Gibbs Free Energy & Equilibrium Constant (HL) (DP IB Chemistry)
Revision Note
Gibbs Free Energy & Equilibrium Constant
When ΔG < 0 for a reaction at constant temperature and pressure, the reaction is spontaneous
When a reversible reaction reaches equilibrium, the Gibbs free energy is changing as the ratio of reactants to products changes
For non-reversible reactions:
As the amount of products increases, the reaction moves towards completion
This leads to a decrease in Gibbs free energy
For reversible reactions:
As the amount of products increases, the reaction moves towards equilibrium
This causes a decrease in Gibbs free energy
At the point of equilibrium, Gibbs free energy is at its lowest as shown on the graph:
Gibbs free energy and equilibrium relationship
Gibbs free energy changes as the reaction proceeds
In section 1 of the graph, the forward reaction is favoured and the reaction proceeds towards a minimum value
Having reached a point of equilibrium, the Gibbs free energy increases
This is when the reaction becomes non-spontaneous (section 2)
The reverse reaction now becomes spontaneous and the Gibbs free energy again reaches the minimum value, so heads back towards equilibrium
The reaction will be spontaneous in the direction that results in a decrease in free energy (becomes more negative)
When the equilibrium constant, K, is determined for a given reaction, its value indicates whether the products or reactants are favoured at equilibrium
ΔG is an indication of whether the forward or backward reaction is favoured
Free energy graph for a spontaneous reaction
Free energy graph for a non-spontaneous reaction
The quantitative relationship between standard Gibbs free energy change, temperature and the equilibrium constant is represented by:
∆Gθ = -RT ln K
The rearrangement of this equation makes it possible to:
Calculate the equilibrium constant
Deduce the position of equilibrium for the reaction
ln K = -∆GθRT
The reaction quotient, Q, is calculated using the same equation as the equilibrium constant expression, but with non-equilibrium concentrations of reactants and products
It is a useful concept because the size of Q can tell us how far a reaction is from equilibrium and in which direction the reaction proceeds
So how is the reaction quotient related to Gibbs free energy?
They are related by the following expression:
ΔG = ΔGθ + RT ln Q
At non-equilibrium conditions ΔG and ΔGθ are not the same; ΔG is the driver that pushes a reaction toward equilibrium
When a reaction reaches equilibrium, Q = K and ΔG = 0, so
0 = ΔGθ + RT ln K
ΔGθ = - RT ln K
Worked Example
Calculating Kc
Ethanoic acid and ethanol react to form the ester ethyl ethanoate and water as follows:
CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)
At 25 oC, the free energy change, ΔGꝊ, for the reaction is -4.38 kJ mol-1. (R = 8.31 J K-1 mol-1)
Calculate the value of Kc for this reaction
Using your answer to part (1), predict and explain the position of the equilibrium
Answers
Answer 1:
Step 1: Convert any necessary values
ΔGꝊ into J mol-1:
-4.38 x 1000 = -4380 J mol-1
T into Kelvin
25 + 273 = 298 K
Step 2: Write the equation:
ΔGꝊ = -RT lnK
Step 3: Substitute the values:
-4380 = -8.31 x 298 x lnKc
Step 4: Rearrange and solve the equation for Kc:
ln K = -4380 ÷ (-8.31 x 298)
ln K = 1.77
K = e1.77
K = 5.87
Answer 2:
From part (1), the value of Kc is 5.87
Therefore, the equilibrium lies to the right / products side because the value of Kc is positive
Worked Example
Finding ΔG
Sulfur dioxide reacts with oxygen to form sulfur trioxide in the following reversible reaction:
2SO2 (g) + O2 (g) + 2SO3 (g)
ΔGθ is = -142 kJmol-1
In an experiment, the concentrations of [SO2], [O2], [SO3], were found to be 0.100 mol dm-3, 0.200 mol dm-3, and 0.950 mol dm-3 respectively at 1455 K. R = 8.31 J K mol-1
Calculate the value of ΔG at this temperature.
Answer
Step 1- Write the Q expression
Step 2 - Solve Q
Q = 451.25
Step 3 - Substitution
ΔG = ΔGθ + RT ln Q
ΔG = -142 + (8.31 x 1455 x 451.25)/1000
ΔG = -142 + 73.9 = -68.1 kJ mol-1
Remember to divide by 1000, because R is in J mol-1K-1 not kJ mol-1K-1
Examiner Tips and Tricks
These equations are given in the data booklet
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