If $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -X$,and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -Y$,calculate $\Delta_f H$ for $CO_{(g)}$ formation.

According to Hess's Law,the enthalpy change of a reaction depends on which of the following?

Given
$(i) \, 2Fe_2O_{3(s)} \to 4Fe_{(s)} + 3O_{2(g)}$
$\Delta _rG^o = + 1487.0 \, kJ \, mol^{-1}$
$(ii) \, 2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$
$\Delta _rG^o = - 514.4 \, kJ \, mol^{-1}$
Free energy change,$\Delta _rG^o$ for the reaction
$2Fe_2O_{3(s)} + 6CO_{(g)} \to 4Fe_{(s)} + 6CO_{2(g)}$ will be ..... $kJ \, mol^{-1}$

Given that,$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$ and $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$. The enthalpy of formation of $CO$ will be:

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 given data $:$
$(a) \ HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O \quad \Delta H = -69.01 \ kJ \ mol^{-1}$
$(b) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O \quad \Delta H = -72.79 \ kJ \ mol^{-1}$
Choose the correct statement $:$