# Question 13.11: Calculating Equilibrium Concentrations from Unequal Initial ......

Calculating Equilibrium Concentrations from Unequal Initial Concentrations

Calculate the equilibrium concentrations of $H_2, I_2$, and HI at 700 K if the initial concentrations are [ $H_2$ ] = 0.100 M and [ $I_2$ ] = 0.200 M. The equilibrium constant $K_c$ for the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ is 57.0 at 700 K.

STRATEGY

This problem is similar to Worked Example 13.10 except that the initial concentrations of $H_2$ and $I_2$ are unequal. Again, we follow the steps in Figure 13.6

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

Step 2 Again, define x as the concentration of $H_2$ that reacts. Set up a table of concentrations under the balanced equation:

$\begin{array}{lccccc} & \mathrm{H}_2(g) & + & \mathrm{I}_2(g) & \rightleftharpoons & 2 \mathrm{HI}(g) \\\hline \text { Initial concentration }(\mathrm{M}) & 0.100 & & 0.200 & & 0 \\\text { Change }(M) & -x & & -x & & +2 x \\\text { Equilibrium concentration }(M) & (0.100-x) & & (0.200-x) & & 2 x\end{array}$

Step 3 Substitute the equilibrium concentrations into the equilibrium equation:

$K_{\mathrm{c}}=57.0=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(2 x)^2}{(0.100-x)(0.200-x)}$

Because the right side of this equation is not a perfect square, we must put the equation into the standard quadratic form, $ax^2$ + bx + c = 0, and then solve for x using the quadratic formula (Appendix A.4):

$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

Rearranging the equilibrium equation gives

(57.0) (0.0200 – 0.300x +$x^2$ ) = 4$x^2$

or  53.0$x^2$ – 17.1x + 1.14 = 0

Substituting the values of a, b, and c into the quadratic formula gives two solutions:

$x=\frac{17.1 \pm \sqrt{(-17.1)^2-4(53.0)(1.14)}}{2(53.0)}=\frac{17.1 \pm 7.1}{106}=0.228 \text { and } 0.0943$

Now decide which of the two solutions is physically possible. Discard the solution that uses the positive square root (x = 0.228) because the $H_2$ concentration can’t change by more than its initial value (0.100 M). Therefore, choose the solution that uses the negative square root (x = 0.0943).

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

\begin{aligned}& {\left[\mathrm{H}_2\right]=0.100-x=0.100-0.0943=0.006 \mathrm{M}} \\& {\left[\mathrm{I}_2\right]=0.200-x=0.200-0.0943=0.106 \mathrm{M}} \\& {[\mathrm{HI}]=2 x=(2)(0.0943)=0.189 \mathrm{M}}\end{aligned}

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

$K_{\mathrm{c}}=57.0=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(0.189)^2}{(0.006)(0.106)}=56.2$

The calculated value of $K_c$ (56.2), which should be rounded to one significant figure (6 × 10¹ ), agrees with the value given in the problem (57.0).

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