When $0.5 \ g$ of sulphur is burnt to $SO_2$,$4.6 \ kJ$ of heat is liberated. What is the enthalpy of formation of sulphur dioxide in $kJ \ mol^{-1}$?

The enthalpies of formation of $CO_{2(g)}$,$H_2O_{(g)}$,and $C_2H_{4(g)}$ are $-393.7$,$-241.8$,and $52.3 \ kJ \ mol^{-1}$ respectively. What will be the enthalpy of combustion of ethylene at $298 \ K$ and $1 \ atm$ pressure in $kJ \ mol^{-1}$?

Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$

Which of the following reactions represents the standard heat of formation of the product?

The enthalpies of combustion of $C$(graphite) and $C$(diamond) are $-393.8 \ kJ \ mol^{-1}$ and $-395.3 \ kJ \ mol^{-1}$ respectively. The enthalpy of conversion of $C$(graphite) to $C$(diamond) is

Calculate the heat change for the reaction $4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(l)$ in $kJ$. The enthalpies of formation of $NH_3(g)$ and $H_2O(l)$ at $298 \ K$ are $-46.0 \ kJ \ mol^{-1}$ and $-286.0 \ kJ \ mol^{-1}$ respectively.