Answer
0.173 mol of $H_2$
Work Step by Step
- Calculate all the concentrations:
$$[H_2O] = ( 0.300 )/(10.0) = 0.0300 M$$
$$[CO] = ( 0.300 )/(10.0) = 0.0300 M$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ H_2 ][ CO_2 ]}{[ H_2O ][ CO ]}$$
2. At equilibrium, these are the concentrations of each compound:
$[ H_2O ] = 0.0300 \space M - x$
$[ CO ] = 0.0300 \space M - x$
$[ H_2 ] = 0 \space M + x$
$[ CO_2 ] = 0 \space M + x$
$$1.87 = \frac{( x)( x)}{(0.0300 - x)(0.0300 - x)}$$
x = 0.0173
$[H_2] = 0.0173 M$
$n(H_2) = 0.0173 M \times 10.0 L = 0.173 \space mol$
