Chemical equilibrium is the state of a chemical reaction, where the concentrations of the reactants and products have no net change over time. Usually, this state results when the forward chemical reactions proceed at the same rate as their reverse reactions.
Having looked at factors that aﬀect the rate of a reaction, we now need to ask some important questions. Does a reaction always proceed in the same direction or can it be reversible? In other words, is it always true that a reaction proceeds from reactants to products, or is it possible that sometimes, the reaction will reverse and the products will be changed back into the reactants? And does a reaction always run its full course so that all the reactants are used up, or can a reaction reach a point where reactants are still present, but there does not seem to be any further change taking place in the reaction? The following demonstration might help to explain this.
Demonstration : Liquid-vapour phase equilibrium
Apparatus and materials: 2 beakers; water; bell jar
- Half ﬁll two beakers with water and mark the level of the water in each case.
- Cover one of the beakers with a bell jar.
- Leave the beakers and, over the course of a day or two, observe how the water level in the two beakers changes. What do you notice?
Note: You could speed up this demonstration by placing the two beakers over a bunsen burner to heat the water. In this case, it may be easier to cover the second beaker with a glass cover.
You should notice that in the beaker that is uncovered, the water level drops quickly because of evaporation. In the beaker that is covered, there is an initial drop in the water level, but after a while evaporation appears to stop and the water level in this beaker is higher than that in the one that is open. Note that the diagram below shows the situation ate time=0.
In the ﬁrst beaker, liquid water becomes water vapour as a result of evaporation and the water level drops. In the second beaker, evaporation also takes place. However, in this case, the vapour comes into contact with the surface of the bell jar and it cools and condenses to form liquid water again. This water is returned to the beaker. Once condensation has begun, the rate at which water is lost from the beaker will start to decrease. At some point, the rate of evaporation will be equal to the rate of condensation above the beaker, and there will be no change in the water level in the beaker.
This can be represented as follows: liquid⇔ vapour In this example, the reaction (in this case, a change in the phase of water) can proceed in either direction. In one direction there is a change in phase from liquid to vapour. But the reverse can also take place, when vapour condenses to form water again.
Open and closed systems
An open system is one in which matter or energy can ﬂow into or out of the system. In the liquid-vapour demonstration we used, the ﬁrst beaker was an example of an open system because the beaker could be heated or cooled (a change in energy), and water vapour (the matter) could evaporate from the beaker.
A closed system is one in which energy can enter or leave, but matter cannot. The second beaker covered by the bell jar is an example of a closed system. The beaker can still be heated or cooled, but water vapour cannot leave the system because the bell jar is a barrier. Condensation changes the vapour to liquid and returns it to the beaker. In other words, there is no loss of matter from the system.
Deﬁnition: Open and closed systems An open system is one whose borders allow the movement of energy and matter into and out of the system. A closed system is one in which only energy can be exchanged, but not matter.
A reversible reaction is a chemical reaction that can proceed in both the forward and reverse directions. In other words, the reactant and product of one reaction may reverse roles.
Some reactions can take place in two directions. In one direction the reactants combine to form the products. This is called the forward reaction. In the other, the products react to form reactants again. This is called the reverse reaction. A special double-headed arrow is used to show this type of reversible reaction:
XY + Z ⇔ X + Y Z
So, in the following reversible reaction:
H2(g) + I2(g) ⇔ 2HI(g) The forward reaction is H2(g) + I2(g) → 2HI(g). The reverse reaction is 2HI(g)→ H2(g) + I2(g).
Using the same reversible reaction that we used in an earlier example:
H2(g) + I2(g) ⇔ 2HI(g)
The forward reaction is:
H2 + I2 → 2HI
The reverse reaction is:
2HI → H2 + I2
When the rate of the forward reaction and the reverse reaction are equal, the system is said to be in equilibrium.
Although it is not always possible to observe any macroscopic changes, this does not mean that the reaction has stopped. The forward and reverse reactions continue to take place and so microscopic changes still occur in the system. This state is called dynamic equilibrium. In the liquid-vapour phase equilibrium demonstration, dynamic equilibrium was reached when there was no observable change in the level of the water in the second beaker even though evaporation and condensation continued to take place.
There are, however, a number of factors that can change the chemical equilibrium of a reaction. Changing the concentration, the temperature or the pressure of a reaction can aﬀect equilibrium. These factors will be discussed in more detail later in this chapter.
The equilibrium constant
The equilibrium constant (Kc), relates to a chemical reaction at equilibrium. It can be calculated if the equilibrium concentration of each reactant and product in a reaction at equilibrium is known.
