Question

What is the standard Enthalpy of the formation of ${\mathrm{NH}}_3$ gas?

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right){∆}_{\mathrm{r}}\mathrm{H}°=-92.2\mathrm{kJ}{\mathrm{mol}}^{-1}$

Answer 1

Step 1: Standard enthalpy of formation:

• The standard enthalpy of formation of a substance is the enthalpy change that occurs when 1 mole of the substance is formed from its constituent elements in their standard states."

Step 2: Chemical reaction for the formation of ammonia gas:

• The chemical reaction for the formation of ammonia gas is depicted below:

$$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right) \\ \left(\mathrm{Nitrogen}\right)\left(\mathrm{Hydrogen}\right)\left(\mathrm{Ammonia}\right) \end{gathered}$$

• The formation enthalpy of ammonia is:

$\left(\mathrm{Enthalpy}\mathrm{of}\mathrm{formation}\right){∆}_{\mathrm{f}}\mathrm{H}=-92.2\mathrm{kJ}{\mathrm{mol}}^{-1}$

Step 3: Calculation for standard enthalpy of formation of ammonia gas:

• The enthalpy change is the enthalpy of formation when one mole of ammonia gas is created from its constituents.
• It is one-half of the enthalpy change associated with the production of two moles of ammonia.
• As a result, the enthalpy of ammonia gas formation is $\left(\frac{92.2}{2}\right)=-46.1\mathrm{kJ}{\mathrm{mol}}^{-1}$.

Thus, the standard enthalpy of formation of ammonia is $-46.1\mathrm{kJ}{\mathrm{mol}}^{-1}$.