Answer
The molar concentration of $H_2$ at this equilibrium mixture is: $1.1 \times 10^{-3} \space M$.
Work Step by Step
1. Write the equilibrium constant expression:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ HI ]^{ 2 }}{[ H_2 ][ I_2 ]}$$
2. Solve for the missing concentration:
$$ [H_2] = \frac{[ HI ]^{ 2 }}{[ I_2 ]\times K_c}$$
3. Evaluate the expression:
$$ [H_2] = \frac{( 0.030 )^{ 2 }}{( 0.015 )\times (54)} $$
$$[H_2] = 1.1 \times 10^{-3} \space M$$
