Observe the following reaction:
$2 A_{2(g)} + B_{2(g)} \xrightarrow{T(K)} 2 A_2 B_{(g)} + 600 \ kJ$
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of $A_2 B_{(g)}$ is:

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then:

Using the data provided,calculate the bond energy $(kJ \ mol^{-1})$ of a $C \equiv C$ bond in $C_{2}H_{2}$. (Take the bond energy of a $C-H$ bond as $350 \ kJ \ mol^{-1}$)
$2C_{(s)} + H_{2(g)} \longrightarrow C_{2}H_{2(g)} \quad \Delta H = 225 \ kJ \ mol^{-1}$
$2C_{(s)} \longrightarrow 2C_{(g)} \quad \Delta H = 1410 \ kJ \ mol^{-1}$
$H_{2(g)} \longrightarrow 2H_{(g)} \quad \Delta H = 330 \ kJ \ mol^{-1}$

The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

State and prove Hess's law of constant heat summation.

Given
$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta_{r}H^{\theta} = -92.4 \, kJ \, mol^{-1}$
What is the standard enthalpy of formation of $NH_{3}$ gas?