Standard enthalpy of formation of ${\mathrm{N}}_2{\mathrm{O}}_5$ is -100 kcal/mol and standard entropy of ${\mathrm{N}}_2,{\mathrm{O}}_2$ and ${\mathrm{N}}_2{\mathrm{O}}_5$ are 35, 40 and 115 kcal respectively, then ${\mathrm{\Delta }}_r{G}^{\circ }$ of following reaction at 227° C will be:

$2{\mathrm{N}}_2+5{\mathrm{O}}_2⟶2{\mathrm{N}}_2{\mathrm{O}}_5$