Calculate the standard enthalpy change for the following reaction: $2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(l)$. Given: $\Delta_f H^{\circ}(C_2H_6) = -85 \text{ kJ mol}^{-1}$, $\Delta_f H^{\circ}(CO_2) = -390 \text{ kJ mol}^{-1}$, $\Delta_f H^{\circ}(H_2O) = -285 \text{ kJ mol}^{-1}$. (in $\text{ kJ}$)

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)

At $298 \, K$,the bond energies of $C-H, C-C, C=C$,and $H-H$ are $414, 347, 615$,and $435 \, kJ \, mol^{-1}$ respectively. What will be the enthalpy change for the reaction ${H_2}C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ in $kJ \, mol^{-1}$?

Comment on the thermodynamic stability of $NO_{(g)}$,given:
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}; \Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$

On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.

If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be: