Chapter 17.1, Problem 5PP

Solution

Interpretation:

- The equilibrium constant for the ratio of concentration of product and reactant is to be calculated.

Concept introduction:

- The ratio of concentration of reactant and product is called an equilibrium constant of a reaction denoted by $\text{'}{K}_C\text{'}$.

For example

$aA+bB\rightleftharpoons cC$

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

$K_{eq}=0.210$

The ratio of molar concentration of product and molar concentration of reactant with each concentration term raised to the power equal to the number of moles in the balanced chemical equation

$P+Q\rightleftharpoons R+S$

$K_{eq}=\frac{[R][S]}{[P][Q]}$

For calculating the $K_{eq}$ of the reaction

$N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$

$K_{eq}=\frac{{[N{O}_2(g)]}^2}{[N_2{O}_4(g)]}$

$[N{O}_2]=0.0627mol/L$

$[N_2{O}_4(g)]=0.0185mol/L$

Put the values for finding $K_{eq}$

$\begin{array}{l}{K}_{eq}=\frac{{[0.0627]}^{2}mol/L}{0.0185mol/L}\\ K_{eq}=\frac{0.0039}{0.0185}\\ K_{eq}=0.210\end{array}$

The value of equilibrium constant $K_{eq}$ is 0.210.

The value of Equilibrium constant for this reaction is 0.210.