g)2NH3(g)] as...

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