The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are, –890.3 kJ mol^–1 , –393.5 kJ mol^–1, and –285.8 kJ mol^–1 respectively. The enthalpy of formation of CH₄(g) is-

The enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^-1. The amount of heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas would be:

The enthalpy of formation of ${\mathrm{CO}}_{\left(\mathrm{g}\right)},$ ${\mathrm{CO}}_{2\left(\mathrm{g}\right)},$ ${\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)} , \mathrm{and}$ ${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}$ $$are –110 kJ ${\mathrm{mol}}^{-1}$, – 393 kJ ${\mathrm{mol}}^{-1}$, 81 kJ ${\mathrm{mol}}^{-\mathrm{1,}}$ and 9.7 kJ \(\text{mol}^{- 1}\) respectively. The value of \(\left(\Delta\right)_{r} H\) for the following reaction would be:

\(\mathrm{N_{2} O_{4 \left(g\right)} + 3 CO{\left(g\right)} \rightarrow N_{2} O_{\left(g\right)} + 3CO_{2 \left(g\right)}}\)

 ${\mathrm{N}}_2 + 3 {\mathrm{H}}_2 \to 2{\mathrm{NH}}_{3 ;} {∆}_{\mathrm{r}}{\mathrm{H}}^{°} = -92.4 \mathrm{kJ} {\mathrm{mol}}^{-1}$. The standard enthalpy of formation of ${\mathrm{NH}}_3$ gas in the above reaction would be:

The standard enthalpy of the formation of CH₃OH_(l) from the following data is:

${∆}_{\mathrm{vap}}\mathrm{H}°$ $\left({\mathrm{CCl}}_4\right)$ $=$ $30.5$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{f}}\mathrm{H}°$ $\left({\mathrm{CCl}}_4\right)$ $=$ $-$ $135.5$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{a}}\mathrm{H}°$ $\left(\mathrm{C}\right)$ $=$ $715.0$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{a}}\mathrm{H}°$ $\left({\mathrm{Cl}}_2\right)\hspace{0.17em}=$ $242$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

The enthalpy change for the reaction

 ${\mathrm{CCl}}_4$ $\left(\mathrm{g}\right)$ $\to$ $\mathrm{C}\left(\mathrm{g}\right)$ $+$ $4\mathrm{Cl}$ $\left(\mathrm{g}\right)$ would be: