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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The enthalpy of neutralisation of \[N{{H}_{4}}OH\]and \[C{{H}_{3}}COOH\] is \[-10.5\text{ }kcal\text{ }mo{{l}^{-1}}\] and enthalpy of neutralisation of \[C{{H}_{3}}COOH\] with strong base is\[-12.5\,kcal\text{ }mo{{l}^{-1}}\]. The enthalpy of ionisation of \[N{{H}_{4}}OH\] will be

    A) \[3.2\text{ }kcal\text{ }mo{{l}^{-1}}\]              

    B) \[2.0\text{ }kcal\text{ }mo{{l}^{-1}}\]

    C) \[3.0\text{ }kcal\text{ }mo{{l}^{-1}}\]              

    D) \[4.0\text{ }kcal\text{ }mo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

    [b] \[{{\Delta }_{neu}}H\]for strong base and strong acid \[=-13.7\text{ }kcal\,e{{q}^{-1}}\] \[\Delta {{H}_{ion}}(C{{H}_{3}}COOH)\] \[=-12.5-\left( -13.7 \right)=1.2\text{ }kcalmo{{l}^{-1}}\] \[\Delta {{H}_{ion}}(N{{H}_{4}}OH)\] \[=-10.5-(-13.7)-\Delta {{H}_{ion}}(C{{H}_{3}}COOH)\] \[=13.7-10.5-1.2\] \[=2\text{ }kcal\,mo{{l}^{-1}}\]


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