Answer
$[H_2] = 1.6 \times 10^{-3} \space M$
Work Step by Step
1. Calculate the concentration:
$$Molarity = \frac{0.45 \space mol}{3.0 \space L} = 0.15 \space M$$
2. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2S ]& [ H_2 ]& [ S_2 ]\\
Initial& 0.15 & 0 & 0 \\
Change& -2 x& 2 x& +x\\
Equilibrium& 0.15 -2 x& 0 2 x& 0 +x\\
\end{vmatrix}$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ H_2 ] ^{ 2 }[ S_2 ]}{[ H_2S ] ^{ 2 }}$$
3. At equilibrium, these are the concentrations of each compound:
$ [ H_2S ] = 0.15 \space M - 2x$
$ [ H_2 ] = 0 \space M + 2x$
$ [ S_2 ] = 0 \space M + x$
$$9.30 \times 10^{-8} = \frac{(2x)^2(x)}{(0.15-2x)^2}$$
4. Solve for x:
$x = 8.0 \times 10^{-4}$
$ [ H_2 ] = 2(8.0 \times 10^{-4}) = 1.6 \times 10^{-3} \space M$
