The molar enthalpies of combustion of $C_2H_{2(g)},$ $C$ (graphite) and $H_{2(g)}$ are $-1300,$ $-394$ and $-286 \ kJ \ mol^{-1},$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is.......$kJ \ mol^{-1}$

The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

The heat of neutralization of a strong acid by a strong base is nearly equal to

The enthalpy of formation of ethane $(C_2H_6)$ from ethylene $(C_2H_4)$ by the addition of hydrogen,where the bond energies of $C-H$,$C-C$,$C=C$,and $H-H$ are $414 \ kJ/mol$,$347 \ kJ/mol$,$615 \ kJ/mol$,and $435 \ kJ/mol$ respectively,is $........$ $kJ/mol$.

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,$\Delta H = -453.5 \ cal$. This points out that