Question

$\underset{\mathrm{Nitrogen}\mathrm{gas}}{{\mathrm{N}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{3{\mathrm{H}}_2\left(\mathrm{g}\right)}\rightleftharpoons \underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3\left(\mathrm{g}\right)}$ for the reaction initially the mole ratio was $1:3$ of ${\mathrm{N}}_2:{\mathrm{H}}_2$. At equilibrium $50\%$ of each has reacted. If the equilibrium pressure is $\mathrm{P}$, the partial pressure of Ammonia at equilibrium is ?

Answer 1

Step 1: Analyze the reaction

• As, $50\%$of Nitrogen (${\mathrm{N}}_2$), Hydrogen (${\mathrm{H}}_2$) has reacted
• So, the reaction between Nitrogen and Hydrogen to form Ammonia (${\mathrm{NH}}_3$), proceeds as follows:

$$\begin{gathered} \underset{\mathrm{Nitrogen}\mathrm{gas}}{{\mathrm{N}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{3{\mathrm{H}}_2\left(\mathrm{g}\right)}\rightleftharpoons \underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3\left(\mathrm{g}\right)} \\ \mathrm{At}\mathrm{initial}:\mathrm{a}3\mathrm{a}0 \\ \mathrm{At}\mathrm{equilibrium}:\mathrm{a}-\left(\frac{\mathrm{a}}{2}\right)3\mathrm{a}-\left(\frac{3\mathrm{a}}{2}\right)2\left(\frac{\mathrm{a}}{2}\right) \end{gathered}$$

Step 2: Calculate total moles at equilibrium

• According to the reaction, the total number of moles at equilibrium can be calculated as

$$\begin{gathered} =\mathrm{a}-\left(\frac{\mathrm{a}}{2}\right)+3\mathrm{a}-\left(\frac{3\mathrm{a}}{2}\right)+2\left(\frac{\mathrm{a}}{2}\right) \\ =\mathrm{a}-\frac{\mathrm{a}}{2}+3\mathrm{a}-\frac{3\mathrm{a}}{2}+\frac{2\mathrm{a}}{2} \\ =\mathrm{a}-\frac{\mathrm{a}}{2}+3\mathrm{a}-\frac{\mathrm{a}}{2} \\ =3\mathrm{a} \end{gathered}$$

Step 3: Calculate Partial Pressure of Ammonia at equilibrium

• Now we have to calculate partial pressure of Ammonia at equilibrium

$\mathrm{Partial}\mathrm{pressure}\mathrm{of}{\mathrm{NH}}_3=\left(\frac{\mathrm{mole}\mathrm{of}{\mathrm{NH}}_3}{\mathrm{Total}\mathrm{moles}}\right)\mathrm{x}\mathrm{Total}\mathrm{pressure}$

• Now, by substituting the values, we get

${\mathrm{P}}_{{\mathrm{NH}}_3}=\left(\frac{\mathrm{a}}{3\mathrm{a}}\right)\mathrm{x}\mathrm{P}$

• Hence, the partial pressure of Ammonia at equilibrium is $\frac{\mathrm{P}}{3}$.