Chapter 15, Problem 79AP

One mole of ${\text{N}}_{\text{2}}$ and three moles of ${\text{H}}_{\text{2}}$ are placed in a flask at $\text{375°C}$. Calculate the total pressure of the system at equilibrium if the mole fraction of ${\text{NH}}_{\text{3}}$ is 0.21. The $K_p$ for the reaction is $4.31\times {10}^{-4}.$

Interpretation Introduction

Interpretation:

The total pressure of the system at equilibrium is to be calculated.

Concept introduction:

According to Dalton’s law, the total pressure of the mixture oftwo or more nonreactive gases is equal to the individual partial pressures of the gases present in the mixture. The pressure applied by a gas in a mixture of gases is known asitspartial pressure.

Answer to Problem 79AP

Solution: $5.0\times {10}^{1}atm$

Explanation of Solution

Given information:

The $K_P$ for the reaction is $4.31\times {10}^{-4}$.

One mole of $N_2$ and three moles of $H_2$ are placed in a flask at ${375}^{o}C$.

The mole fraction of $N{H}_3$ is $0.21$.

Thetable for the number of moles ininitialchange and equilibrium for thereaction is given as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{cccccc}& {\text{N}}_{\text{2}}& \text{+}& {\text{3H}}_{\text{2}}& \rightleftharpoons & {\text{2NH}}_{\text{3}}\\ \text{Initial}\text{(mol):}& \text{1}& & \text{3}& & \text{0}\\ \text{Change}\text{(mol):}& -x& & -\text{3}x& & \text{2}x\end{array}}\\ \begin{array}{cccc}\text{Equilibrium}\text{(mol)}& \text{(1}-x\text{)}& \text{(3}-\text{3}x\text{)}& \end{array}2x\end{array}$

The mole fraction of ammonia in the reaction is calculated by the following expression:

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_{\text{3}}\text{=}\frac{\text{mol}\text{of}{\text{NH}}_{\text{3}}}{\text{total}\text{number}\text{of}\text{moles}}$

Substitute the values of number of moles of ammonia and total number of moles in the above equation as follows:

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{NH}}_{\text{3}}=\frac{\text{2}x}{\left(\text{1}-x\right)+\left(\text{3}-\text{3}x\right)+\left(\text{2}x\right)}\\ =\frac{\text{2}x}{\text{4}-\text{2}x}\end{array}$

Substitute $0.21$ for mole fraction of ammonia in the above equation to calculate the value of $x$ as:

$\begin{array}{c}\text{0}\text{.21}=\frac{\text{2}x}{\text{4}-\text{2}x}\\ \left(\text{4}-\text{2}x\right)\left(\text{0}\text{.21}\right)=\text{2}x\\ \text{0}\text{.84}-\text{0}\text{.42}x=\text{2}x\\ \text{0}\text{.84}=\text{2}\text{.42}x\end{array}$

On solving further,

$\begin{array}{l}x=\frac{\text{0}\text{.84}}{\text{2}\text{.42}}\\ x=\text{0}\text{.35}\end{array}$

Substitute the value of $x$ into the following mole fraction equations,

The mole fractions of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ arecalculated as follows:

For ${\text{N}}_{\text{2}}$,

$\begin{array}{c}{X}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{1}-x}{\text{4}-\text{2}x}\\ \text{=}\frac{\text{1}-\text{0}\text{.35}}{\text{4}-\text{2}\left(\text{0}\text{.35}\right)}\\ =\frac{0.65}{3.3}\\ \text{=}\text{0}\text{.20}\end{array}$

For ${\text{H}}_{\text{2}}$,

$\begin{array}{c}{X}_{{\text{H}}_{\text{2}}}=\frac{\text{3}-\text{3}x}{\text{4}-\text{2}x}\\ =\frac{\text{3}-\text{3}\left(\text{0}\text{.35}\right)}{\text{4}-\text{2}\left(\text{0}\text{.35}\right)}\\ =\frac{\text{1}\text{.95}}{\text{3}\text{.3}}\\ =\text{0}\text{.59}\end{array}$

The partial pressures of each component are equal to the product of the respectivemole fractions and the total pressure. Hence, the expressions for partial pressures of the three gases are given as follows:

$\begin{array}{c}{P}_{{\text{NH}}_{\text{3}}}=0.\text{21}{P}_{\text{T}}\\ P_{{\text{N}}_{\text{2}}}=\text{0}\text{.20}{P}_{\text{T}}\\ P_{{\text{H}}_{\text{2}}}=\text{0}\text{.59}{P}_{\text{T}}\end{array}$

The equilibrium constant expression is given as follows:

$K_p\text{=}\frac{P_{{\text{NH}}_{\text{3}}}^{\text{2}}}{P_{{\text{H}}_{\text{2}}}^{\text{3}}{P}_{{\text{N}}_{\text{2}}}}$

Here, $K_p$ is the equilibrium constant of partial pressure, $P_{{\text{NH}}_{\text{3}}}$ is the partial pressure of the product that is ammonia gas, $P_{{\text{H}}_{\text{2}}}$ is the partial pressure of hydrogen gas, and $P_{{\text{N}}_{\text{2}}}$ is the partial pressure of nitrogen gas.

Substitute the values of $K_p,P_{{\text{NH}}_{\text{3}}},P_{{\text{N}}_2}$, and $P_{{\text{H}}_{\text{2}}}$ in above expression and solve as follows:

$\begin{array}{c}\text{4}{\text{.31×10}}^{\text{-4}}\text{=}\frac{{\left(\text{0}\text{.21}\right)}^{\text{2}}{P}_{\text{T}}^{\text{2}}}{{\left(\text{0}\text{.59}{P}_{\text{T}}\right)}^{\text{3}}\left(\text{0}\text{.20}{P}_{\text{T}}\right)}\\ \text{4}{\text{.31×10}}^{\text{-4}}\text{=}\frac{\text{0}\text{.0441}}{\text{0}\text{.0410}{P}_{\text{T}}^{\text{2}}}\\ \text{4}{\text{.31×10}}^{\text{-4}}\text{=}\frac{\text{1}\text{.07}}{P_{\text{T}}^{\text{2}}}\\ P_{\text{T}}^{\text{2}}\text{=}\frac{\text{1}\text{.07}}{\text{4}{\text{.31×10}}^{\text{-4}}}\end{array}$

On solving further,

$P_{\text{T}}=\text{5}{\text{.0×10}}^{\text{1}}\text{atm}$

Conclusion

The total pressure of the system at equilibrium is $\text{5}{\text{.0×10}}^{\text{1}}\text{atm}$.