Find the value $\Delta H_f^o[NH_{3(g)}]$ if $\Delta H_r$ for the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is $-\,183.6 \ kJ/mol$ of $N_{2(g)}$.

Given:
$(A) \ 2 \ CO_{(g)} + O_{2(g)} \rightarrow 2 \ CO_{2(g)} \quad \Delta H_1^\theta = -x \ kJ \ mol^{-1}$
$(B) \ C \ (\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H_2^\theta = -y \ kJ \ mol^{-1}$
The $\Delta H^\theta$ for the reaction $C \ (\text{graphite}) + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$ is:

Given that:
$2C_{(s)} + 2O_{2(g)} \to 2CO_{2(g)}$; $\Delta H = -787 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}$; $\Delta H = -286 \ kJ$
$C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \to 2CO_{2(g)} + H_2O_{(l)}$; $\Delta H = -1301 \ kJ$
Calculate the heat of formation of acetylene $(C_2H_{2(g)})$ in $kJ$.

According to the following figure,the magnitude of the enthalpy change of the reaction $A+B \rightarrow M+N$ in $kJ \ mol^{-1}$ is equal to ...... .

When the aqueous solution of $0.5 \, mole$ of $HNO_3$ is mixed with $0.3 \, mole$ of $OH^{-}$ solution,what will be the liberated heat in $kJ$? (Enthalpy of neutralization is $= 57.1 \, kJ \, mol^{-1}$)

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. The heat of combustion at constant pressure is $(-321.30 - x R) \ kJ$. The value of $x$ is . . . . . . .