Chapter 17, Problem 38A

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 ratio of the concentration of the products to the concentration of the reactants. 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 38A

The value of equilibrium constant $\left(K\right)$ for the reaction is $\text{3}{\text{.4×10}}^{\text{-4}}\text{.}$

Explanation of Solution

The given chemical reaction is,

  ${\text{2CO}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2CO}\left(\text{g}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)$

Given data,

The equilibrium concentration of $\left[\text{CO}\right]$ is $\text{0}\text{.11}\text{M}\text{.}$

The equilibrium concentration of $\left[{\text{O}}_{\text{2}}\right]$ is $\text{0}\text{.055}\text{M}\text{.}$

The equilibrium concentration of $\left[{\text{CO}}_{\text{2}}\right]$ is $1.4\text{M}\text{.}$

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 equilibrium constant.

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

  $\text{K=}\frac{{\left[\text{CO}\right]}^{\text{2}}\left[{\text{O}}_{\text{2}}\right]}{{\left[{\text{CO}}_{\text{2}}\right]}^{\text{2}}}$

Substitute the values of equilibrium concentration in the above expression,

  $\begin{array}{l}K=\frac{{\left(\text{0}\text{.11}\right)}^{\text{2}}\left(\text{0}\text{.055}\right)}{{\left(\text{1}\text{.4}\right)}^{\text{2}}}\\ K=\text{3}{\text{.39×10}}^{\text{-4}}\\ K=\text{3}{\text{.4×10}}^{\text{-4}}\end{array}$

The value of equilibrium constant for the reaction is $\text{3}{\text{.4×10}}^{\text{-4}}\text{.}$

Conclusion

The value of equilibrium constant $\left(K\right)$ for the reaction is $\text{3}{\text{.4×10}}^{\text{-4}}\text{.}$