The lowest value of heat of neutralization is obtained for

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 $\text{ kJ}$)

In order to decompose $9 \ g$ of water,$142.5 \ kJ$ of heat is required. Hence,the enthalpy of formation of water is ...... $kJ/mol$.

If the standard enthalpy change $\left(\Delta_{r} H^\theta\right)$ for the reaction $H_{2(g)} + Br_{2(l)} \rightarrow 2 HBr_{(g)}$ is $-72.8 \ kJ$,the standard enthalpy of formation $\left(\Delta_{f} H^\theta\right)$ of $HBr_{(g)}$ (in $kJ \ mol^{-1}$) is

At $25^{\circ}C$,the heats of combustion for $CH_{4(g)}$,$C_{(s)}$,and $H_{2(g)}$ are $-212.4 \, kcal$,$-94.0 \, kcal$,and $-68.4 \, kcal$ respectively. The heat of formation for $CH_{4(g)}$ in $kcal$ is:

The standard heats of combustion of $C_{(s)}$,$S_{(s)}$,and $CS_{2(\ell)}$ are $-393.3 \, kJ \, mol^{-1}$,$-293.72 \, kJ \, mol^{-1}$,and $-1108.76 \, kJ \, mol^{-1}$ respectively. The standard heat of formation of $CS_{2(\ell)}$ in $kJ \, mol^{-1}$ is: