If we know the concentrations of all the substances at equilibrium, we can calculate the equilibrium constant. However, it is sometimes impossible to experimentally know all the concentrations of the substances present at equilibrium. If we know the initial concentration of the reactants and have at least one other information, then it is possible to predict algebraically all the concentrations at equilibrium. This is done using an Initial - Change - Equilibrium (ICE) table.
For example, if we consider the following reaction: |I_{2(g)} + H_{2(g)} \rightleftharpoons 2\; HI_{(g)}|. The initial concentration of the two reactants is 1.0 mol/L, whereas there are 1.57 mol/L of products when equilibrium is reached. Place the known data or values in an ICE table:
|
Reaction |
|I_{2(g)}| + | |H_{2(g)}| | |\rightleftharpoons| | |2\ HI_{(g)}| |
| Initial | |1.0\ mol/L| | |1.0\ mol/L| | |0\ mol/L| | |
| Change | ? | ? | ? | |
|
Equilibrium |
? | ? | |1.57\ mol/L| |
Since the changes in concentration are within the stoichiometric ratios, we can then determine the missing equilibrium concentrations. With the given values, the change in the products can be determined:
| Reaction | |I_{2(g)}| + | |H_{2(g)}| | |\rightleftharpoons| | |2\ HI_{(g)}| |
| Initial | |1.0\ mol/L| | |1.0\ mol/L| | |0\ mol/L| | |
| Change | ? | ? | |\color{red}{+ 1.57\ mol/L}| | |
|
Equilibrium |
? | ? | |1.57\ mol/L| |
The change in concentrations is proportional to the coefficients in the balanced equation. Thus, in the example studied, as there are half as many reactants as products in the balanced equation, the change in concentrations will also be half as great. The change in the reactants can be determined from the change in the product HI by simple cross multiplication. For example: |\frac{H_{2(g)}}{?}=\frac{2 HI_{(g)}}{1.57 mol/L}|. Remember that the change in concentration of the reactants is negative because they are consumed during the reaction, whereas the change in concentration of the products is positive because they are formed during the reaction. The result is
| Reaction | |I_{2(g)}| + | |H_{2(g)}| | |\rightleftharpoons| | |2\ HI_{(g)}| |
| Initial | |1.0\ mol/L| | |1.0\ mol/L| | |0\ mol/L| | |
| Variation | |\color{red}{-0.785\ mol/L}| | |\color{red}{- 0.785\ mol/L}| | |\color{red}{+ 1.57\ mol/L}| | |
|
Equilibrium |
|\color{red}{0.215\ mol/L}| | |\color{red}{0.215\ mol/L}| | |1.57\ mol/L| |
These given values can then be used to calculate the equilibrium constant.
|K_c = \displaystyle \frac{\left[HI\right]^2}{\left[I_2\right] \cdot\left[H_2\right]} = \displaystyle \frac{(1.57)^2}{(0.215)(0.215)} = 53.32|
A systematic approach is needed to solve problems involving equilibria:
In a volume of 2 L, there are 8 moles of |NH_{3}|, 48 g of |N_{2}| and 10 g of |H_{2}| at equilibrium. Determine the value of |K_{c}| in this reaction.
|N_{2(g)} + 3 H_{2(g)} \rightleftharpoons 2 NH_{3(g)}|
1. Expression of the equilibrium constant
|\displaystyle K_{c}=\frac{\left[NH_{3(g)}\right]^{2}}{\left[N_{2(g)}\right]\cdot\left[H_{2(g)}\right]^{3}}|
2. Molar concentrations at equilibrium
|\displaystyle NH_{3}:\frac{8\; moles}{2\ L}=\frac{4\; moles}{1L}=4.0\ M|
|\displaystyle N{}_{2}:\frac{48\ g}{2\ L}=\frac{24\ g}{1\ L}=\frac{24\ g}{28\ g/mol\; de\; N_{2}}=0.86M|
|\displaystyle H{}_{2}:\frac{10\ g}{2\ L}=\frac{5\ g}{1\ L}=\frac{5\ g}{2\ g/mol\; de\; H_{2}}=2.5\ M|
3. Calculating the equilibrium constant
|\displaystyle K_{c}=\frac{[4.0\ M]^{2}}{[0.86\ M]\cdot[2.5\ M]^{3}}=1.19|
Consider the following system: |2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}|
At a certain temperature, 20 moles of |NH_{3}| are introduced into a 1 L distilling flask. At equilibrium there are 12 moles of |H_{2}| . Determine the value of |K_{c}| for this system.
