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Question 13.10: Calculating Equilibrium Concentrations from Equal Initial Co......

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?


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

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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:

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:


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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