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

  • question_answer
    The enthalpy of atomisation of \[C{{H}_{4}}\] and \[{{C}_{2}}{{H}_{6}}\] are 360 and \[620\text{ }kcal\text{ }mo{{l}^{-1}}\] respectively. The \[C-C\] bond energy is expected to be

    A) \[210\text{ }kcal\text{ }mo{{l}^{-1}}\]             

    B) \[~80\text{ }kcal\text{ }mo{{l}^{-1}}\]

    C) \[130\text{ }kcal\text{ }mo{{l}^{-1}}\]             

    D) \[180\text{ }kcal\text{ }mo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

    [b] Atomisation of methane \[C{{H}_{4}}\left( g \right)\xrightarrow{{}}C\left( g \right)+4H\left( g \right);\] \[\Delta H=360\text{ }kcal\] \[\therefore \text{ }C-H\] bond energy \[=\frac{360}{4}\] \[=90\text{ }kcal\text{ }mo{{l}^{-1}}\] \[{{C}_{2}}{{H}_{6}}(g)\xrightarrow{{}}2C(g)+6H(g);\] \[\Delta H=620\text{ }kcal\] or \[{{H}_{C-C}}+6{{H}_{C-H}}=620\] \[\therefore {{H}_{C-C}}=620-6{{H}_{C-H}}\] \[=620-6\times 90=80\text{ }kcal\text{ }mo{{l}^{-1}}\]


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