Enthalpy change of the reaction $4H_{(g)} \to 2H_{2(g)}$ is $-869.6\, kJ$. The bond dissociation energy of $H-H$ bond is .....$kJ$

The values of heat of formation of $SO_2$ and $SO_3$ are $-298.2 \ kJ$ and $-395.2 \ kJ$ respectively. The heat of reaction for the following reaction will be $... \ kJ$
$SO_2 + \frac{1}{2}O_2 \to SO_3$

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 $kJ \ mol^{-1}$.

The heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be:

Heat of transition is the heat evolved or absorbed when a substance is converted from

Find the bond enthalpy of $N-H$ bond in ammonia by using the change in enthalpy for the reaction given below: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta H = -23 \ kcal$. Given bond energies: $N \equiv N = 226 \ kcal/mol$,$H-H = 103 \ kcal/mol$.