Chapter 17.1, Problem 6PP

Solution

Interpretation:

Equilibrium constant is the ratio of concentration of product and reactant is to be calculated.

Concept introduction:

- Equilibrium constant is a numerical expression which expresses the relation between the molar concentration of reactant and product in a balanced chemical equation

Consider a reaction $aA+bB\rightleftharpoons cC+dD$

$K_C=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

The value of $K_{eq}$ for the reaction is 3.92.

Equilibrium constant $K_{eq}$ is the ratio of the molar concentration of product and reactant with each concentration term raised to the power equal to the number of moles in the balanced chemical equation.

Calculate of $K_{eq}$ the reaction is as follows.

$CO(g)+3H_2(g)\rightleftharpoons C{H}_4(g)+H_{2}O(g)$

$K_{eq}=\frac{[C{H}_4(g)][H_{2}O(g)]}{[CO(g)]{[H_2(g)]}^3}$

Now, insert the value of each component.

$\begin{array}{l}{K}_{eq}=\frac{(0.0387mol/L)(0.0387mol/L)}{(0.0613mol/L){(0.1839mol/L)}^3}\\ K_{eq}=\frac{0.00149}{0.0613\times 0.0062}\\ K_{eq}=\frac{0.00149}{0.00038}\\ K_{eq}=\frac{149}{38}\\ K_{eq}=3.92\end{array}$

The equilibrium constant for the following reaction is 3.92.

The value of $K_{eq}$ for the reaction is 3.92.