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EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    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)  \[10.24\times {{10}^{3}}\]

    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,          \[64=\frac{160}{{{k}_{b}}}\] \[{{k}_{b}}=\frac{160}{64}=2.5\]


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