Chapter 17, Problem 23A

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 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 23A

The value of equilibrium constant $\left(K\right)$ for the reaction is $\text{6}{\text{.9×10}}^{\text{-4}}\text{.}$

Explanation of Solution

The given chemical reaction is,

  $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$

Given data,

The equilibrium concentration of $\left[N_2\right]$ is $\text{0}\text{.041}\text{M}\text{.}$

The equilibrium concentration of $\left[O_2\right]$ is $\text{0}\text{.0078}\text{M}\text{.}$

The equilibrium concentration of $\left[NO\right]$ is $\text{4}\text{.7}\times {\text{10}}^{-4}\text{M}\text{.}$

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,

  $K=\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}$

Substitute the values of equilibrium concentration in the above expression,

  $\begin{array}{l}K=\frac{{\left(\text{4}{\text{.7×10}}^{\text{-4}}\right)}^{\text{2}}}{\left(\text{0}\text{.041}\right)\left(\text{0}\text{.0078}\right)}\\ K=\text{6}{\text{.9×10}}^{\text{-4}}\end{array}$

The value of equilibrium constant for the reaction is $\text{6}{\text{.9×10}}^{\text{-4}}\text{.}$

Conclusion

The value of equilibrium constant $\left(K\right)$ for the reaction is $\text{6}{\text{.9×10}}^{\text{-4}}\text{.}$