# Question 595cf

May 30, 2015

The equilibrium concentration of ${H}_{2}$ will be $1.86 \cdot {10}^{- 3} \text{M}$.

Before doing any actual calculations, take a look at the value of the equilibrium constant, ${K}_{c}$.

Since ${K}_{c}$ is smaller than 1, the equilibrium will lie to the left, meaning that, at equilibrium, the mixture will contain more reactant than products.

This implies that the equilibrium concentration of the products will be significantly smaller than that of the reactant, $H I$.

$\text{ } \textcolor{red}{2} H {I}_{\left(g\right)} r i g h t \le f t h a r p \infty n s {H}_{2 \left(g\right)} + {I}_{2 \left(g\right)}$
I.....0.150.............0............0
C....(-$\textcolor{red}{2}$x).............(+x)........(+x)
E....0.150-2x.........x...........x

The equilibrium constant is equal to

${K}_{c} = \frac{\left[{H}_{2}\right] \cdot \left[{I}_{2}\right]}{{\left[H I\right]}^{\textcolor{red}{2}}} = \frac{x \cdot x}{0.150 - 2 x} ^ 2 = {x}^{2} / {\left(0.150 - 2 x\right)}^{2}$

You acn solve this by taking the square root of both sides of the equation to get

$\sqrt{{x}^{2} / {\left(0.150 - 2 x\right)}^{2}} = \sqrt{{K}_{c}}$

$\frac{x}{0.15 - 2 x} = \sqrt{1.62 \cdot {10}^{- 4}} = 0.01273$

This is equivalent to

$1.02546 x = 0.00191 \implies x = \frac{0.00191}{1.02546} = 0.00186$

Since $x$ is equal to the equilibrium concentrations of ${H}_{2}$ and ${I}_{2}$, the answer will be

[H_2] = color(green)(1.86 * 10^(-3)"M"#

Notice that the initial prediction turned out to be correct. The equilibrium concentrations of the products are indeed smaller than that of the reactant.