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Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    Heat or formation of \[{{H}_{2}}O(g)\] at \[1\,\,atm\] and \[{{25}^{o}}C\] is\[-243\,\,kJ\]. \[\Delta E\] for the reaction \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(g)\]at \[{{25}^{o}}C\] is

    A) \[-243\,\,kJ\]                    

    B) \[-241.8\,\,kJ\]

    C) \[241.8\,\,kJ\]                  

    D) \[243\,\,kJ\]

    Correct Answer: B

    Solution :

    \[\Delta H=\Delta E+\Delta nRT,\,\,\Delta n=-1/2\] \[\therefore \]\[-243=\Delta E+(-1/2)\times 8.314\]\[\times 298\times {{10}^{-3}}\] \[\therefore \]  \[E=-241.76\,\,kJ\]


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