Answer
$$n_{H_2} = n_{I_2} = 0.0332 \space mol$$
$$n_{HI} = 0.234 \space mol$$
Work Step by Step
1. Calculate all the concentrations:
$$[H_2] = ( 0.150 )/(3.25) = 0.0462 M$$
$$[I_2] = ( 0.150 )/(3.25) = 0.0462 M$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HI ] ^{ 2 }}{[ H_2 ][ I_2 ]}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ H_2 ] = 0.0462 \space M - x$
$ [ I_2 ] = 0.0462 \space M - x$
$ [ HI ] = + 2x$
$$50.2 = \frac{(2x)^2}{(0.0462 - x)(0.0462 - x)}$$
3. Solving for x:
$x_ 1= 0.0360$
$x_2 = 0.0644$
But $x$ cannot be greater than $0.0462$, because that would result in a negative concentration.
Therefore: x = 0.0360
4. Find the equilbrium concentrations and number of moles.
$ [ H_2 ] = 0.0462 \space M - 0.0360 \space M = 0.0102 \space M$
$ [ I_2 ] = 0.0462 \space M - 0.0360 \space M = 0.0102 \space M$
$ [ HI ] = + 2x = 2(0.0360) = 0.0720 \space M$
$$n_{H_2} = n_{I_2} = 3.25 \space L \times \frac{0.0102 \space mol}{1 \space L} = 0.0332 \space mol$$
$$n_{HI} = 3.25 \space L \times \frac{0.0720 \space mol}{1 \space L} = 0.234 \space mol$$
