Chapter 15, Problem 104AP

The equilibrium constant $K_p$ for the following reaction is $\text{4}\text{.31 }\times {\text{ 10}}^{\text{-4}}\text{ at 375°C:}$

${\text{N}}_{\text{2}}{\text{(g)+3H}}_{\text{2}}\text{(}g\text{)}\stackrel{}{\rightleftarrows }2{\text{NH}}_3(g)$

In a certain experiment a student starts with 0.862 atm of N₂ and 0.373 atm of H₂ in a constant-volume vessel at $\text{375°C}\text{.}$ Calculate the partial pressures of all species when equilibrium is reached.

Interpretation Introduction

Interpretation:

The partial pressure of the species present in the given equilibrium reaction is to be calculated with given $K_P$ value.

Concept introduction:

Equilibrium constant for gas phase reactions is written in terms of the partial pressures of the species present in the reactionas the concentration of gases is directly proportional to the partial pressures. The equilibrium constant in terms of partial pressure is denoted by $K_P$ and whether the reactants are present in the solid or liquid phase, is not represented in the $K_P$ expression.

Equilibrium constant is the ratio of the concentration of reactants and products present in the chemical reaction.

For a general reaction: $a\text{A}\text{+}b\text{B}\rightleftharpoons c\text{C}\text{+}d\text{D}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_C=\frac{{\left[\text{C}\right]}^c{}_{eqm}{\left[\text{D}\right]}^d{}_{eqm}}{{\left[\text{A}\right]}^a{}_{eqm}{\left[\text{B}\right]}^b{}_{eqm}}$

Here, $K_C$ is the equilibrium constant and $C$ in $K_C$ stands for the concentration. $\left[\text{A}\right]$, $\left[\text{B}\right]$, $\left[\text{C}\right]$, and $\left[\text{D}\right]$ are the equilibrium concentration of reactants A and B and product C and D, respectively.

A and B are reactants, C is products, and $x,y$, and $z$ are their respective stoichiometric coefficients.

For the general reaction: $a\text{A}\text{+}b\text{B}\rightleftharpoons c\text{C}\text{+}d\text{D}$

The general formula for writing equilibrium expression for the reaction is given as:

$K_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

Here, $p$ in $K_p$ stands for the pressure. $P_A,P_B,P_C,\text{and}{P}_D$ are the equilibrium partial pressure of reactants A and B and products C and D.

A and B are reactants, C and D are products, and $a,b,c$, and $d$ are their respective stoichiometric coefficients.

Equilibrium constants of gas phase reaction are written in terms of partial pressures because concentration of gases is directly proportional to partial pressures.

The relationship between $K_P$ and $K_C$ is given as:

$\begin{array}{c}{K}_P=K_C{\left(RT\right)}^{\Delta n}\\ K_c=\frac{K_P}{{\left(RT\right)}^{\Delta n}}\end{array}$

Answer to Problem 104AP

Solution: $P_{{\text{N}}_2}=0.860\text{atm,}{P}_{{\text{H}}_2}=0.366\text{atm}$, and $P_{{\text{NH}}_3}=4.40\times {10}^{-3}\text{atm}$.

Explanation of Solution

Given information: $K_p=4.31\times {10}^{-4}$

The given balanced equation is as follows:

${\text{N}}_2\left(g\right)+{\text{H}}_2\left(g\right)\rightleftharpoons 2{\text{NH}}_3\left(g\right)$

The initial change equilibrium table for the reaction is as follows:

$\begin{array}{cccc}& {\text{N}}_2& {\text{H}}_2& {\text{NH}}_3\\ \text{Initial}& 0.862& 0.373& 0\\ \text{Change}& -x& -3x& +2x\\ \text{Equilibrium}& \left(0.862-x\right)& \left(0.373-3x\right)& 2x\end{array}$

The equilibrium constant expression for the reaction is given as follows:

$K_p=\frac{{\left(P_{{\text{NH}}_3}\right)}^2}{\left(P_{{\text{N}}_2}\right){\left(P_{{\text{H}}_2}\right)}^3}$

Here, $K_p$ is the equilibrium constant of pressure, $P_{{\text{N}}_2}$ is the partial pressure of ${\text{N}}_2$, $P_{{\text{H}}_2}$ is the partial pressure of ${\text{H}}_2$, and $P_{{\text{NH}}_3}$ is the partial pressure of ${\text{NH}}_3$.

Substitute the values of $K_p$, $P_{{\text{N}}_2}$, $P_{{\text{H}}_2}$, and $P_{{\text{NH}}_3}$ in the above equation,

$4.31\times {10}^{-4}=\frac{{\left(2x\right)}^2}{\left(0.862-x\right){\left(0.373-3x\right)}^3}$

As the value of $x$ is too small, therefore, the expression $\left(0.862-x\right)$ is nearly equal to $0.862$ and the expression $\left(0.373-3x\right)$ is nearly equal to $0.373$.

$\begin{array}{c}4.31\times {10}^{-4}=\frac{{\left(2x\right)}^2}{\left(0.862\right){\left(0.373\right)}^3}\\ x=\sqrt{4.82\times {10}^{-6}}\\ x=2.20\times {10}^{-3}\end{array}$

The partial pressure of $N_2$ is calculated as follows:

$\begin{array}{c}{P}_{{\text{N}}_2}=\left(0.862-2.20\times {10}^{-3}\right)\text{atm}\\ =0.860\text{atm}\end{array}$

The partial pressure of $H_2$ is calculated as:

$\begin{array}{c}{P}_{{\text{N}}_2}=\left(0.373-3\times 2.20\times {10}^{-3}\right)\text{atm}\\ =0.366\text{atm}\end{array}$

The partial pressure of $N{H}_3$ is calculated as follows:

$\begin{array}{c}{P}_{{\text{NH}}_3}=2\times 2.20\times {10}^{-3}\text{atm}\\ =4.40\times {10}^{-3}\text{atm}\end{array}$

Conclusion

The partial pressures of ${\text{N}}_2$, ${\text{H}}_2$, and $N{H}_3$ in the given reaction are $0.860\text{atm}$, $0.366\text{atm}$, and $4.40\times {10}^{-3}\text{atm}$, respectively.