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

D

Explanation

Given : `4NO_(2(g))+O_(2(g))rarr 2N_(2)O_(5(g)) , Delta H=-111 kJ`
`N_(2)O_(5(s))rarr N_(2)O_(5(g)) , Delta H = 54 kJ`
`4NO_(2(g))+O_(2(g))rarr 2N_(2)O_(5(g)) , Delta H = ?`
`eq(iii)=eq(i)-2xx eq(ii)`
`Delta H_(3)=-111-2xx54 =-219 kJ`