CET Karnataka Medical CET - Karnataka Medical Solved Paper-2012

  • question_answer
    The equilibrium constant of a reaction is \[0.008\] at 298 K. The standard free energy change of the reaction at the same temperature is

    A) \[+11.96\text{ }kJ\]       

    B) \[-11.96\text{ }kJ\]

    C)  \[-5.43\text{ }kJ\]

    D)  \[-8.46\text{ }kJ\]

    Correct Answer: A

    Solution :

    :  \[\Delta {{G}^{o}}=-2.303RT\log K\] \[\Delta {{G}^{o}}=-2.303\times 8.314\times 218\log (0.008)\] \[\Delta {{G}^{o}}=-5705.84\times -2.09\] \[\Delta {{G}^{o}}=11964.65J\] or \[11.96kJ\]


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