Gibbs Free Energy & Equilibrium Constant (HL) (DP IB Chemistry): Revision Note
Gibbs Free Energy & the 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

- 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
- Where: - ∆G = standard Gibbs free energy change (kJ mol-1) 
- R = gas constant (8.31 J K-1 mol-1) 
- T = temperature (K) 
- ln K = natural logarithm equilibrium constant 
 
- The rearrangement of this equation makes it possible to: - Calculate the equilibrium constant 
- Deduce the position of equilibrium for the reaction 
 
- 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 
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 > 1
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 = -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-1 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 ln451.25)/1000
ΔG = -142 + 73.9 = -68.1 kJ mol-1
Remember to divide by 1000, because R is in J K-1 mol-1 not kJ K-1 mol-1
Examiner Tips and Tricks
These equations are given in the IB Chemistry data booklet (Section 1)
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