## Chapter 13

## Q. 13.11

**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

## Verified Solution

**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:

**Step 3** Substitute the equilibrium concentrations into the equilibrium equation:

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 ) = 4x^2

or 53.0x^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 } 0.0943Now 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:

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

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).