Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

Based on Hess's law calculations,what is the average $S-O$ bond energy in $SO_3$ if $\Delta H_f^o$ of $SO_3$ is $-270 \ kJ \ mol^{-1}$. Given: Bond energy of $O=O$ is $495 \ kJ \ mol^{-1}$,heat of sublimation for $S_{(s)}$ is $277 \ kJ \ mol^{-1}$,and bond energy of $S=O$ is not provided,but we assume the formation reaction: $S_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}$. Use the atomization energy of $S_{(s)} = 277 \ kJ \ mol^{-1}$ and $O=O = 495 \ kJ \ mol^{-1}$. Calculate the average $S-O$ bond energy in $SO_3$.

Bond enthalpies of $A_{2}$,$B_{2}$ and $AB$ are in the ratio $2:1:2$. If the enthalpy of formation of $AB$ is $-100 \ kJ \ mol^{-1}$,the bond enthalpy of $B_{2}$ is:

Calculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
$CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_{1.8\,^{\circ}C} = -15.26\,K\,cal$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_{1.8\,^{\circ}C} = -68.37\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_{1.8\,^{\circ}C} = -151.80\,K\,cal$

If $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$,then the heat of formation of $CO$ is

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 ..... $kJ \, mol^{-1}$