Question

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)`

Answer 1

`Deltan_(g)=(n_(P)-n_(R))_(g)=2-3/2=+1/2 mol`
`DeltaH=DeltaU+Deltan_(g)RT`
`=-742.7 kJ mol^(-1)+(+1/2 mol)`
`(8.314 xx10^(-3) kJ mol^(-1)K^(-1))(298 K)`
`=-742.7+1.2 kJ mol^(-1)=-741.5 kJ mol^(-1)`