The enthalpies of combustion of diamond and graphite are $-395.4 \, kJ$ and $-393.5 \, kJ$ respectively. The enthalpy of transformation of diamond to graphite is ..... $kJ$.

Calculate the standard enthalpy of combustion of carbon monoxide $(CO)$ given that $\Delta_f H^{\circ}(CO) = -110 \text{ kJ mol}^{-1}$ and $\Delta_f H^{\circ}(CO_2) = -393 \text{ kJ mol}^{-1}$.

Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

The $H_2O_{(g)}$ molecule dissociates as:
$(i)$ $H_2O_{(g)} \to H_{(g)} + OH_{(g)}; \Delta H = 490 \ kJ$
$(ii)$ $OH_{(g)} \to H_{(g)} + O_{(g)}; \Delta H = 424 \ kJ$
The average bond energy (in $kJ$) for water is

$Fe_2O_{3(s)} + \frac{3}{2} C_{(s)} \to \frac{3}{2} CO_{2(g)} + 2Fe_{(s)}$
$\Delta H^o = +234.1 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
$\Delta H^o = -393.5 \ kJ$
Use these equations and $\Delta H^o$ values to calculate $\Delta H^o$ for this reaction:
$4Fe_{(s)} + 3O_{2(g)} \to 2Fe_2O_{3(s)}$
..... $kJ$

Given: $H_2 + 1/2 O_2 \rightarrow H_2O : \Delta H = -68.4 \ \text{kcal}$,$C + O_2 \rightarrow CO_2 : \Delta H = -94.0 \ \text{kcal}$,and $C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O : \Delta H = -327.0 \ \text{kcal}$. Calculate the heat of formation of $C_2H_5OH$ in $\text{kcal}$.