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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    The enthalpy change\[(\Delta H)\]for the neutralization of M\[HCl\] by caustic potash in dilute solution at 298 K is:

    A)  68 kJ                                    

    B)  65 kJ

    C)   57.3 kJ                                

    D)  50 kJ

    Correct Answer: C

    Solution :

                     \[\underset{\begin{smallmatrix}  (strong \\  acid \end{smallmatrix}}{\mathop{HCl}}\,+\underset{\begin{smallmatrix}  (strong \\  base) \end{smallmatrix}}{\mathop{KOH}}\,KCl+{{H}_{2}}O\] In this reaction,\[HCl\]is the .strong acid and\[KOH\]is the base and it has been found that the heat of neutralization of a strong acid with a strong base is always constant i. e., 57.3 kJ.


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