Solve this:
Q.4. Consider the reaction
$4N{O}_2\left(g\right)+O_2\left(g\right)\to 2N_2{O}_5\left(g\right), {∆}_{r}H=-111 kJ$.
If $N_2{O}_5$(s) is formed instead of $N_2{O}_5$(g) in the above reaction, the ${∆}_{r}H$ value will be
(Given, $∆H$ of sublimation for $N_2{O}_5$ is 54 kJ $mo{l}^{-1}$)
(a) - 165 kJ
(b) + 54 kJ
(c) + 219 kJ
(d) - 219 kJ