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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Mock Test - Chemical Thermodynamics

  • question_answer
    The value of enthalpy change (\[\Delta H\]) for the reaction \[{{C}_{2}}{{H}_{2}}OH(l)+3{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(l)\]a \[27{}^\circ C\]is -1366.5 kJ\[mo{{l}^{-1}}\]. The value of internal energy change for the above reaction at this temperature will be

    A) -1371.5 kJ        

    B) -1369.0 kJ

    C) -1364.0 kJ       

    D) -1361.5 kJ

    Correct Answer: C

    Solution :

    [c] Relation between \[\Delta H\](enthalpy change) and \[\Delta E\](internal energy change) is \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] Where \[\Delta {{n}_{g}}\]=(moles of gaseous products)-(moles of gaseous reactants) For the given reaction, \[\Delta {{n}_{g}}=2-3=-1\] \[\Rightarrow -1366.5=\Delta E-1\times 8.314\times {{10}^{-3}}\times 300\] \[\therefore \Delta E=-1364.0kJ\,mo{{l}^{-1}}\]


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