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11th Class Physics Thermodynamical Processes Question Bank Critical Thinking

  • question_answer
    \[{{C}_{\text{(graphite)}}}+{{O}_{2}}(g)\,\to \,C{{O}_{2}}(g)\]                 \[\Delta H=-94.05\,kcal\,mo{{l}^{-1}}\] \[{{C}_{\text{(diamond)}}}+{{O}_{2}}(g)\,\to \,C{{O}_{2}}(g);\,\]\[\Delta H=-94.50\,kcal\,mo{{l}^{-1}}\] therefore                                            [DPMT 2001]

    A)                 \[{{C}_{\text{(graphite)}}}\,\to \,{{C}_{\text{(diamond)}}}\]; \[\Delta H_{298K}^{o}=-450\,cal\,mo{{l}^{-1}}\]

    B)                 \[{{C}_{\text{(diamond)}}}\,\to \,{{C}_{\text{(graphite)}}};\] \[\Delta H_{298K}^{o}=+450\,cal\,mo{{l}^{-1}}\]

    C)                 Graphite is the stabler allotrope

    D)                 Diamond is harder than graphite

    Correct Answer: C

    Solution :

               Heat energy is also involved when one allotropic form of an element is converted in to another. graphite is  the stabler allotrope because the heat of transformation of \[{{C}_{(diamond)}}\to {{C}_{(graphite)}}\].                    (i) \[{{C}_{(dia)}}+{{O}_{2(g)}}=C{{O}_{2(g)}}\Delta H=-94.5kcal\]                    (ii) \[{{C}_{(graphite)}}+{{O}_{2(g)}}=C{{O}_{2(g)}}\Delta H=-94.0\,k\,cal\]                    \[\Delta {{H}_{transformation}}=-94.5-(-94.0)\]                                 \[=-0.5k\,cal\].


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