The standard enthalpies of formation for $CH_{4(g)}$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-75$,$-393.5$,and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$ is ......... $kJ$.

Given the following thermochemical equations:
$(i) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H = -68.39 \, kcal$
$(ii) \ K_{(s)} + H_2O_{(l)} + aq \rightarrow KOH_{(aq)} + \frac{1}{2}H_{2(g)} ; \Delta H = -48.0 \, kcal$
$(iii) \ KOH_{(s)} + aq \rightarrow KOH_{(aq)} ; \Delta H = -14.0 \, kcal$
Calculate the heat of formation of $KOH_{(s)}$.

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

If bond enthalpies of $N \equiv N$,$H - H$ and $N - H$ bonds are $x_1, x_2$ and $x_3$ respectively,$\Delta H_f^o$ for $NH_3$ will be

The heat change $\Delta H$ for the reaction $2CO + O_2 \to 2CO_2; \Delta H = -135 \ kcal$ is called

The heat of combustion of methane is $-809 \, kJ \, mol^{-1}$. What is the calorific value of methane in $kJ \, g^{-1}$?