Answer
$ [ CO ] = 0212 M$
$ [ H_2O ] = 0.212 M$
$ [ CO_2 ] = 0.288 M$
$ [ H_2 ] = 0.288 M$
Work Step by Step
- Calculate all the concentrations:
$$[CO] = ( 0.500 )/(1.00) = 0.500 M$$
$$[H_2O] = ( 0.500 )/(1.00) = 0.500 M$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CO ]& [ H_2O ]& [ CO_2 ]& [ H_2 ]\\
Initial& 0.500 & 0.500 & 0 & 0 \\
Change& -x& -x& +x& +x\\
Equilibrium& 0.500 -x& 0.500 -x& x& x\\
\end{vmatrix}$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CO_2 ][ H_2 ]}{[ CO ][ H_2O ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ CO ] = 0.500 \space M - x$
$ [ H_2O ] = 0.500 \space M - x$
$ [ CO_2 ] = x$
$ [ H_2 ] = x$
$$1.845 = \frac{( x)(x)}{(0.500 - x)(0.500 - x)}$$
$x_1 = 0.288$
$x_2 = 1.90$
x cannot be greater than 0.500, because the concentration would become negative.
$x = 0.288$
$ [ CO ] = 0.500 \space M - 0.288 M = 0212 M$
$ [ H_2O ] = 0.500 \space M - 0.288 M = 0.212 M$
$ [ CO_2 ] = 0.288 M$
$ [ H_2 ] = 0.288 M$
