Consider the reaction      $4{\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right); {\mathrm{\Delta }}_{\mathrm{r}}H=-111\mathrm{kJ}{\mathrm{mol}}^{-1}$

If ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)$ is formed instead of ${\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)$ in the above reaction, the ${\mathrm{\Delta }}_{r}H$ value will be (Given: $\mathrm{\Delta }H$ of sublimation for ${\mathrm{N}}_2{\mathrm{O}}_5$ is $54\mathrm{kJ}{\mathrm{mol}}^{-1}$)