Chapter 15, Problem 33QP

Consider the equilibrium:

$2\text{NOBr(}g\text{)}\stackrel{}{\rightleftarrows }2\text{NO(}g\text{)+Br(}g\text{)}$

If nitrosyl bromide ( $\text{NOBr}$ ) is 34 percent dissociated at $\text{25°C}$ and the total pressure is 0.25 atm. calculate $K_p$ and $K_c$ for the dissociation at this temperature.

Interpretation Introduction

Interpretation:

The values of equilibrium constants, $K_C$ and $K_P$, of the given reaction with total pressure are to be calculatedat $\text{25}°\text{C}$.

Concept introduction:

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

The equilibrium constant in terms of partial pressure is denoted by $K_P$, and the reactants present in solid or liquid phase are not present 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: $\text{xA}\text{+}\text{yB}\rightleftarrows \text{zC}$

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

$K_C=\frac{{\left[\text{C}\right]}^Z{}_{eqm}}{{\left[\text{A}\right]}^X{}_{eqm}{\left[\text{B}\right]}^Y{}_{eqm}}$

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

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

For the general reaction: $a\text{W}\text{+}b\text{X}\to c\text{Y}\text{+}d\text{Z}$

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

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

Here, $p$ in $K_p$ stands for the pressure. $P_W,P_X,P_Y,and{P}_Z$ are the equilibrium partial pressure of reactants W and X and products Y and Z.

W and X are reactants, Y and Z 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:

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

The expressions for the partial pressures at equilibrium are given as:

$P_A=\left[\text{A}\right]RT$

Where $\left[\text{A}\right]$ is the molar concentration of A.

Answer to Problem 33QP

Solution: The $K_C$ value is $3.9\times {10}^{-4}$, and the $K_P$ value is $9.6\times {10}^{-3}$.

Explanation of Solution

Given Information: The given reaction is as follows:

$\text{2NOBr}\left(g\right)\rightleftarrows \text{2NO}\left(g\right){\text{+Br}}_{\text{2}}\left(g\right)$

The percent dissociation of nitrosyl bromide is $34\%$.

The total pressure is $0.25\text{atm}$.

The initial pressure of $NOBr$ is considered to be $x$.

The expressions for the partial pressures at equilibrium are given as follows:

$P_A=\left[\text{A}\right]RT$

Where $\left[\text{A}\right]$ is the molar concentration of A.

For $NOBr$,

$\begin{array}{c}{P}_{\text{NOBr}}=\left(1-0.34\right)x\\ =0.66x\end{array}$

Here, $P_{\text{NOBr}}$ is the partial pressure of $\text{NOBr}$.

$P_A=\left[\text{A}\right]RT$

For $\text{NO}$,

$\begin{array}{c}{P}_{\text{NO}}=\left(1-0.66\right)x\\ =0.34x\end{array}$

Here, $P_{\text{NO}}$ is the partial pressure of $\text{NO}$.

For ${\text{Br}}_2$,

$\begin{array}{c}{P}_{{\text{Br}}_2}=\frac{0.34x}{2}\\ =0.17x\end{array}$

Here, $P_{{\text{Br}}_2}$ is the partial pressure of ${\text{Br}}_2$.

The sum of the partial pressure of the three gases is equal to the total pressure.

$\begin{array}{c}0.66x+0.34x+0.17x=0.25\text{atm}\\ x=\frac{0.25\text{atm}}{1.17}\\ =0.214\text{atm}\end{array}$

The equilibrium pressures are calculated as follows:

For $NOBr$,

$\begin{array}{c}{P}_{\text{NOBr}}=0.66\left(0.214\text{atm}\right)\\ =0.141\text{atm}\end{array}$

For $\text{NO}$,

$\begin{array}{c}{P}_{NO}=0.34\left(0.214atm\right)\\ =0.073atm\end{array}$

For ${\text{Br}}_2$,

$\begin{array}{c}{P}_{B{r}_2}=0.17\left(0.214atm\right)\\ =0.036atm\end{array}$

For the general reaction: $a\text{W}\text{+}b\text{X}\to c\text{Y}\text{+}d\text{Z}$

The $K_P$ expression is given as follows:

$K_p=\frac{P_{\text{Y}}^c{P}_{\text{Z}}^d}{P_{\text{W}}^a{P}_{\text{X}}^b}$

$K_P=\frac{{\left(P_{\text{NO}}\right)}^2{P}_{{\text{Br}}_2}}{{\left(P_{\text{NOBr}}\right)}^2}$

Substitute the values of $P_{\text{NO}},P_{{\text{Br}}_2}$, and $P_{\text{NOBr}}$ in the equation given above.

$\begin{array}{c}{K}_p=\frac{{\left(0.073\right)}^2\left(0.036\right)}{{\left(0.141\right)}^2}\\ =9.6\times {10}^{-3}\end{array}$

The equilibrium constant of partial pressure is $9.6\times {10}^{-3}$.

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

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

Here, $K_P$ is the equilibrium constant of partial pressure, $K_C$ is the equilibrium constant of concentration, $R$ is the ideal gas constant, $T$ is the temperature, and $\Delta n$ is the difference between the number of moles of products and reactants.

For this reaction, $\Delta n=+1$. So, the relation between $K_P$ and $K_C$ becomes,

$K_C=\frac{K_P}{RT}$

Substitute the values of $K_p$, $R$, $T$, and $\Delta n$ in the equation given above.

$\begin{array}{c}{K}_c=\frac{9.6\times {10}^{-3}}{\left(0.0821\times 298\right)}\\ =3.9\times {10}^{-4}\end{array}$

The equilibrium constant is $3.9\times {10}^{-4}$.

Conclusion

The value of equilibrium constant, $K_c$, is $9.6\times {10}^{-3}$, and the value of equilibrium constant, $K_P$, is $3.9\times {10}^{-4}$.