In the reaction, $\mathrm{H}_2 \mathrm{O}(l) E \longrightarrow \mathrm{H}_2 \mathrm{O}(s)$ at $0^{\circ} \mathrm{C}$ and 1 atom, the internal energy change is $-41 \mathrm{~kJ} / \mathrm{mol}$. What will be the value of molar enthalpy change?

The correct order of " $\Delta H_f^{\circ}$ " values of diamond (I), graphite (II) and fullerene (III) is

An air bag on adiabatic expansion undergoes $5 \%$ increase in its volume. The percentage change in pressure is $\left[\gamma_{\text {air }}=1.4\right]$

Given,

$$ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H=-285 \mathrm{~kJ} $$

$$ \begin{aligned} & \mathrm{N}_2 \mathrm{O}_5(g)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_3(l) ; \Delta H=-76.6 \mathrm{~kJ} \\ & \mathrm{~N}_2(g)+3 \mathrm{O}_2(g)+\mathrm{H}_2(g) \longrightarrow 2 \mathrm{HNO}_3(l) ; \\ & \Delta H=-348.2 \mathrm{~kJ} \end{aligned} $$

Calculate the $\Delta \mathrm{H}$ of $2 \mathrm{~N}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g})$.