Chapter 5, Problem 5I.14E

(a)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume $2.00L$ has to be calculated.  The reaction is given below.

  $C{l}_2\left(g\right)\rightleftharpoons 2Cl\left(g\right),K_c=1.7\times {10}^{-5}$

Explanation of Solution

The required reaction is given below.

  $C{l}_2\left(g\right)\rightleftharpoons 2Cl\left(g\right),K_c=1.7\times {10}^{-5}$

The expression for equilibrium constant for the above reaction can be written as shown below,

  $K_c=\frac{{\left[Cl\right]}^2}{\left[C{l}_2\right]}$

Given that, $5.0mmolC{l}_2\left(g\right)$ was sealed into a reaction vessel of volume $2.0L.$  The concentration of $C{l}_2\left(g\right)$ can be calculated as given below.

  $\left[C{l}_2\right]=\frac{moles}{volume}=\frac{5\times {10}^{-3}mol}{2.00L}=0.0025M$

An equilibrium table can be set up as given below.

  $\begin{array}{ccc}Composition(M)& C{l}_2& Cl\\ Initialconcentration& 0.0025& 0\\ Changeinconcentration& -x& +2x\\ Equilibriumconcentration& 0.0025-x& 2x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_c=\frac{{\left(2x\right)}^2}{\left(0.0025-x\right)}$

By assuming $x\ll 0.0025$, it can be written that $0.0025-x\simeq \mathrm{0.0025.}$

Now, the above equilibrium expression can be solved for $x.$

  $\begin{array}{l}{K}_c=\frac{{\left(2x\right)}^2}{\left(0.0025-x\right)}\\ \\ 1.7\times {10}^{-5}=\frac{4x^2}{\left(0.0025\right)}\\ \\ x^2=\frac{1.7\times {10}^{-5}\times 0.0025}{4}\\ \\ x=1.03\times 10^{-4}\\ \\ \left[C{l}_2\right]=0.0025M\\ \\ \left[Cl\right]=2x=2\times \text{1}{\text{.03×10}}^{\text{-4}}=2.06\times 10^{-4}M\end{array}$

Therefore, the equilibrium concentrations of $C{l}_{2}andCl$ in a reaction vessel are $0.0025Mand2.06\times 10^{-4}M$ respectively.

(b)

Interpretation Introduction

Interpretation:

The equilibrium composition of the mixture in a reaction vessel of volume $2.00L$ has to be calculated.  The reaction is given below.

  $B{r}_2\left(g\right)\rightleftharpoons 2Br\left(g\right),K_c=1.7\times {10}^{-5}$

Explanation of Solution

The required reaction is given below.

  $B{r}_2\left(g\right)\rightleftharpoons 2Br\left(g\right),K_c=1.7\times {10}^{-5}$

The expression for equilibrium constant for the above reaction can be written as shown below,

  $K_c=\frac{{\left[Br\right]}^2}{\left[B{r}_2\right]}$

Given that, $5.0molB{r}_2\left(g\right)$ was sealed into a reaction vessel of volume $2.0L.$  The concentration of $B{r}_2\left(g\right)$ can be calculated as given below.

  $\left[B{r}_2\right]=\frac{moles}{volume}=\frac{5mol}{2.00L}=2.5M$

An equilibrium table can be set up as given below.

  $\begin{array}{ccc}Composition(M)& B{r}_2& Br\\ Initialconcentration& 2.5& 0\\ Changeinconcentration& -x& +2x\\ Equilibriumconcentration& 2.5-x& 2x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_c=\frac{{\left(2x\right)}^2}{\left(2.5-x\right)}$

By assuming $x\ll 2.5$, it can be written that $2.5-x\simeq \mathrm{2.5.}$

Now, the above equilibrium expression can be solved for $x.$

  $\begin{array}{l}{K}_c=\frac{{\left(2x\right)}^2}{\left(2.5-x\right)}\\ \\ 1.7\times {10}^{-5}=\frac{4x^2}{\left(2.5\right)}\\ \\ x^2=\frac{1.7\times {10}^{-5}\times 2.5}{4}\\ \\ x=2.30\times 10^{-3}\\ \\ \left[B{r}_2\right]=2.5M\\ \\ \left[Br\right]=2x=2\times \text{2}{\text{.30×10}}^{\text{-3}}=4.6\times 10^{-3}M\end{array}$

Therefore, the equilibrium concentrations of $B{r}_{2}andBr$ in a reaction vessel are $2.5Mand4.6\times 10^{-3}M$ respectively.

(c)

Interpretation Introduction

Interpretation:

The one which is thermodynamically more stable relative to its atoms at $1200K$ has to be determined.

Explanation of Solution

The percentage of dissociation of $F_{2}andB{r}_2$ can be calculated as given below.

  $\begin{array}{l}ForC{l}_2:\\ \%ofdissociation=\frac{dissociatedamount}{originalamount}\times 100\\ \\ \%ofdissociation=\frac{1.03\times {10}^{-4}}{0.0025}\times 100\\ \\ \%ofdissociation=4.12\%\\ \\ ForB{r}_2:\\ \%ofdissociation=\frac{2.30\times {10}^{-3}}{2.5}\times 100\\ \\ \%ofdissociation=0.092\%\end{array}$

The percentage of dissociation of $B{r}_2$ into its atoms is almost negligible than that of $C{l}_2.$  Therefore, $B{r}_2$ is thermodynamically more stable relative to its atoms at $1200K.$