Equilibrium Question 623
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 [KCET 1989]
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}} $
