Question

You have learned how to treat systems that have small equilibrium constants by making approximations to simplify the math. What if the system has a very large equilibrium constant? What can you do to simplify the math for this case? Use the same example from the text, but change the value of the equilibrium constant to 1.610⁵ and rework the problem. Why can you not use approximations for the case in which K = 1.6?

Short answer

You need to rework the problem from the text at K = 1.610⁵. Also you need to explain why you can’t use approximations for the case in which K = 1.6.

Step-by-step solution

Example reaction

The example reaction from the text is

$2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

Here, K = 1.610⁵

^{$\mathrm{Again},\mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Cl}}_2\right]}{{\left[\mathrm{NOCl}\right]}^2}$}

Calculation of concentrations of reactants and products at equilibrium

Now, you can write,

$\begin{array}{cccc}& 2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons & 2\mathrm{NO}\left(\mathrm{g}\right)+& {\mathrm{Cl}}_2\left(\mathrm{g}\right)\\ \mathrm{initial}:& 0.5& 0& 0\\ \mathrm{change}:& -2\mathrm{x}& +2\mathrm{x}& +\mathrm{x}\\ \mathrm{equilibrium}:& (0.5-2\mathrm{x})& 2\mathrm{x}& \mathrm{x}\end{array}$

Hence,

$\begin{array}{l}\mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Cl}}_2\right]}{{\left[\mathrm{NOCl}\right]}^2}\\ 1.6\times {10}^5=\frac{{\left(2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{(0.5-2\mathrm{x})}^2}\end{array}$

As, K is very high, therefore, you can approximate as $0.5-2\mathrm{x}\approx -2\mathrm{x}$

Now,

$\begin{array}{c}1.6\times {10}^5=\frac{{\left(2\mathrm{x}\right)}^2\left(\mathrm{x}\right)}{{(0.5-2\mathrm{x})}^2}\\ 1.6\times {10}^5=\frac{4{\mathrm{x}}^3}{4{\mathrm{x}}^2}[(0.5-2\mathrm{x})\approx -2\mathrm{x}]\\ \mathrm{x}=1.6\times {10}^5\end{array}$

K is very high, so you can’t use the approximation like before. As you know, a 4% error is acceptable for approximation.