Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

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 $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

For the reaction $H_{2} + I_{2} \rightarrow 2 HI$,provide the potential energy profile diagram and prove that it is an exothermic reaction.

The enthalpy changes for the following reactions are given:
$Cl_{2(g)} = 2Cl_{(g)}, 242.3 \, kJ \, mol^{-1}$; $I_{2(g)} = 2I_{(g)}, 151.0 \, kJ \, mol^{-1}$
$ICl_{(g)} = I_{(g)} + Cl_{(g)}, 211.3 \, kJ \, mol^{-1}$; $I_{2(s)} = I_{2(g)}, 62.76 \, kJ \, mol^{-1}$
Given that the standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is $...... \, kJ \, mol^{-1}$.

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

For which of the following reactions will the value of $\Delta H$ be positive?