Chapter 6, Problem 96CP

(a)

Interpretation Introduction

Interpretation: Value of $K_{\text{p}}$ in indicated reaction is to be determined.

  ${\text{N}}_{\text{2}}{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_{\text{2}}\left(g\right)$

Concept introduction: Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Answer to Problem 96CP

Value of $K_{\text{p}}$ is 0.16.

Explanation of Solution

Given Information: Total pressure at equilibrium is $\text{1}\text{.5 atm}$ at $25°\text{C}$ . ${\text{N}}_{\text{2}}{\text{O}}_4$ reached equilibrium at $16\text{ \%}$ decomposition.

Given reaction occurs as follows:

  ${\text{N}}_{\text{2}}{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_{\text{2}}\left(g\right)$

Since after $12.5\text{ \%}$ change reaction reached equilibrium thus ICE table for the above reaction can be drawn as follows:

  $\begin{array}{cccc}\text{Equation}& 2{\text{N}}_{\text{2}}{\text{O}}_4& \rightleftharpoons & 2{\text{NO}}_2\\ \text{Initial}& x& & 0\\ \text{Change}& -0.16x& & +0.32x\\ \text{Equilibrium}& \left(0.84\right)x& & \left(0.32\right)x\end{array}$

Total equilibrium moles are calculated as follows:

  $\begin{array}{c}\text{Total moles =}\left(0.84\right)x+\left(0.32\right)x\\ =\left(0.84+0.32\right)x\end{array}$

The formula to calculate mole fraction is as follows:

  $\chi =\frac{{\text{Moles of NO}}_2}{\text{Total moles}}$

Moles of ${\text{NO}}_2$ is $\left(0.84\right)x$ .

Total Moles is $\left(0.84+0.32\right)x$ .

Substitute the value in above formula.

  $\begin{array}{c}\chi =\frac{{\text{Moles of NO}}_2}{\text{Total moles}}\\ =\frac{\left(0.84\right)x}{\left(0.84+0.32\right)x}\\ =0.72413\end{array}$

The formula to calculate mole fraction is as follows:

  $\chi =\frac{{\text{Moles of N}}_2{\text{O}}_4}{\text{Total moles}}$

Moles of ${\text{N}}_2{\text{O}}_4$ is $\left(0.32\right)x$ .

Total Moles is $\left(0.84+0.32\right)x$ .

Substitute the value in above formula.

  $\begin{array}{c}\chi =\frac{{\text{Moles of N}}_2{\text{O}}_4}{\text{Total moles}}\\ =\frac{\left(0.32\right)x}{\left(0.84+0.32\right)x}\\ =0.27586\end{array}$

The formula to calculate partial pressure is as follows:

  $P=\chi \cdot P_{\text{Total}}$

For ${\text{NO}}_2$

The value of $\chi$ is 0.72414.

The value of $P_{\text{total}}$ is $\text{1}\text{.5 atm}$ .

Substitute the value in above formula.

  $\begin{array}{c}P=\chi \cdot P_{\text{Total}}\\ =\left(0.72414\right)\left(\text{1}\text{.5 atm}\right)\\ =1.08621\text{ atm}\end{array}$

For ${\text{N}}_2{\text{O}}_4$

The value of $\chi$ is 0.27586.

The value of $P_{\text{total}}$ is $\text{1}\text{.5 atm}$ .

Substitute the value in above formula.

  $\begin{array}{c}P=\chi \cdot P_{\text{Total}}\\ =\left(0.27586\right)\left(\text{1}\text{.5 atm}\right)\\ =0.41379\text{ atm}\end{array}$

Expression for $K_{\text{p}}$ of given reaction is as follows:

  $K_{\text{p}}=\frac{{\left(P_{{\text{NO}}_{\text{2}}}\right)}^2}{\left(P_{{\mathrm{N}}_2{O}_4}\right)}$

Value of $P_{{\text{NO}}_{\text{2}}}$ is $0.41379\text{ atm}$ .

Value of $P_{{\mathrm{N}}_2{O}_4}$ is $1.08621\text{ atm}$ .

Substitute the value in above equation.

  $\begin{array}{c}{K}_{\text{p}}=\frac{{\left(P_{{\text{NO}}_{\text{2}}}\right)}^2}{\left(P_{{\mathrm{N}}_2{O}_4}\right)}\\ =\frac{{\left(0.41379\text{ atm}\right)}^2}{\left(1.08621\text{ atm}\right)}\\ =0.16\end{array}$

Hence, value of $K_{\text{p}}$ is 0.16.

(b)

Interpretation Introduction

Interpretation:Equilibrium pressure of ${\text{N}}_{\text{2}}{\text{O}}_4$ , ${\text{NO}}_2$ in indicated reaction is to be determined.

