Chapter 19, Problem 19.74P

(a)

Interpretation Introduction

Interpretation:

The possibility of having an equilibrium reaction mixture for which $P_{HI}=P_{H_2}=P_{I_2}$ at $553K$ has to be predicted.

  $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right),K_P=85at553K$

Explanation of Solution

The reaction is given below.

  $\begin{array}{l}{H}_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)\\ \\ K_P=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=85at{280}^{\circ }C\end{array}$

If $P_{HI}=P_{H_2}=P_{I_2}$ at $553K$, the value of $K_P$ will be $1$, but its value is $85at553K$ which is a constant.

Therefore, it is not possible to have an equilibrium reaction mixture for which $P_{HI}=P_{H_2}=P_{I_2}at553K.$

(b)

Interpretation Introduction

Interpretation:

The composition of the equilibrium reaction mixture has to be calculated.

  $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right),K_P=85at553K$

Concept Introduction:

Relationship between $K_{P}and{K}_C:$

  $K_P=K_C{\left(RT\right)}^{\Delta {\nu }_{gas}}$

Where, $\Delta {\nu }_{gas}$ is the difference between the stoichiometric coefficients of the products and reactants in gaseous phase.

Explanation of Solution

The reaction is given below.

  $\begin{array}{l}{H}_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)\\ \\ K_P=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=85at553K\end{array}$

The reverse of the above reaction is required.

  $\begin{array}{l}2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)\\ \\ K^{\text{'}}{}_P=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=\frac{1}{K_P}=\frac{1}{85}=0.012\end{array}$

The value of $K_C$ for the above reaction can be calculated as given below.

  $\begin{array}{c}\Delta {\nu }_{gas}=2-2=0\\ \\ K_P=K_C{\left(RT\right)}^{\Delta {\nu }_{gas}}\\ \\ K_P=K_C{\left(RT\right)}^0\\ \\ K_C=0.012\end{array}$

The initial concentration of $HI\left(g\right)$ can be calculated as given below.

  $\left[HI\right]=\frac{Moles}{Volume}=\frac{\frac{5.0g}{127.911g\cdot mo{l}^{-1}}}{2.00L}=0.02M$

A concentration table can be set up as given below.

  $\begin{array}{cccccccc}Concentration\left(M\right)& 2HI\left(g\right)& \rightleftharpoons & H_2\left(g\right)& +& I_2\left(g\right)& & \\ Initial& 0.02& & 0& & 0& & \\ Change& -2x& & +x& & +x& & \\ Equilibrium& 0.02-2x& & x& & x& & \end{array}$

Substitute the equilibrium concentrations into the equilibrium – constant expression as shown below.

  $\begin{array}{c}{K}_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=0.012\\ \\ \frac{{\left(x\right)}^2}{{\left(0.02-x\right)}^2}=0.012\\ \\ 0.952x^2+\left(9.6\times {10}^{-4}\right)x-\left(4.8\times {10}^{-6}\right)=0\\ \\ x=0.0017\end{array}$

The composition of the equilibrium reaction mixture is given below.

  $\begin{array}{l}\left[HI\right]=\text{0}\text{.02}-2x=\left(0.02\right)-\left(2\times 0.0017\right)=0.016M\\ \\ \left[H_2\right]=x=1.7\times {10}^{-3}M\\ \\ \left[I_2\right]=x=1.7\times {10}^{-3}M.\end{array}$