Given two processes:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.

The enthalpy of neutralization of $NH_4OH$ with $HCl$ is $-51.40 \, kJ/equiv$. The enthalpy of dissociation of $NH_4OH$ is ..... $kJ$.

The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is

The bond enthalpies of $C-C, C=C, H-H$ and $C-H$ bonds are $360, 600, 400$ and $410 \ kJ \ mol^{-1}$ respectively. What is the heat of hydrogenation of ethylene?

The bond dissociation enthalpy of $X_2$,$\Delta H_{\text{bond}}^{\circ}$,calculated from the given data is $...$ $kJ \ mol^{-1}$. (Nearest integer)
$M^{+}X^{-}_{(s)} \rightarrow M^{+}_{(g)} + X^{-}_{(g)} \quad \Delta H_{\text{lattice}}^{\circ} = 800 \ kJ \ mol^{-1}$
$M_{(s)} \rightarrow M_{(g)} \quad \Delta H_{\text{sub}}^{\circ} = 100 \ kJ \ mol^{-1}$
$M_{(g)} \rightarrow M^{+}_{(g)} + e^{-}_{(g)} \quad \Delta H_{i}^{\circ} = 500 \ kJ \ mol^{-1}$
$X_{(g)} + e^{-}_{(g)} \rightarrow X^{-}_{(g)} \quad \Delta H_{\text{eg}}^{\circ} = -300 \ kJ \ mol^{-1}$
$M_{(s)} + \frac{1}{2}X_{2(g)} \rightarrow M^{+}X^{-}_{(s)} \quad \Delta H_{f}^{\circ} = -400 \ kJ \ mol^{-1}$
[Given : $M^{+}X^{-}$ is a pure ionic compound and $X$ forms a diatomic molecule $X_2$ in gaseous state]

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