Calculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$

Which of the following equations is correct for the heat of sublimation?

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:

Calculate the enthalpy of formation of sucrose $(C_{12}H_{22}O_{11})$ in $kJ \, mol^{-1}$ using the following data:
$(i) \, C_{12}H_{22}O_{11} + 12O_2 \to 12CO_2 + 11H_2O, \, \Delta H = -5200.7 \, kJ \, mol^{-1}$
$(ii) \, C + O_2 \to CO_2, \, \Delta H = -394.5 \, kJ \, mol^{-1}$
$(iii) \, H_2 + \frac{1}{2}O_2 \to H_2O, \, \Delta H = -285.8 \, kJ \, mol^{-1}$

Consider the reaction $2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$. The magnitude of enthalpy change for the reaction in $\text{kJ mol}^{-1}$ is . . . . . . . (Nearest integer). Given: $\Delta_f H^\circ(H_2S) = -20.1 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(H_2O) = -286.0 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(SO_2) = -297.0 \text{ kJ mol}^{-1}$

$C + \frac{1}{2} O_2 \to CO; \Delta H = -42 \ kJ$
$CO + \frac{1}{2} O_2 \to CO_2; \Delta H = -24 \ kJ$
The heat of formation of $CO_2$ is ..... $kJ$.