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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 11

  • question_answer
    The heat of formation of \[C{{H}_{3}}OC{{H}_{3(g)}}\]is [Given: \[B.E{{.}_{H-H}}=103\]kcal, \[B.E{{.}_{C-H}}=87\]kcal \[B.E{{.}_{C-O}}=70\]kcal, \[B.E{{.}_{O=O}}=177\]kcal; Heat of vaporisation of 1 gram atom of carbon = 125 kcal.]

    A)  -14.5 kcal       

    B) - 15.4 kcal

    C) + 14.5 kcal      

    D) + 15.4 kcal

    Correct Answer: A

    Solution :

    [a] : The formation of \[C{{H}_{3}}OC{{H}_{3}}\]may be represented as \[2{{C}_{(s)}}+3{{H}_{2(g)}}+\frac{1}{2}{{O}_{2(g)}}\xrightarrow[{}]{{}}\] \[C{{H}_{3}}-O-C{{H}_{3(g)}};\Delta H=?\] It involves vaporisation of 2 gram atoms of solid carbon and breaking of 3 moles of H - H bonds and \[\frac{1}{2}\]mole of \[O=O\] bond resulting in the formation of \[6\text{ }C-H\] bonds and \[2\text{ }C-O\] bonds. Thus, the heat of formation of \[C{{H}_{3}}-O-C{{H}_{3}}\] is given by \[\Delta H=(2\times 125+3\times 103+\frac{1}{2}\times 177)\] \[-(6\times 87+2\times 70)\] \[=-14.5\,\text{kcal}\]


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