Assume each reaction is carried out in an open container. For which reaction will $\mathrm{\Delta H}=\mathrm{\Delta E}$?

The factor of $\mathrm{\Delta G}$ values is important in metallurgy. The $∆\mathrm{G}$ values for the following reactions at $800°\mathrm{C}$ are given as: 

${\mathrm{S}}_2\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2{\mathrm{SO}}_2\left(\mathrm{g}\right); \mathrm{\Delta G}=-544 \mathrm{kJ}$

$2\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{S}}_2\left(\mathrm{g}\right)⟶2\mathrm{ZnS}\left(\mathrm{s}\right); \mathrm{\Delta G}=-293 \mathrm{kJ}$

$2\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2\mathrm{ZnO}\left(\mathrm{s}\right); \mathrm{\Delta G}=-480 \mathrm{kJ}$

The $∆\mathrm{G}$ for the reaction, $2\mathrm{ZnS}\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)⟶2\mathrm{ZnO}\left(\mathrm{g}\right)+2{\mathrm{SO}}_2\left(\mathrm{g}\right)$ will be:

Assertion: The enthalpy of both graphite and diamond is taken to be zero, being elementary substances.

Reason: The enthalpy of formation of an elementary substance in any state is taken as zero.

If at $298 \mathrm{K}$, the bond energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ bonds are, respectively $414, 347, 615$ and $435 \mathrm{kJ} {\mathrm{mol}}^{-1}$, the value of enthalpy change for the reaction.