Equilibrium Question 863

Question: In a chemical equilibrium, the rate constant of the backward reaction is $ 7.5\times {10^{-4}} $ and the equilibrium constant is 1.5. So the rate constant of the forward reaction is

Options:

A) $ 5\times {10^{-4}} $

B) $ 2\times {10^{-3}} $

C) $ 1.125\times {10^{-3}} $

D) $ 9.0\times {10^{-4}} $

Show Answer

Answer:

Correct Answer: C

Solution:

$ K_{c}=\frac{K_{f}}{K_{b}} $

$ K_{f}=K_{c}\times K_{b}=1.5\times 7.5\times {10^{-4}} $

$ =1.125\times {10^{-3}} $



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