Chapter 17.2, Problem 2RQ

Interpretation Introduction

Interpretation:

The equilibrium constant value for the given reaction is to be calculated.

Concept Introduction: For an equilibrium reaction as follows:

  $aA+bB\rightleftharpoons cC+dD$

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

Here, $\left[A\right]$ , $\left[B\right]$ , $\left[C\right]$ and $\left[D\right]$ is concentration of A, B, C and D respectively and a, b, c and d are stoichiometric coefficient.

Answer to Problem 2RQ

The value of equilibrium constant for the given reaction is $\text{2}{\text{.354×10}}^{13}$ .

Explanation of Solution

Give information:

The given reaction is

  ${\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{3Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NCl}}_{\text{3}}\left(\text{g}\right)$

The molar concentration of $\left[{\text{N}}_{\text{2}}\right]\text{=}\text{0}\text{.000104M}$

The molar concentration of $\left[{\text{Cl}}_{\text{2}}\right]\text{=}\text{0}\text{.000201M}$

The molar concentration of $\left[{\text{NCl}}_{\text{3}}\right]\text{=}\text{0}\text{.141M}$

Solution:

The equilibrium constant is given as

  $\begin{array}{l}{\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{3Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{2NCl}}_{\text{3}}\left(\text{g}\right)\\ \text{K}\text{=}\frac{{\left[{\text{NCl}}_{\text{3}}\right]}^{\text{2}}}{\left[{\text{N}}_{\text{2}}\right]{\left[{\text{Cl}}_{\text{2}}\right]}^{\text{3}}}\\ \text{=}\frac{{\left(\text{0}\text{.141}\right)}^{\text{2}}}{\left(\text{0}\text{.000104}\right)\text{×}{\left(\text{0}\text{.000201}\right)}^{\text{3}}}\\ \text{=}\frac{\text{0}\text{.019881}}{\text{8}{\text{.4454×10}}^{\text{-16}}}\\ \text{=}\text{2}{\text{.354×10}}^{13}\end{array}$

Conclusion

The calculated value of equilibrium constant for the given reaction is $\text{2}{\text{.354×10}}^{13}$ .