Chapter 19, Problem 19.82P

Interpretation Introduction

Interpretation:

The percentage of $I_2\left(g\right)$ that will be converted to $HI\left(g\right)$ for the given reaction has to be calculated.

  $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right),K_P=54.4$

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.

  $\begin{array}{l}{H}_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)\\ \\ K_P=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=54.4\end{array}$

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

  $\begin{array}{c}\Delta {\nu }_{gas}=2-2=0\\ \\ K_P=K_C{\left(RT\right)}^{\Delta {\nu }_{gas}}\\ \\ K_P=K_C{\left(RT\right)}^0\\ \\ K_C=54.4\end{array}$

The initial concentration of $H_2\left(g\right)and{I}_2\left(g\right)$ can be calculated as given below.

  $\begin{array}{l}\left[H_2\right]=\frac{Moles}{Volume}=\frac{0.200mole}{1.00L}=0.2M\\ \\ \left[I_2\right]=\frac{Moles}{Volume}=\frac{0.200mole}{1.00L}=0.2M\end{array}$

A concentration table can be set up as given below.

  $\begin{array}{cccccccc}Concentration\left(M\right)& H_2\left(g\right)& +& I_2\left(g\right)& \rightleftharpoons & 2HI\left(g\right)& & \\ Initial& 0.2& & 0.2& & 0& & \\ Change& -x& & -x& & +2x& & \\ Equilibrium& 0.2-x& & 0.2-x& & 2x& & \end{array}$

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

  $\begin{array}{c}{K}_c=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}=54.4\\ \\ \frac{{\left(2x\right)}^2}{{\left(0.2-x\right)}^2}=54.4\\ \\ 50.4x^2-21.76x+2.176=0\\ \\ \text{x}\text{=}0.274,0.157\end{array}$

The value $0.274$ will give negative concentration value; therefore, this value is rejected.  The value $0.157$ is accepted.

The composition of the equilibrium reaction mixture is given below.

  $\begin{array}{l}\left[H_2\right]=\text{0}\text{.2}-x=0.2-0.157=0.043M\\ \\ \left[I_2\right]=0.2-x=0.2-0.157=0.043M\\ \\ \left[HI\right]=2x=2\times 0.157=0.314M\end{array}$

The percentage of $I_2\left(g\right)$ that will be converted to $HI\left(g\right)$ for the given reaction can be calculated as given below.

  $\begin{array}{c}\%I_2\left(g\right)convertedtoHI\left(g\right)=\frac{Change}{Initial}\times 100\\ \\ =\frac{0.157}{0.2}\times 100\\ \\ =78.5\%\end{array}$

Therefore, the percentage of $I_2\left(g\right)$ that will be converted to $HI\left(g\right)$ is $78.5\%.$