Consider the reaction,
`4NO_(2)(g)+O_(2)(g)rarr2N_(2)O_(5)(g),Delta_(r )H=-111 kJ.` If `N_(2)O_(5)(s)` is formed instead of `N_(2)O_(5)(g)` in the above reaction, the `Delta_(r )H` value will be
(Given, `DeltaH` of sublimation for `N_(2)O_(5)` is `54 kJ mol^(-1))`

Options

A. `-165 kJ`
B. `+54 kJ`
C. `+219 kJ`
D. `-219 kJ`

Answer

A

Explanation

From Hess's law.
`DeltaH_(f)+DeltaH_("sub")=DeltaH_("reaction")`
`DeltaH_(f)=DeltaH_("reaction")-DeltaH_("sub")`
`=-111 kJ - (54 kJ)=-111-54 kJ`
Thus, the enthalpy of formation,
`DeltaH_(f)` for `N_(2)O_(5) (s)` is `-165 kJ`