Question

The value of $K_c=4.24\text{at}800K$ for the reaction, $CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)$ Calculate equilibrium concentrations of $C{O}_2,H_2,CO$ and $H_{2}Oat800K,$ if only $CO$ and $H_{2}O$ are present initially at concentrations of $0.10M$ each.

Answer 1

Step 1 Equilibrium constant

For a general reversible reaction $aA+bB\rightleftharpoons cC+dD,$the concentrations in an

equilibrium mixture are related by the following equilibrium equation.

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

Here $x$ is the amount of $C{O}_2$ and $H_2$ at equilibrium.

Now, equilibrium constant $K_c$ can be written as,

$K_c=\frac{x^2}{(0.1-x{)}^2}=4.24$
$\Rightarrow x^2=4.24(0.01+x^2-0.2x)$

$x^2=0.0424+4.24x^2-0.848x$

$3.24x^2-0.848x+0.0424=0$

$a=3.24,b=-0.848,c=0.0424$

We will neglect value $0.194$ as it will give concentration of the reactant which is more than initial concentration.

Hence the equilibrium concentrations are,

$\left[C{O}_2\right]=\left[H_2\right]=x=0.067M$

$\left[CO\right]=\left[H_{2}O\right]=0.1-0.067=0.033M$

Final answer:

The equilibrium concentrations are,

$\left[C{O}_2\right]=\left[H_2\right]=0.067M$

$\left[CO\right]=\left[H_{2}O\right]=0.033M$