The enthalpy of formation of `CO(g),CO_(2)(g),N_(2)O(g) and N_(2)O_(4)(g)` is -110, -393, +811 and 10kJ/mol respectively.
for the reaction, `N_(2)O_(4)(g)+3CO(g)toN_(2)O(g)+3CO_(2)(g) DeltaH_(r)(kJ//mol)` is-

Options

A. `-212`
B. `+212`
C. `+48`
D. `-48`

Answer

D

Explanation

`N_(2)O_(4)(g)+3CO(g)toN_(2)O(g)+3CO_(2)(g)`
`DeltaH_("reaction")=sum^("Heat of formation of product")-sum^("Heat of formation of reactant")`
`DeltaH_("reaction")=[DeltaH_(f)(N_(2)O)+3xxDeltaH_(f)(CO_(2))]-[DeltaH_(f)(N_(2)O_(4))+3xxDeltaH_(f)(CO)]`
`DeltaH_("reaction")=[+811+3(-393)]-[10+3(-110)]`
`=[811-1179]-[-320]=-368+320`
`=48kJ*mol^(-1)`.