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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    Following method of extracting\[Zn\]is based on thermodynamics                 \[A:2ZnS+3{{O}_{2}}\xrightarrow{{}}2ZnO+2S{{O}_{2}}\]                 \[B:ZnO+C\xrightarrow{{}}Zn+CO\] If\[\Delta G_{f}^{\text{o}}\](standard free energies of formation , in\[kJmo{{l}^{-1}}\])of\[ZnS=-205.4\], \[ZnO=-318.0,\,\,S{{O}_{2}}=-300.4\]and of\[CO=-137.3\] Free energy changes of the above reaction\[A\]and\[B\](respectively) will be

    A) \[-826.4\,\,kJ,\,\,+180.9\,\,kJ\]

    B) \[+826.4\,\,kJ,\,\,-180.9\,\,kJ\]

    C) \[-826.4\,\,kJ,\,\,-180.9\,\,kJ\]

    D) \[+826.4\,\,kJ,\,\,+180.9\,\,kJ\]

    Correct Answer: A

    Solution :

    For reaction A \[\Delta {{G}^{o}}=2\Delta {{G}^{o}}_{f}(ZnO)+2\Delta {{G}^{o}}_{f}(S{{O}_{2}})-2{{G}^{o}}_{f}(ZnS)\]\[=2[-3182-300.4+205.4]\] \[=-826.4\,\,kJ\]                               \[(\Delta {{G}^{o}}_{f}(element)=0)\] For reaction B \[\Delta {{G}^{o}}=\Delta {{G}^{o}}_{f}(CO)-\Delta {{G}^{o}}_{f}(ZnO)\] \[=-137.3+3282\] \[=180.9\,\,kJ\]


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