Answer
$$K_c = 27.2$$
Work Step by Step
1. Determine each molarity:
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
- Calculate the amount of moles:
$$ 26.9 \space g \times \frac{1 \space mol}{ 28.01 \space g} = 0.960 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.960 \space mol}{ 5.19 \space L} = 0.185 \space M $$
$ H_2 $ : ( 1.008 $\times$ 2 )= 2.016 g/mol
- Calculate the amount of moles:
$$ 2.34 \space g \times \frac{1 \space mol}{ 2.016 \space g} = 1.16 \space mol$$
- Calculate the molarity:
$$ \frac{ 1.16 \space mol}{ 5.19 \space L} = 0.224 \space M $$
$ CH_3OH $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 32.04 g/mol
- Calculate the amount of moles:
$$ 8.65 \space g \times \frac{1 \space mol}{ 32.04 \space g} = 0.270 \space mol$$
- Calculate the molarity:
$$ \frac{ 0.270 \space mol}{ 5.19 \space L} = 0.0520 \space M $$
2. At equilibrium, these are the concentrations of each compound:
$ [ CO ] = 0.185 \space M - x$
$ [ H_2 ] = 0.224 \space M - 2x$
$ [ CH_3OH ] = 0 \space M + x$
3. Using the concentration of $ CH_3OH $ at equilibrium, find x:
$ 0 + x = 0.0520 $
$x = 0.0520 $
$ [ CO ] = 0.185 \space M - 0.0520 =0.133 $
$ [ H_2 ] = 0.224 \space M - 2*( 0.0520 ) = 0.120 $
$ [ CH_3OH ] = 0 \space M + 0.0520 =0.0520 $
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ CH_3OH ]}{[ CO ][ H_2 ] ^{ 2 }}$$
4. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.0520 )}{( 0.133 )( 0.120 )^{ 2 }} = 27.2$$
