Question

The equilibrium constant for the reaction $ {{H}_{2}}O(g)+CO(g)\rightleftharpoons {{H}_{2}}(g)+C{{O}_{2}}(g) $ 64. The rate constant for the forward reaction is 160, the rate constant for the backward reaction is:

• A. 0.4
• B. 2.5
• C. 6.2
• D. $ 10.24\times {{10}^{3}} $

Answer 1

Answer: (B)

$ {{H}_{2}}O+CO\xrightarrow{{}}{{H}_{2}}+C{{O}_{2}} $ $ {{K}_{c}}=64,{{k}_{f}}=160,{{k}_{b}}=? $ We know that, Equilibrium constant $ {{K}_{c}} $ $ \text{=}\,\frac{\text{Rate}\,\text{constant}\,\text{of}\,\text{forward}\,\text{reaction}}{\text{Rate}\,\text{constant}\,\text{of}\,\text{backward}\,\text{reaction}} $ $ {{\text{K}}_{c}}=\frac{{{k}_{f}}}{{{k}_{b}}} $ So, $ 64=\frac{160}{{{k}_{b}}} $ $ {{k}_{b}}=\frac{160}{64}=2.5 $