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KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    Calculate the standard free energy change for the formation of methane at 298 K. The value of \[{{\Delta }_{r}}H{}^\circ \] for \[C{{H}_{4}}(g)\] is \[-\,74.81\,kJ\,mo{{l}^{-\,1}}\]and S values for C (graphite) \[{{H}_{2}}(g)\] and \[C{{H}_{4}}(g)\] are \[5.70,\]\[130.7\] and \[186.3J{{K}^{-\,1}}mo{{l}^{-\,1}}\] respectively.

    A) \[-\,80.8\]

    B) \[-\,98.91\]

    C) \[-\,50.74\]

    D) \[-\,40.4\]

    Correct Answer: C

    Solution :

    For the reaction;
    \[C+2{{H}_{2}}\xrightarrow{{}}C{{H}_{4}}\]
    Given,    \[\Delta H{}^\circ =-74.81\,\,kJ\,mo{{l}^{-1}}\]
    \[\Delta S_{m}^{o}\]can be calculated as:
    \[\Delta S_{m}^{o}=S_{m(product)}^{o}-S_{m\,(reactant)}^{o}\]
    \[=S_{m}^{o}C{{H}_{4}}(g)-[S_{m}^{o}{{C}_{(graphite)}}+2S_{m}^{o}{{H}_{2}}(g)]\]
    \[=[186.3-(5.70+2\times 130.7)]J{{K}^{-1}}mo{{l}^{-1}}\]
    \[=-80.8J{{K}^{-1}}mo{{l}^{-1}}\]
    \[=-80.8\times {{10}^{-3}}J{{K}^{-1}}mo{{l}^{-1}}\]
    Since \[\Delta G{}^\circ =\Delta H{}^\circ -T\Delta S{}^\circ \]
    \[=-74.81-[(298)\times (-80.8\times {{10}^{-3}})]\]
    \[=-74.81+24.07\,\,kJ\,mo{{l}^{-1}}\]
    \[=-50.74\,kJ\,mo{{l}^{-1}}\]


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