The bond enthalpies of $H_2$,$X_2$ and $HX$ are in the ratio of $2 : 1 : 2$. If the enthalpy for formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is ..... $kJ \ mol^{-1}$

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

What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)

In the following reaction,how much enthalpy change in $kJ$ is associated with the formation of $5.67 \ mol$ of $HCl$ gas? $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. The heat of combustion at constant pressure is $(-321.30 - x R) \ kJ$. The value of $x$ is . . . . . . .