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InterviewSolution
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The heat of combustion of naphthalene `(C_(10)H_(8)(s))` at constant volume was found to be - 5133 k J `mol^(-1)` . Calculate the value of enthalpy change. |
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Answer» Correct Answer - `-5138kJ mol^(-1)` `C_(10)H_(g)(s) +12O_(2)(g) rarr 10CO_(2)(g)+4H_(2)O(l), DeltaU = - 5133 k J mol^(-1), Deltan_(g) =10-12=-2` `DeltaH = DeltaU + Deltan_(g) RT = - 5132 kJ mol^(-1) + ( - 2 mol) ( 8.314 xx 10^(-3)kJK^(-1) mol^(-1)) ( 298K)` `= -5133kJ mol^(-1) -5 kJ mol^(-1) = - 5138 kJ mol^(-1)` |
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