The equilibrium constant measures the extent of a chemical reaction. At equilibrium there is no tendency for the reversible reaction to go forwards or backwards. Gibbs free energy change is zero. This section explores the link between K, the equilibrium constant and ΔG, Gibbs free energy change.
Syllabus reference R2.3.7
Reactivity 2.3.7 - The equilibrium constant and Gibbs energy change, ΔG, can both be used to measure the position of an equilibrium reaction. (HL)
Guidance
- Calculations using the equation ΔG⦵ = −RT lnK. The equation is given in the data booklet.
Tools and links
- Reactivity 1.4 - How can Gibbs energy be used to explain which of the forward or backward reaction is favoured before reaching equilibrium?
Gibbs free energy and entropy
Gibbs free energy is related to the entropy of the entire universe. It is a measure of the amount of chemical potential energy of a system that is available to do useful work. In doing so, the entropy of the universe would increase and thermodynamic feasibility of the process is confirmed.
Consider a reversible reaction from the point of view of the reactants, i.e. the forward reaction. If it is to be possible, then Gibbs free energy change must be negative. However, once equilibrium is attained the forward reaction is no longer possible and Gibbs free energy for a further hypothetical reaction must be positive.
The same argument can be applied to the reverse reaction. Initially, Gibbs free energy change is negative, but once equilibrium is attained it would become positive for a further hypothetical reaction.
Hence, at equilibrium the Gibbs free energy change in either direction is positive, i.e. no reaction is possible under the ambient conditions. At equilibrium, Gibbs free energy chage for any possible change to the system is positive. At this point, the entropy of the universe (and hence the system) is at a maximum and any further change would reduce it - fundamentally impossible.
To summarise, Gibbs free energy at equilibrium is zero, equilibrium represents a Gibbs free energy well, from which any further change is impossible (under the current conditions)
^ top
Gibbs free energy and the equilibrium constant
It is possible to relate the Gibbs free energy change of the initial hypothetical reaction in an equilibrium to the value of the equilibrium constant for the same reaction.
The formula is given in the data booklet:
ΔG = -RTlnk
Where: R is the universal gas constant, T is the absolute temperature and k is the equilibrium constant. "ln" means the natural logarithm (the number expressed as an exponent of the natural number base 'e', 2.718).
Determine an equilibrium constant.
Given that the value for Gibbs free energy change = 60 kJ mol-1, determine the value of the equilibrium constant at 300K
60 kJ = 6.0 x 104 J
R = 8.31 J K-1 mol-1
T = 300
6.0 x 104 = 8.31 x 300 (lnk)
lnk = 24.1, k = e24.1
k = 2.83 x 1010
The large value of the equilibrium constant, k, shows us the the equilibrim position lies almost totally on the side of the products, i.e. the reaction goes almost to completion.
^ top
Worked examples
Q722-01 Consider the endothermic
reaction below:
5CO(g) + I2O5(g) ⇋
5CO2(g) + I2(g)
According to Le Chatelier's principle, which change would result in an increase in the amount of CO2?
- Increasing the temperature
- Decreasing the temperature
- Increasing the pressure
- Decreasing the pressure
Answer
|
As it is an endothermic reaction in the forward direction an increase
in temperature pushes the equilibrium to the side of the products. There
is no change in the number of moles and so the equilibrium is unaffected
by pressure change.
The correct response is A.
|
Q722-02 The compounds N
2O
4
and NO
2 produce an equilibrium mixture according to the equation below:
N2O4 (g) ⇋
2NO2(g) ........ ∆H = +57.2 kJmol-1
An increase in the equilibrium concentration of NO2 can be produced
by increasing which of the following factors:
- I. Pressure
- II. Temperature
Answer
|
Increasing the temperature pushes the reaction in the direction of
ENDOthermic change, ie to the right hand side.
Increasing the pressure pushes the reaction in the direction of the
fewer number of moles of gas, ie to the left hand side in this case.
|
Q722-03 For the following system
at equilibrium, which change will shift the position of equilibrium to the right?
CH3COOH(l) + C2H5OH(l) ⇋
CH3COOC2H5(l) + H2O(l) .....
∆H < 0
- Adding a catalyst
- Increasing the pressure
- Removing water
- Increasing the temperature
Answer
|
Remove from one side of an equilibrium and you pull the reaction towards
that same side. If you remove one of the reacting components of the
REVERSE reaction, then the rate of the reverse reaction becomes less
than that of the forward reaction. The system then reacts faster in
the forward direction, making more of the right hand side.
|
Q722-04 Which change will shift
the position of equilibrium to the right in this reaction?
N2(g) + 3H2(g) ⇋
2NH3(g) ... ∆H = - 92 kJ
- Increasing the temperature.
- Decreasing the pressure.
- Adding a catalyst.
- Removing the ammonia from the equilibrium mixture.
Answer
|
The sign of ∆H is negative therefore it is exothermic in the
forward direction. Increasing temperature favours the direction of ENDOthermic
change ie to the left hand side in this case. II does not go
to the right hand side.
