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

By Hess's law, `DeltaH+108 kJ =-111 kJ`
or `DeltaH=(-111-108)=-219 kJ`