Chapter 17, Problem 68A

Interpretation Introduction

Interpretation:

The number of molecules of H₂O,CO,H₂ and CO₂ present in the equilibrium have to be calculated using trial and error method.

Concept Introduction:

Chemical equilibrium: Chemical equilibrium is a state of a chemical reaction when the rate of forward reaction is equal to the rate of backward reaction. The chemical reaction follows the law of mass action. The law of mass action states that the rate of a chemical reaction is directly proportional to the product of active masses of the reactants.

Answer to Problem 68A

At equilibrium the container contains 4 molecules of H₂O,2molecules of CO,4molecules of H₂ and 4molecules of CO₂.

Explanation of Solution

Data given: K_eq = 2.0

The given chemical equation is-

  $\begin{array}{l}{H}_{2}O(g)+CO(g)\rightleftharpoons H_2(g)+C{O}_2(g)\\ K_{eq}=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}\end{array}$

Initially 8molecules of H₂O and6molecules of COare present in the container.

Case I:

If 2molecules of CO react:

The container will contain6molecules of H₂O,4molecules of CO,2molecules of H₂ and 2molecules of CO₂.

  $\begin{array}{l}{K}_{eq}=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}\\ =\frac{2.0\times 2.0}{6.0\times 4.0}=\frac{1}{6}\end{array}$

Case I is not right.

Case II:

If 3molecules of COreact:

The container will contain5molecules of H₂O,3molecules of CO,3molecules of H₂ and 3molecules of CO₂.

  $\begin{array}{l}{K}_{eq}=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}\\ =\frac{3.0\times 3.0}{5.0\times 3.0}=\frac{3}{5}\end{array}$

Case II is not right.

Case III:

If 4molecules of CO react:

The container will contain 4molecules of H₂O,2molecules of CO,4molecules of H₂ and 4molecules of CO₂.

  $\begin{array}{l}{K}_{eq}=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}\\ =\frac{4.0\times 4.0}{4.0\times 2.0}=2.0\end{array}$

The equilibrium constant K_eq value calculated in case III matches with the given data.