Chapter 17, Problem 74A

Interpretation Introduction

Interpretation:

The equilibrium constant of a given reaction is to be calculated.

Concept introduction:

The chemical equilibrium constant is the ratio of the concentration of the products to that of reactants is known as equilibrium constant.

Answer to Problem 74A

The equilibrium constant is $\text{3}\text{.342}\text{.}$

Explanation of Solution

Given information: the reaction is ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH + CH}}_{\text{3}}\text{COOH }\rightleftharpoons {\text{ CH}}_{\text{3}}{\text{COOCH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

Concentration of ethanol $\text{= 0}\text{.313M}$

Concentration of acetic acid $\text{= 2}\text{.90M}$

Concentration of water $\text{= 0}\text{.114 M}$

Concentration of ethyl acetate $\text{= 2}\text{.90M}$

The chemical equilibrium constant is the ratio of the concentration of the products to that of reactants is known as equilibrium constant.

$\begin{array}{l}{\text{K}}_{\text{eq`}}\text{ = }\frac{\left[{\text{CH}}_{\text{3}}{\text{COOCH}}_{\text{2}}{\text{CH}}_{\text{3}}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}{\left(\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}\right]\right)}\hfill \\ \text{ = }\frac{\left(\text{2}\text{.90 × 0}\text{.114 }\right)}{\left(\text{0}\text{.313 ×0}\text{.316}\right)}\hfill \\ \text{ = }\frac{\text{0}\text{.3306}}{\text{0}\text{.09891}}\hfill \\ \text{ =3}\text{.342}\hfill \end{array}$

Conclusion

The equilibrium constant is $\text{=3}\text{.342}$