7.3 Equilibrium calculations  (Page 4/11)

Calculation of changes in concentration

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium , we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

1. Determine the direction the reaction proceeds to come to equilibrium.
   1. Write a balanced chemical equation for the reaction.
   2. If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Q _c from the initial concentrations and compare to K _c to determine the direction of change.
2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
   1. Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
   2. Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
3. Solve for the change and the equilibrium concentrations.
   1. Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x , and check any assumptions used to find x .
   2. Calculate the equilibrium concentrations.
4. Check the arithmetic.
   1. Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
      Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
      In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.

Calculation of concentration changes as a reaction goes to equilibrium

Solution

1. Determine the direction the reaction proceeds.

   The balanced equation for the decomposition of PCl ₅ is

   ${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)$

   Because we have no products initially, Q _c = 0 and the reaction will proceed to the right.

2. Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.

   Let us represent the increase in concentration of PCl ₃ by the symbol x . The other changes may be written in terms of x by considering the coefficients in the chemical equation.

   $\begin{array}{llll}{\text{PCl}}_5(g)\hfill & \rightleftharpoons \hfill & {\text{PCl}}_3(g)+\hfill & {\text{Cl}}_2(g)\hfill \\ -x\hfill & & x\hfill & x\hfill \end{array}$

   The changes in concentration and the expressions for the equilibrium concentrations are:

   

3. Solve for the change and the equilibrium concentrations.

   Substituting the equilibrium concentrations into the equilibrium constant equation gives

   $K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{\left[{\text{PCl}}_5\right]}=0.0211$
   $=\frac{(x)(x)}{(1.00-x)}$

   This equation contains only one variable, x , the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.

   $0.0211=\frac{(x)(x)}{(1.00-x)}$
   $0.0211(1.00-x)=x^2$
   $x^2+0.0211x-0.0211=0$

   Appendix B shows us an equation of the form ax ² + bx + c = 0 can be rearranged to solve for x :

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

   In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a , b , and c yields:

   $x=\frac{-0.0211\pm \sqrt{{(0.0211)}^2-4(1)(\mathrm{-0.0211})}}{2(1)}$
   $=\frac{-0.0211\pm \sqrt{(4.45\times {10}^{\mathrm{-4}})+(8.44\times {10}^{\mathrm{-2}})}}{2}$
   $=\frac{-0.0211\pm 0.291}{2}$

   Hence

   $x=\frac{-0.0211+0.291}{2}=0.135$

   or

   $x=\frac{-0.0211-0.291}{2}=\mathrm{-0.156}$

   Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M .

   The equilibrium concentrations are

   $\left[{\text{PCl}}_5\right]=1.00-0.135=0.87M$
   $\left[{\text{PCl}}_3\right]=x=0.135M$
   $\left[{\text{Cl}}_2\right]=x=0.135M$

4. Check the arithmetic.

   Substitution into the expression for K _c (to check the calculation) gives

   $K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{[{\text{PCl}}_5]}=\frac{(0.135)(0.135)}{0.87}=0.021$

   The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K _c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.