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

  • question_answer
    The molar enthalpies of combustion of isobutene and n-butane are \[-2870\text{ }kJ\text{ }mo{{l}^{-1}}\] and \[-2875\text{ }kJ\text{ }mo{{l}^{-1}}\] respectively at 298 K and 1 atm. Calculate \[\Delta H{}^\circ \] for the conversion of 1 mole of n-butane to 1 mole of isobutane

    A) \[-8\text{ }kJ\text{ }mo{{l}^{-1}}\]     

    B) \[+8\text{ }kJ\text{ }mo{{l}^{-1}}\]

    C) \[-5748\text{ }kJ\text{ }mo{{l}^{-1}}\]  

    D)   \[+5748\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    [a] \[Isobutane+oxygen\to C{{O}_{2}}+{{H}_{2}}O\] \[\Delta H=-2870\,kJmo{{l}^{-1}}\]       ......(i) \[n-butane+oxygen\to C{{O}_{2}}+{{H}_{2}}O\] \[\Delta H=-2878\,kJmo{{l}^{-1}}\]       ....(ii) (iii) (i); n-butane - Isobutane, \[\Delta H=\left( -2878+2870 \right)\] \[=-8\,kJmo{{l}^{-1}}\].


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