1. Expression of the equilibrium constant
|K_{c}=\displaystyle \frac{\left[N_{2(g)}\right]\cdot\left[H_{2(g)}\right]^{3}}{\left[NH_{3(g)}\right]^{2}}|
2. Determine the equilibrium concentrations using the ICE table
| | |2 NH_{3(g)}| | |\rightleftharpoons| | |1N_{2(g)}| + | |3 H_{2(g)}| |
| Initial | |20\ mol/L| | |0\ mol/L| | |0\ mol/L| | |
| Change | |\color{red}{-8\ mol/L}| | |\color{red}{+4\ mol/L}| | |\color{red}{+12\ mol/L}| | |
| Equilibrium | |\color{red}{12\ mol/L}| | |\color{red}{4\ mol/L}| | |12\ mol/L| |
3. Calculating the equilibrium constant
|K_{c}=\displaystyle \frac{[N_{2}]\cdot[H_{2}]^{3}}{[NH_{3}]^{2}}=\frac{[4.0\ M]\cdot[12.0\ M]^{3}}{[12.0\ M]^{2}}=48|
A 1.0 L vessel is filled with 0.5 moles of HI at 448 ºC. The value of the equilibrium constant |K_{c}| for the reaction |I_{2(g)} + H_{2(g)} \rightleftharpoons 2\; HI_{(g)}| at this temperature is 50.5. What are the concentrations of |I_{2}|, |H_{2}| and |HI| at equilibrium?
1. Expression of the equilibrium constant
|K_{c}=\displaystyle \frac{\left[HI_{(g)}\right]^{2}}{\left[I_{2(g)}\right]\cdot\left[H_{2(g)}\right]}|
2. Determine the equilibrium concentrations using the ICE table
| | |I_{2(g)}| + | |H_{2(g)}| | |\rightleftharpoons|
|
|2 HI_{(g)}| |
| Initial | |0\ mol/L| | |0\ mol/L| | |0.5\ mol/L| | |
| Change |
|\color{red}{+ x\ mol/L}| |
|\color{red}{+ x\ mol/L}| |
|\color{red}{- 2x\ mol/L}| |
|
|
Equilibrium |
|\color{red}{x\ mol/L}| |
|\color{red}{x\ mol/L}| |
|\color{red}{(0.5 - 2x)\ mol/L}| |
3. Calculation of equilibrium concentrations
|K_{c}=\displaystyle \frac{\left[HI_{(g)}\right]^{2}}{\left[I_{2(g)}\right]\cdot\left[H_{2(g)}\right]}=\frac{[0.5-2x]^{2}}{[x][x]}=50.5|
The equation then needs to be solved in order to insulate the x.
|\displaystyle \frac{(0.5-2x)(0.5-2x)}{x^2}=50.5|
|\displaystyle \frac{0.25-2x+4x^2}{x^2}=50.5|
|0.25-2x+4x^2=50.5x^2|
|-46.5x^2-2x+0.25=0|
This equation is of the second degree and will require the use of the formula for finding the zeros of a quadratic equation to be solved. This formula will give us two values for x. We would reject a negative value or a value greater than the initial concentrations in the case of the reactants.
4. Equilibrium concentrations
|[I_{2}]| = x = 0.055 mol/L
|[H_{2}]| = x = 0.055 mol/L
|[HI]| = 0.5 - 2(0.055) = 0.39 mol/L