Calculating the equilibrium constant
Consider the following generalised reaction which takes place in a closed container at a constant temperature:
A + B ⇔ C + D
As we already know that the rate of the forward reaction is directly proportional to the concentration of the reactants. In other words, as the concentration of the reactants increases, so does the rate of the forward reaction. This can be shown using the following equation:
Rate of forward reaction ∝ [A][B]
Rate of forward reaction = k1[A][B]
Similarly, the rate of the reverse reaction is directly proportional to the concentration of the products. This can be shown using the following equation:
Rate of reverse reaction ∝ [C][D]
Rate of reverse reaction = k2[C][D]
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This can be shown using the following equation:
k1[A][B] = k2[C][D]
K1/(k2 ) = ([A][B] )/([C][D])
or, if the constants k1 and k2 are simpliﬁed to a single constant, the equation becomes:
kc = ([C][D] )/([A][B])
A more general form of the equation for a reaction at chemical equilibrium is:
aA + bB ⇔ cC + dD
where A and B are reactants, C and D are products and a, b, c, and d are the coeﬃcients of the respective reactants and products. A more general formula for calculating the equilibrium constant is therefore:
kc = ([C]c [D]d)/([A]a[B]b )
It is important to note that if a reactant or a product in a chemical reaction is in either the liquid or solid phase, the concentration stays constant during the reaction. Therefore, these values can be left out of the equation to calculate kc. For example, in the following reaction:
C(s) + H2O(g) ⇔ CO(g) + H2(g)
kc = ([CO][H2])/([H2O])
calculations on equilibrium constant
Question: For the reaction: S(s) + O2(g) ⇔ SO2(g)
- Write an equation for the equilibrium constant.
- Calculate the equilibrium concentration of O2 if Kc=6 and [SO2] = 3mol.dm−3 at equilibrium.
Answer Step 1 : Write the equation for kc
kc = ([SO2] )/([O2])
(Sulfur is left out of the equation because it is a solid and its concentration stays constant during the reaction)
Step 2 : Re-arrange the equation so that oxygen is on its own on one side of the equation
[O2] = [SO2] kc
Step 3 : Fill in the values you know and calculate [O2]
[O2] =(3mol.dm-3 )/6
Worked Example 2
Question: Initially 1.4 moles of NH3(g) is introduced into a sealed 2.0 dm−3 reaction vessel. The ammonia decomposes when the temperature is increased to 600K and reaches equilibrium as follows:
2NH3(g) ⇔ N2(g) + 3H2(g) When the equilibrium mixture is analysed, the concentration of NH3(g) is 0.3 mol.dm−3
- Calculate the concentration of N2(g) and H2(g) in the equilibrium mixture.
- Calculate the equilibrium constant for the reaction at 900 K.
Answer Step 1 : Calculate the number of moles of NH3 at equilibrium.
c =n V
n = c×V = 0.3×2 = 0.6mol
Step 2 : Calculate the number of moles of ammonia that react (are ’used up’) in the reaction.
Moles used up = 1.4 – 0.6 = 0.8 moles
Step 3 : Calculate the number of moles of product that are formed.
Remember to use the mole ratio of reactants to products to do this. In this case, the ratio of NH3:N2:H2 = 2:1:3.
Therefore, if 0.8 moles of ammonia are used up in the reaction, then 0.4 moles of nitrogen are produced and 1.2 moles of hydrogen are produced.
Using the values in the table, calculate [N2] and [H2]
[N2] = (n )/V
= 0.4/2 = 0.2 mol.dm-3
[H2] = (n )/V
= 0.6 mol.dm−3
Step 6 : Calculate kc
kc = ([H2]3[N2] )/([NH3]2)
= ((0.6)3(0.2) )/((0.3)2)
- Write the equilibrium constant expression, Kc for the following reactions: (a) 2NO(g) + Cl2(g) ⇔ 2NOCl (b) H2(g) + I2(g) ⇔ 2HI(g)
- The following reaction takes place: Fe3+(aq) + 4Cl− ⇔ FeCl− 4 (aq) Kc for the reaction is 7.5 × 10−2 mol.dm−3. At equilibrium, the concentration of FeCl− 4 is 0.95 × 10−4 mol.dm−3 and the concentration of free iron (Fe3+)is 0.2 mol.dm −3. Calculate the concentration of chloride ions at equilibrium.
- Ethanoic acid (CH3COOH) reacts with ethanol (CH3CH2OH) to produce ethyl ethanoate and water. The reaction is: CH3COOH + CH3CH2OH → CH3COOCH2CH3 + H2O At the beginning of the reaction, there are 0.5 mols of ethanoic acid and 0.5 mols of ethanol. At equilibrium, 0.3 mols of ethanoic acid was left unreacted. The volume of the reaction container is 2 dm3. Calculate the value of Kc.
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