$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:

The mutual heat of neutralisation of $40 \, g$ of $NaOH$ and $60 \, g$ of $CH_3COOH$ will be

The enthalpy of reaction for the reaction: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ is $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$. What will be the standard enthalpy of formation of $H_{2}O_{(l)}$?

The standard enthalpy of formation of $NH_3$ is $-46.0 \, kJ/mol$. If the enthalpy of formation of $H_2$ from its atoms is $-436 \, kJ/mol$ and that of $N_2$ is $-712 \, kJ/mol$,the average bond enthalpy of $N-H$ bond in $NH_3$ is......$kJ/mol$

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