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.
`-111 = Delta H + 54`
`Delta H = -165 kJ`