Answer
Partial pressure $\left(\mathrm{NH}_{3}\right)$ is equal to 1.36 $\mathrm{atm}$
Partial pressure $\left(\mathrm{N}_{2}\right)$ is equal to 0.34 atm
Work Step by Step
In this excercise we have a vessel of volume $\mathbf{V}=22.4 \mathrm{dm}^{3}$ and it contains 2.0
mol $\mathrm{H}_{2}$ and 1.0 $\mathrm{mol} \mathrm{N}_{2}$ at the temperature of $\mathrm{T}=273.15 \mathrm{K}$ initially.
When these two molecules react they form $\mathrm{NH}_{3}$
It is shown with the reaction:
$3 \mathrm{H}_{2}+\mathrm{N}_{2} \longrightarrow 2 \mathrm{NH}_{3}$
We have to calculate partial pressures of the gases in the final mixture and
the total pressure
Firstly we will calculate the moles of nitrogen that react with 2 moles of
hydrogen:
$$
2.0 \mathrm{mol} \mathrm{H}_{2} \cdot \frac{1 \mathrm{mol} \mathrm{N}_{2}}{3 \mathrm{mol} \mathrm{H}_{2}}=0.667 \mathrm{mol} \mathrm{N}_{2}
$$
Moles of nitrogen left after the reaction can be calculated as: total
moles of nitrogen minus moles of nitrogen reacted
So:
$$
\begin{aligned} \text { Mol of } \mathrm{N}_{2} \text { left } &=1.0 \mathrm{mol}-0.667 \mathrm{mol} \\ &=0.333 \mathrm{mol} \end{aligned}
$$
Moles of ammonia that are formed in this reaction are based on the moles of hydrogen reacted
$$
\begin{aligned} \text { Mole of } \mathrm{NH}_{3} \text { formed } &=2.0 \mathrm{mol} ] \mathrm{H}_{2} \cdot \frac{2 \mathrm{mol} \mathrm{NH}_{3}}{3 \mathrm{molH}_{2}} \\ &=1.333 \mathrm{mol} \mathrm{NH}_{3} \end{aligned}
$$
And now total moles of gas in the conteiner:
$0.333 \mathrm{mol}+1.333 \mathrm{mol}=1.666 \mathrm{mol}$
_______________________________________________________________
Total pressure of the final mixture would be calculated by using ideal gas
equation:
$p V=n R T$
$p=\frac{n R T}{V}$
These symbols are:
$\mathbf{p}$ = total pressure
$\mathbf{n}=1.666$ mol - total number of moles
$\mathbf{R}=0.0821 \mathrm{dm}^{3}$ atmmol $^{-1} \mathrm{K}^{-1}$ - gas constant
$\mathbf{T}=273.15 \mathrm{K}$ - temperature
$\mathbf{V}=22.4 \mathrm{dm}^{3}$ - volume
So total pressure in final mixture is:
$\begin{aligned} p &=\frac{n R T}{V} \\ &=\frac{1.666 \mathrm{mol} \cdot 0.0821 \mathrm{dm}^{3} \mathrm{atmmol}^{-1} \mathrm{K}^{-1} \cdot 273.15 \mathrm{K}}{22.4 \mathrm{dm}^{3}} \\ &=1.7 \mathrm{atm} \end{aligned}$
Mole fraction of $\mathrm{N}_{2}$ and $\mathrm{NH}_{3} :$
$$
\begin{aligned} \text { Mole fractionof } \mathrm{N}_{2} &=\frac{0.333 \mathrm{mol}}{1.666 \mathrm{mol}} \\ &=0.20 \end{aligned}
$$
$$\begin{aligned} \text { Mole fractionof } \mathrm{NH}_{3} &=\frac{1.333 \mathrm{mol}}{1.666 \mathrm{mol}} \\ &=0.80 \end{aligned}$$
The partial pressure of a gas in the mixture will be calculated when we multiply
product of mole fraction of the gas and total pressure of the mixture:
$\begin{aligned} \text { Partial pressure of } \mathrm{N}_{2}, p_{\mathrm{N}_{2}} &=\chi_{N_{2}} p_{\text { total }} \\ &=0.20 \cdot 1.7 \mathrm{atm} \\ &=0.34 \mathrm{atm} \end{aligned}$
$\begin{aligned} \text { Partial pressure of } \mathrm{NH}_{3}, p_{\mathrm{N}_{2}} &=\chi_{N H_{3}} p_{\text { total }} \\ &=0.80 \cdot 1.7 \mathrm{atm} \\ &=1.36 \mathrm{atm} \end{aligned}$
