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

`-743.939 kJ`

Explanation

Enthalyp change for a raction `(DeltaH)` is given by the expression,
`DeltaH=DeltaU+Deltan_(g)RT`
Where, `DeltaU`= change in internal energy
`Deltan_(g)` = change in number of moles
For the given reaction,
`Deltan_(g)=sumn_(g) ("products")-sumn_(g)("reactants")`
`=(2-2.5)` moles
`Deltan_(g)=-0.5` moles
And,
`DeltaU=-742.7 kJ"mol"^(-1)`
T=298 K
`R=8.314xx10^(-3)kJ "mol"^(-1)`
Substituting the values in the expression of `DeltaH :`
`DeltaH=(-742.7 kJ "mol"^(-1))+(-0.5 "mol")(298 K) (8.314xx10^(-3)KJ "mol"^(1)K^(-1))`
`=-742.7-1.2`
`DeltaH=-743.9 kJ "mol"^(-1)`