1.

The heat of combustion of naphthalene (C_(10)H_(8)(s)) at constant volume was found to be - 5133 k J mol^(-1) . Calculatethe value of enthalpy change.

Answer»


Solution :`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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