103 kJ + 2HBr(g) = H₂(g) + Br₂(g) K = 5.4 x 10-20 What is the equilibrium constant for the reaction when it is altered as below? H₂(g) + Br₂(g) 103 kJ + 2HBr(g) K' = [?] x 10?] Coefficient (green) Exponent (yellow) Enter

Chemistry for Engineering Students
4th Edition
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Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.18PAE: The reaction, 3 H2(g) + N2(g) (g), has the fol lowing equilibrium constants at the temperatures...
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103 kJ + 2HBr(g) = H₂(g) + Br₂(g)
K = 5.4 x 10-20
What is the equilibrium constant for the
reaction when it is altered as below?
H₂(g) + Br₂(g) = 103 kJ + 2HBr(g)
K' = [? ] × 10¹?]
Coefficient (green)
Exponent (yellow)
Enter
Transcribed Image Text:103 kJ + 2HBr(g) = H₂(g) + Br₂(g) K = 5.4 x 10-20 What is the equilibrium constant for the reaction when it is altered as below? H₂(g) + Br₂(g) = 103 kJ + 2HBr(g) K' = [? ] × 10¹?] Coefficient (green) Exponent (yellow) Enter
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