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BCECE Medical BCECE Medical Solved Papers-2013

  • question_answer
    From the following reactions at 298 K, [a] \[Ca{{C}_{2}}(s)+2{{H}_{2}}O\,(l)\xrightarrow{{}}Ca{{(OH)}_{2}}(g)\]\[+{{C}_{2}}{{H}_{2}}(g)\,D{{H}^{o}}(kJ\,mo{{l}^{-1}})-127.9\] [b] \[Ca\,(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CaO(s)-635.1\] [c] \[CaO(s)+{{H}_{2}}O(I)\xrightarrow{{}}Ca{{(OH)}_{2}}\]  \[-(s)-65.2\] [d] \[C\,\,(s)+{{O}_{2}}(s)\xrightarrow{{}}C{{O}_{2}}(s)-393.5\] [e] \[{{C}_{2}}{{H}_{2}}(g)+\frac{5}{2}{{O}_{2}}(g)\xrightarrow{{}}2C{{O}_{2}}(g)\] \[+{{H}_{2}}O\,(e)-1299.58\] Calculate the heat of formation of \[Ca{{C}_{2}}(s)\] at 298 K.

    A)  \[-59.82\,KJ\,\,mo{{l}^{-1}}\]   

    B)  \[+59.82\,KJ\,\,mo{{l}^{-1}}\]

    C)  \[-190.22\,KJ\,\,mo{{l}^{-1}}\]

    D)  \[+190.22\,KJ\,\,mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    We have to calculate \[\Delta {{H}^{o}}\] for the following reaction in which one mole of \[Ca{{C}_{2}}(s)\] is formed form its elements.  .                 \[Ca(s)+2C\,(s)\xrightarrow{{}}Ca{{C}_{2}}(s)\] This is obtained when we operate.                 \[(B)+(C)+z(D)-(E)-a\] \[\therefore \Delta H=(-635.1)+(-65.2)+2\,(-393.5)\] \[-(-1299.58)-(-127.9)\,kJ\,mo{{l}^{-1}}\] \[=-59.82\,kJ\,mo{{l}^{-1}}\]


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