A) 0.4
B) 2.5
C) 6.2
D) \[10.24\times {{10}^{3}}\]
Correct Answer: B
Solution :
\[{{H}_{2}}O+CO\xrightarrow{{}}{{H}_{2}}+C{{O}_{2}}\] \[{{K}_{c}}=64,{{k}_{f}}=160,{{k}_{b}}=?\] 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, \[64=\frac{160}{{{k}_{b}}}\] \[{{k}_{b}}=\frac{160}{64}=2.5\]You need to login to perform this action.
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