The standard enthalpy of formation of $NH_3$ is $-46.0 \, kJ \, mol^{-1}$. If the enthalpy of atomization of $H_2$ is $436 \, kJ \, mol^{-1}$ and that of $N_2$ is $712 \, kJ \, mol^{-1}$,what is the average bond enthalpy of the $N-H$ bond in $NH_3$ in $kJ \, mol^{-1}$?

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

Which of the following pairs has a heat of neutralisation equal to $13.7 \, Kcal$?

The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.

From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:

$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$	$\Delta H = 42.09 \ kJ \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$	$\Delta H = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)}$	$\Delta H = 436 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2 O_{(g)}$	$\Delta H = 496 \ kJ \ mol^{-1}$

The heat of neutralization of a strong dibasic acid by a dilute solution of $NaOH$ is approximately ....... $Kcal/equivalent$.