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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2001

  • question_answer
    The forward rate of reaction is \[2.5\times {{10}^{-4}}\]at equilibrium for a reaction and backward rate of reaction is\[7.5\times {{10}^{-4}}\]. Then velocity constant is:

    A)  \[2\]                                    

    B)         \[3\]

    C)         \[1/3\]                                

    D)         none of these

    Correct Answer: C

    Solution :

    Equilibrium constant \[\frac{rate\text{ }constant\text{ }of\text{ }forward\text{ }reaction}{~rate\text{ }constant\text{ }of\text{ }backward\text{ }reaction}\] \[E{{q}^{m}}\] constant \[=\frac{2.5\times {{10}^{-4}}}{7.5\times {{10}^{-4}}}=\frac{1}{3}\]


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