  ${\text{N}}_{\text{2}}{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_{\text{2}}\left(g\right)$

Concept introduction: Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Answer to Problem 96CP

Equilibrium pressure of ${\text{N}}_{\text{2}}{\text{O}}_4$ and ${\text{NO}}_2$ is $\text{0}\text{.82 atm}$ and $0.36\text{ atm}$ respectively.

Explanation of Solution

Given reaction occurs as follows:

  ${\text{N}}_{\text{2}}{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_{\text{2}}\left(g\right)$

ICE table in terms of pressure for the above reaction can be drawn as follows:

  $\begin{array}{cccc}\text{Equation}& {\text{N}}_{\text{2}}{\text{O}}_4& \rightleftharpoons & 2{\text{NO}}_2\\ \text{Initial}\left(\text{atm}\right)& 1& & 0\\ \text{Change}\left(\text{atm}\right)& -x& & \\ \text{Equilibrium}\left(\text{atm}\right)& 1-x& & 2x\end{array}$

Expression for $K_{\text{p}}$ of given reaction is as follows:

  $K_{\text{p}}=\frac{{\left(P_{{\text{NO}}_{\text{2}}}\right)}^2}{\left(P_{{\mathrm{N}}_2{O}_4}\right)}$

Value of $P_{{\text{NO}}_{\text{2}}}$ is $2x$ .

Value of $P_{{\mathrm{N}}_2{O}_4}$ is $1-x$ .

Value of $K_{\text{p}}$ is 0.16

Substitute the value in above equation.

  $\begin{array}{c}{K}_{\text{p}}=\frac{{\left(P_{{\text{NO}}_{\text{2}}}\right)}^2}{\left(P_{{\mathrm{N}}_2{O}_4}\right)}\\ 0.16=\frac{{\left(2x\right)}^2}{\left(1-x\right)}\\ x=0.18\end{array}$

Simplify to obtain the value of $x$ as $0.18\text{ atm}$ .

Equilibrium pressure $P_{{\text{NO}}_{\text{2}}}$ is calculated as follows:

  $\begin{array}{c}{P}_{{\text{NO}}_{\text{2}}}=\left(2\right)\left(0.18\right)\text{ atm}\\ =0.36\text{ atm}\end{array}$

Equilibrium pressure $P_{{\mathrm{N}}_2{O}_4}$ is calculated as follows:

  $\begin{array}{c}{P}_{{\mathrm{N}}_2{O}_4}=\left(1-0.18\right)\text{ atm}\\ =\text{0}\text{.82 atm}\end{array}$

Hence, equilibrium pressure of ${\text{N}}_{\text{2}}{\text{O}}_4$ and ${\text{NO}}_2$ is $\text{0}\text{.82 atm}$ and $0.36\text{ atm}$ respectively.

(c)

Interpretation Introduction

Interpretation: Moles percentage of original ${\text{N}}_{\text{2}}{\text{O}}_4$ dissociated at new equilibrium is to be determined.

  ${\text{N}}_{\text{2}}{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_{\text{2}}\left(g\right)$

Concept introduction: Chemical equilibrium is taken into consideration if rate of forward and backward reactions become equal. At this stage, both reactants and products have constant concentration. It can be studied in terms of pressure also. Equilibrium constant in pressure is denoted by $K_{\text{p}}$ .

Answer to Problem 96CP

Percent of ${\text{N}}_{\text{2}}{\text{O}}_4$ dissociated is $18\text{ \%}$ .

Explanation of Solution

ICE table in terms of pressure for the above reaction can be drawn as follows:

  $\begin{array}{cccc}\text{Equation}& {\text{N}}_{\text{2}}{\text{O}}_4& \rightleftharpoons & 2{\text{NO}}_2\\ \text{Initial}\left(\text{atm}\right)& P_0& & 0\\ \text{Change}\left(\text{atm}\right)& -y& & +2y\\ \text{Equilibrium}\left(\text{atm}\right)& 0.82& & 0.36\end{array}$

From the table net pressure for ${\text{NO}}_2$ is calculated as follows:

  $\begin{array}{c}2y=0.36\\ y=0.18\text{ atm}\end{array}$

Similarly net pressure for ${\text{N}}_{\text{2}}{\text{O}}_4$ is calculated as follows:

  $\begin{array}{c}{P}_0-y=0.82\\ P_0=0.82+y\\ =0.82+0.18\\ =1\text{ atm}\end{array}$

Therefore extent of ${\text{N}}_{\text{2}}{\text{O}}_4$ that has dissociated is calculated as follows:

  $\begin{array}{c}{\text{Percent dissociation of N}}_{\text{2}}{\text{O}}_4=\left(\frac{0.18}{1}\right)\left(100\text{ \%}\right)\\ =18\text{ \%}\end{array}$

Hence, percent of ${\text{N}}_{\text{2}}{\text{O}}_4$ dissociated is $18\text{ \%}$ .