Based on the following thermochemical equations,find the value of $x$ in $kJ$.
$(i) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} ; \Delta H = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} ; \Delta H = -242 \ kJ$
$(iv) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -x \ kJ$

The standard enthalpy of formation of $CO_{(g)}$, $CO_{2(g)}$, $N_2O_{(g)}$ and $N_2O_{4(g)}$ are respectively $-110$, $-393$, $81$ and $-10 \ kJ \ mol^{-1}$. The enthalpy change (in $kJ$) of the following reaction is $N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$

Calculate $\Delta H \ (kJ/mol)$ for the reaction:
$2FeO_{(s)} + \frac{1}{2} O_{2_{(g)}} \to Fe_2O_{3_{(s)}}$
Given $\Delta H$ values:
$(i)$ $Fe_2O_{3_{(s)}} + 3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ : $492 \ kJ/mol$
$(ii)$ $FeO_{(s)} + C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ : $156 \ kJ/mol$
$(iii)$ $C_{(graphite)} + O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-393 \ kJ/mol$
$(iv)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-283 \ kJ/mol$

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 $\text{ kJ}$)

Determine the enthalpy of formation for $H_2O_{2(l)}$,using the listed enthalpies of reaction:
$N_2H_{4(l)} + 2H_2O_{2(l)} \to N_{2(g)} + 4H_2O_{(l)}$; $\Delta_r H_1^o = -818 \ kJ/mol$
$N_2H_{4(l)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(l)}$; $\Delta_r H_2^o = -622 \ kJ/mol$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta_r H_3^o = -285 \ kJ/mol$

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then: