**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.

Step-by-Step

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**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:

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:

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

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

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