In an exothermic reaction,$\Delta H$ is

The heat released when $0.8 \, g$ of carbon is converted to carbon dioxide is $x \, cal$. The heat released when $0.8 \, g$ of carbon is converted to carbon monoxide is $y \, cal$. If $x > y$,then the heat released when $1.86 \, g$ of carbon monoxide is converted to carbon dioxide will be:

The bond dissociation energies of $X_2$,$Y_2$ and $XY$ are in the ratio of $1 : 0.5 : 1$. $\Delta H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$. The bond dissociation energy of $X_2$ will be $...... \ kJ \ mol^{-1}$.

The enthalpy of combustion of carbon disulfide $(CS_2)$ is $-110.2 \, kJ \, mol^{-1}$. If the enthalpies of formation of $SO_2$ and $CO_2$ are $-297.4 \, kJ \, mol^{-1}$ and $-394.5 \, kJ \, mol^{-1}$ respectively,the enthalpy of formation of carbon disulfide is:

The $\Delta H_f^{\circ}$ values (in $kJ \ mol^{-1}$) for graphite, diamond, and $C_{60}$ are respectively:

On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$