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Gibbs Free Energy & Equilibrium Constant (HL) (HL IB Chemistry)

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Philippa Platt

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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-and-react

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

Gibbs free energy and equilibrium relationship for a spontaneous reaction

 

Free energy graph for a non-spontaneous reaction

free-energy-diagrams-2

 

  • The quantitative relationship between standard Gibbs free energy change, temperature and the equilibrium constant is represented by:

Gθ = -RT ln K{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

The rearrangement of this equation makes it possible to:

  • Calculate the equilibrium constant
  • Deduce the position of equilibrium for the reaction

ln K  = -GθRT{"language":"en","fontFamily":"Times New Roman","fontSize":"18","autoformat":true}

  • 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, ΔG, for the reaction is -4.38 kJ mol-1(R = 8.31 J K-1 mol-1)

  1. Calculate the value of Kc for this reaction
  2. Using your answer to part (1), predict and explain the position of the equilibrium

Answers

Answer 1:

Step 1: Convert any necessary values

    • ΔG into J mol-1:
      • -4.38 x 1000 = -4380 J mol-1

    • T into Kelvin
      • 25 + 273 = 298 K

Step 2: Write the equation:

    • ΔG = -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) rightwards harpoon over leftwards harpoon + 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

Q space equals fraction numerator space open square brackets SO subscript 3 close square brackets squared over denominator open square brackets SO subscript 2 close square brackets squared open square brackets straight O subscript 2 close square brackets end fraction

Q space equals fraction numerator space stretchy left square bracket 0.950 stretchy right square bracket squared over denominator stretchy left square bracket 0.100 stretchy right square bracket squared stretchy left square bracket 0.200 stretchy right square bracket end fraction

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 Tip

These equations are given in the data booklet

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.