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 subscript, then it is possible to predict algebraically all the concentrations at equilibrium. This is done using an Initial - Variation - 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| | |
| Variation | ? | ? | ? | |
|
Equilibrium |
? | ? | |1,57\ mol/L| |
Since the variations in concentrations are within the stoichiometric ratios, we can then determine the missing equilibrium concentrations. With the data or values known, the variation 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| | |
| Variation | ? | ? | |\color{red}{+ 1,57\ mol/L}| | |
|
Equilibrium |
? | ? | |1,57\ mol/L| |
The variation 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 variation in concentrations will also be half as great. The variation in the reactants can be determined from the variation in the product HI by simple crossbreeding. 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 data or 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:
- Write the balanced chemical equation for the reaction at equilibrium.
- Write the expression for the equilibrium constant.
- Construct an ICE table of all the reactants and products and record all the information you know about the initial and equilibrium concentrations.
- Using the balanced equation and the various concentrations, deduce all the missing concentrations or express them in terms of unknowns (variable x).
- If you need to determine the value of the equilibrium constant, you must have all the information on the concentrations you need in order to substitute them into the expression of the constant to estimate its value.
- If a concentration is to be determined, this concentration must be identified by a variable which is directly linked to the expression of the equilibrium constant.
In a volume of 2L, there are 8 moles of |NH_{3}|, 48g of |N_{2}| and 10g 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}{2L}=\frac{4\; moles}{1L}=4,0M|
|\displaystyle N{}_{2}:\frac{48g}{2L}=\frac{24g}{1L}=\frac{24g}{28g/mol\; de\; N_{2}}=0,86M|
|\displaystyle H{}_{2}:\frac{10g}{2L}=\frac{5g}{1L}=\frac{5g}{2g/mol\; de\; H_{2}}=2,5M|
3. Calculating the equilibrium constant
|\displaystyle K_{c}=\frac{[4,0M]^{2}}{[0,86M]\cdot[2,5M]^{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| | |
| Variation | |\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{[4,0M]\cdot[12,0M]^{3}}{[12,0M]^{2}}=48|
A 1.0 L vessel is filled with 0.5 moles of HI at 448ºC. 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| | |
| Variation |
|\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
Pour valider ta compréhension à propos des concentrations à l'équilibre de façon interactive, consulte la MiniRécup suivante :