Decreasing the pressure forces the equilibrium to move towards the
side of greater moles of gas to re-establish the equilibrium concentrations.
Catalysts do NOT affect equilibria.
Removing ammonia disturbs the equilibrium concentrations and the reaction
responds by making more ammonia to re-establish the equilibrium concentration
proportions - this is the correct response.
|
Q722-05
2H2O(l) ⇋
H3O+(aq) +OH-(aq)
The equilibrium constant for the reaction above is 1.0 x 10-14
at 25ºC and 2.1 x 10-14 at 35ºC. What can be concluded
from this information?
- [H3O+] decreases as the temperature is raised
- [H3O+] is greater than [OH-]
at 35ºC
- Water is a stronger electrolyte at 25ºC
- The ionization of water is endothermic.
Answer
|
Kc has a larger value at the higher temperature. Thus the temperature
increases and the proportion of products increases. Temperature increase
favours the direction of endothermic change therefore this reaction
is endothermic in the forward direction.
|
Q722-06 Which change increases
the amount of NH
4+ in the below reaction?
NH3(g) + H2O(l) ⇋
NH4+(aq) + OH-(aq)....
∆H > 0
- Decreasing the temperature
- Decreasing the pressure
- Removing water
- Adding an acid
Answer
|
The sign of ∆H is positive (it is greater than 0) therefore it
is endothermic in the forward direction. Increasing temperature favours
the direction of ENDOthermic change ie to the right hand side in this
case. Decreasing the temperature will move the equilibrium position
to the left hand side.
Decreasing the pressure forces the equilibrium to move towards the
side of greater moles of gas to re-establish the equilibrium concentrations,
in this case the left (notice the state symbols)
Removing water disturbs the equilibrium concentrations and the reaction
responds by making more water to re-establish the equilibrium concentration
proportions. This pulls the reaction to the left to restore the value
of Kc.
Adding acid reacts with the OH- ions from the right hand side disturbing
the equilibrium proportions that are then re-established by the reaction
moving to the right hand side making more NH4+.
|
Q722-07 Consider the equilibrium
reaction:
4NH3(g) + 3O2(g) ⇋
2N2(g) + 6H2O(g) ΔH = -1268
kJ
Which change will cause the reaction to shift to the right?
- Increase the temperature
- Decrease the volume of the container.
- Add a catalyst to speed up the reaction.
- Remove the gaseous water by allowing it to react and be absorbed by KOH.
Answer
|
The reaction is exothermic in the forward direction, thus increasing
the temperature drives the equilibrium in the reverse direction.
Decreasing the volume increases the pressure. The equilibrium has 7
moles of gas on the left and 8 moles of gas on the right, hence increasing
the pressure drives the reaction to the left hand side.
Adding a catalyst has no effect on the position of equilibrium.
Removing the gasous water disturbs the equilibrium and the system must
respond by making more water, i.e. by moving
to the right to re-establish the equilibrium proportions.
|
Q722-08 Which changes will shift
the position of equilibrium to the right in the following reaction?
2CO2(g) ⇋
2CO(g) + O2(g)
- I. adding a catalyst
- II. decreasing the oxygen concentration
- III. increasing the volume of the cylinder
- I and II only
- I and III only
- II and III only
- I, II and III
Answer
|
I - catalysts have no effect on equilibrium.
II - decreasing the oxygen concentration pulls the reaction to the
right to re-establish the equilibrium proportions.
II - increasing the volume decreases the pressure. There are two moles
of gas on the left and three moles of gas on the right, hence a decrease
in pressure moves the reaction towards the side of the most moles (the
right) to re-establish the equilibrium proportions.
Corrct response II and III only
|
Q722-09 The reaction below is
an important step in the production of sulfuric acid. An increase in which of
the following, will increase the ratio of SO
3(g) to SO
2(g)
at equilibrium?
2SO2(g) + O2(g) ⇋
2SO3(g) ΔH = -197 kJ
- Pressure only
- Temperature
- Both pressure and temperature
- Neither pressure nor temperature
Answer
|
Increase pressure moves the reaction in the directin of fewer moles,
in this case to the right. This increases the ratio of SO3(g)
to SO2(g) at equilibrium. Increased temperature moves the
position of equilibrium towards the side of endothermic change. in this
case the left. This reduces the ratio of SO3(g) to SO2(g)
at equilibrium.
Correct response A
|
Q722-10 The equation for a reaction
used in the manufacture of nitric acid is:
4NH3(g) + 5O2(g) ⇋
4NO(g) + 6H2O(g) ΔH =
-900 kJ
Which changes occur when the temperature of the reaction is increased?
| |
Position of equilibrium
|
Value of Kc
|
|
A.
|
shifts to the left
|
increases
|
|
B.
|
shifts to the left
|
decreases
|
|
C.
|
shifts to the right
|
increases
|
|
D.
|
shifts to the right
|
decreases
|
Answer
|
When the temperature is increased the position of equilibrium moves
towards the direction of endothermic change, in this case it shifts
to the left. The value of Kc decreases as there is less product (right
hand side) and more reactants (left hand side) at equilibrium.
Correct response B
|
^ top