Chapter 17, Problem 20A

Interpretation Introduction

Interpretation: The value of equilibrium constant $\left(K\right)$ for the reaction has to be calculated.

Concept Introduction: For a balanced chemical reaction,

  $aA+bB\rightleftharpoons cC+dD$

The equilibrium expression is the ratio of the concentration of the products to the concentration of the reaction. The concentration of each reactant/product is raised to a power corresponding to their coefficient in the balanced equation.

The law of chemical equilibrium is given by the equilibrium expression,

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

The equilibrium constant is represented as $K.$

Answer to Problem 20A

The value of the equilibrium constant $\left(K\right)$ for the reaction is $\text{6}\text{.78}\text{.}$

Explanation of Solution

The given chemical reaction is,

  ${\text{CH}}_{\text{3}}\text{OH}\left(g\right)\rightleftharpoons {\text{CH}}_{\text{2}}\text{O}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)$

Given data,

The equilibrium concentration of $\left[{\text{CH}}_{\text{3}}\text{OH}\right]$ is $\text{0}\text{.00215}\text{M}\text{.}$

The equilibrium concentration of $\left[{\text{CH}}_{\text{2}}\text{O}\right]$ is $\text{0}\text{.441}\text{M}\text{.}$

The equilibrium concentration of $\left[{\text{H}}_{\text{2}}\right]$ is $0.0331\text{M}\text{.}$

For a balanced chemical reaction,

  $aA+bB\rightleftharpoons cC+dD$

Where,

• $A,B,C,D$ are chemical species.
• $a,b,c$ and $d$ are coefficients in the balanced equation.

The equilibrium expression is given as,

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

Where,

  $K$ is the value of the equilibrium constant.

For the given reaction, the expression for the equilibrium constant is written as,

  $K=\frac{\left[C{H}_{2}O\right]\left[H_2\right]}{\left[C{H}_{3}OH\right]}$

Substitute the values of equilibrium concentration in the above expression,

  $\begin{array}{l}K=\frac{\left(\text{0}\text{.441}\right)\left(\text{0}\text{.0331}\right)}{\left(\text{0}\text{.00215}\right)}\\ K=\text{6}\text{.78}\end{array}$

The value of the equilibrium constant $\left(K\right)$ for the reaction is $\text{6}\text{.78}\text{.}$

Conclusion

The value of the equilibrium constant $\left(K\right)$ for the reaction is $\text{6}\text{.78}\text{.}$