The reaction of cyanamide, `NH_(2)CN(s)`, with dioxygen was carried out in a bomb calorimeter, and `DeltaU` was found to be `-742.7 kJ mol^(-1)` at `298 K`. Calculate enthalpy change for the reaction at `298 K`.
`NH_(2)CN(g)+3/2O_(2)(g) rarr N_(2)(g)+CO_(2)(g)+H_(2)O(l)`

Explanation

`t = 298 ,DeltaU=-742.4 kJ mol^(-1)`
`DeltaH=?`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaH=DeltaU+(n_(p)-n_(r)) RT`
`DeltaH=-742.4+(2-3/2)xx8.314xx10^(-3)xx298`
`=-742.4+(0.5xx8.314xx10^(-3)xx298)`
`=-742+1.24`
`=-741.16 "kJ. mol"^(-1)`