Question

the equilibrium constant, ( k_c ), for the following reaction is 77.5 at 600 k.

( co(g)+cl_2(g)
ightleftharpoons cocl_2(g) )

calculate the equilibrium concentrations of reactant and products when 0.452 moles of co and 0.452 moles of ( cl_2 ) are introduced into a 1.00 l vessel at 600 k.

\begin{align}co&=\text{m}\cl_2&=\text{m}\cocl_2&=\text{m}end{align}

Explanation:

Step1: Calculate initial concentrations

Since concentration \(C=\frac{n}{V}\), for \(CO\) and \(Cl_{2}\), \(n = 0.452\) mol and \(V=1.00\) L. So \([CO]_{0}=[Cl_{2}]_{0}=\frac{0.452\space mol}{1.00\space L}=0.452\space M\), and \([COCl_{2}]_{0} = 0\space M\).

Let \(x\) be the change in concentration of \(CO\) (and also of \(Cl_{2}\)) at equilibrium. Then the equilibrium concentrations are: \([CO]=0.452 - x\), \([Cl_{2}]=0.452 - x\), \([COCl_{2}]=x\).

Step2: Write the equilibrium - constant expression

The equilibrium - constant expression for the reaction \(CO(g)+Cl_{2}(g)
ightleftharpoons COCl_{2}(g)\) is \(K_{c}=\frac{[COCl_{2}]}{[CO][Cl_{2}]}\).

Given \(K_{c}=77.5\), we substitute the equilibrium concentrations into the expression: \(77.5=\frac{x}{(0.452 - x)(0.452 - x)}\).

Cross - multiply to get \(77.5(0.452 - x)^{2}=x\).

Expand \((0.452 - x)^{2}=0.204304-0.904x + x^{2}\).

So \(77.5(0.204304-0.904x + x^{2})=x\).

\(15.83356 - 69.05x+77.5x^{2}-x = 0\).

\(77.5x^{2}-70.05x + 15.83356=0\).

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\), where \(a = 77.5\), \(b=-70.05\), \(c = 15.83356\).

\(b^{2}-4ac=(-70.05)^{2}-4\times77.5\times15.83356\)

\(=4907.0025-4907.0024\approx0.0001\)

\(x=\frac{70.05\pm\sqrt{0.0001}}{2\times77.5}=\frac{70.05\pm0.01}{155}\)

We take the smaller root (because if we take the larger root \(x=\frac{70.05 + 0.01}{155}\approx0.452\) which will make \([CO]=0.452 - x\approx0\), but we know that in a non - trivial equilibrium, reactants are not completely consumed). So \(x=\frac{70.05 - 0.01}{155}=\frac{70.04}{155}\approx0.452\) (this is wrong, let's go back to the original equation \(77.5=\frac{x}{(0.452 - x)^{2}}\)

Take the square root of both sides: \(\sqrt{77.5}=\frac{\sqrt{x}}{0.452 - x}\), \(\sqrt{77.5}\approx8.8\)

\(8.8(0.452 - x)=\sqrt{x}\)

Let \(y = \sqrt{x}\), then \(8.8(0.452 - y^{2})=y\)

\(3.9776-8.8y^{2}-y = 0\)

\(8.8y^{2}+y - 3.9776 = 0\)

Using the quadratic formula \(y=\frac{-1\pm\sqrt{1^{2}-4\times8.8\times(- 3.9776)}}{2\times8.8}=\frac{-1\pm\sqrt{1 + 139.67744}}{17.6}=\frac{-1\pm\sqrt{140.67744}}{17.6}=\frac{-1\pm11.86}{17.6}\)

We take the positive root \(y=\frac{-1 + 11.86}{17.6}=\frac{10.86}{17.6}\approx0.617\)

Since \(y=\sqrt{x}\), \(x=y^{2}\approx0.381\)

\([CO]=0.452 - x=0.452-0.381 = 0.071\space M\)

\([Cl_{2}]=0.452 - x=0.071\space M\)

\([COCl_{2}]=x = 0.381\space M\)

Answer:

\([CO]=0.071\space M\)

\([Cl_{2}]=0.071\space M\)

\([COCl_{2}]=0.381\space M\)