## Q. 13.10

Calculating Equilibrium Concentrations from Equal Initial Concentrations

The equilibrium constant $K_c$ for the reaction of $H_2$ with $I_2$ is 57.0 at 700 K:

$\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad K_{\mathrm{c}}=57.0 \text { at } 700 \mathrm{~K}$

If 1.00 mol of $H_2$ is allowed to react with 1.00 mol of $I_2$ in a 10.0 L reaction vessel at 700 K, what are the concentrations of $H_2, I_2$, and HI at equilibrium? What is the composition of the equilibrium mixture in moles?

STRATEGY

We need to calculate equilibrium concentrations from initial concentrations, so we use the method outlined in Figure 13.6.

## Verified Solution

Step 1 The balanced equation is given: $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$

Step 2 The initial concentrations are [ $H_2$ ] = [$I_2$ ] = (1.00 mol)/(10.0 L) = 0.100 M. For convenience, define an unknown, x, as the concentration (mol/L) of $H_2$ that reacts. According to the balanced equation for the reaction, x mol/L of $H_2$ reacts with x mol/L of $I_2$ to give 2x mol/L of HI. This reduces the initial concentrations of $H_2$ and $I_2$ from 0.100 mol/L to (0.100 – x) mol/L at equilibrium. Let’s summarize these results in a table under the balanced equation:

$\begin{array}{lccc} & \ \ \ \ \ \mathbf{H}_2(g) \ \ \ \ \ \ \ \ + & \mathbf{I}_2(g)& \ \ \ \rightleftharpoons 2 \mathrm{HI}(g) \\\hline \text { Initial concentration }(\mathrm{M}) & 0.100 & 0.100 & 0 \\\text { Change }(M) & -x & -x & +2 x \\\text { Equilibrium concentration }(\mathrm{M}) & (0.100-x) & (0.100-x) & 2 x\end{array}$

Step 3 Substitute the equilibrium concentrations into the equilibrium equation for the reaction:

$K_{\mathrm{c}}=57.0=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(2 x)^2}{(0.100-x)(0.100-x)}=\left(\frac{2 x}{0.100-x}\right)^2$

Because the right side of this equation is a perfect square, we can take the square root of both sides:

$\sqrt{57.0}=\pm 7.55=\frac{2 x}{0.100-x}$

Solving for x, we obtain two solutions. The equation with the positive square root of 57.0 gives

\begin{aligned}& +7.55(0.100-x)=2 x \\& 0.755=2 x+7.55 x \\& x=\frac{0.755}{9.55}=0.0791 \ \mathrm{M}\end{aligned}

The equation with the negative square root of 57.0 gives

\begin{aligned}& -7.55(0.100-x)=2 x \\& -0.755=2 x-7.55 x \\& x=\frac{-0.755}{-5.55}=0.136 \ \mathrm{M} \end{aligned}

Because the initial concentrations of $H_2$ and $I_2$ are 0.100 M, x can’t exceed 0.100 M. Therefore, discard x = 0.136 M as chemically unreasonable and choose the first solution, x = 0.0791 M.

Step 4 Calculate the equilibrium concentrations from the calculated value of x:

\begin{aligned}& {\left[\mathrm{H}_2\right]=\left[\mathrm{I}_2\right]=0.100-x=0.100-0.0791=0.021 \mathrm{M}} \\& {[\mathrm{HI}]=2 x=(2)(0.0791)=0.158 \mathrm{M}}\end{aligned}

Step 5 Check the results by substituting them into the equilibrium equation:

$K_{\mathrm{c}}=57.0=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(0.158)^2}{(0.021)(0.021)}=57$

The number of moles of each substance in the equilibrium mixture is obtained by multiplying each concentration by the volume of the reaction vessel:

Moles of $H_2$ = Moles of $I_2$ = (0.021 mol/L) (10.0 L) = 0.21 mol

Moles of HI = (0.158 mol/L) (10.0 L) = 1.58 mol