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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    The value of enthalpy change \[(\Delta H)\]for the reaction\[{{C}_{2}}{{H}_{5}}O{{H}_{(l)}}+3{{O}_{2(g)}}\to 2C{{O}_{2(g)}}+3{{H}_{2}}{{O}_{(l)}}\]at 27°C is  \[-1366.5\text{ }kJ\text{ }mo{{l}^{-1}}.\]The value of internal energy change for the above reaction at this temperature will be :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) -1369.0 kJ        

    B) -1364.0kJ        

    C) -1361.5 kJ        

    D) -1371.5 kJ

    Correct Answer: B

    Solution :

                                    \[{{C}_{2}}{{H}_{5}}OH(\ell )+3{{O}_{2}}(g)\xrightarrow[{}]{{}}2C{{O}_{2}}(g)+3{{H}_{2}}O(\ell )\]                 \[\Delta {{n}_{g}}=2-3=-1\]                 \[\Delta U=\Delta H-\Delta {{n}_{g}}RT\] \[=-1366.5-(-1)\times \frac{8.314}{{{10}^{3}}}\times 300\] \[=-1366.5+0.8314\times 3=-1364KJ\]


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