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Spontaneous Reactions (HL) (HL IB Chemistry)

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Philippa

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Philippa

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Spontaneous Reactions

  • Gibbs free energy provides an effective way of focusing on a reaction system at constant temperature and pressure to determine its spontaneity
  • For a reaction to be spontaneous, Gibbs free energy must have a negative value (ΔG ≤ 0)
  • We can use the Gibbs equation to calculate whether a reaction is spontaneous / feasible or not

ΔG = ΔHreaction – TΔSsystem

    • When ΔG is negative, the reaction is spontaneous / feasible and likely to occur
    • When ΔGis positive, the reaction is not spontaneous / feasible and unlikely to occur
  • Depending on the value for ΔH and ΔS we can determine whether the reaction is spontaneous at a given temperature (T)
  • We can also look at the values for enthalpy change, ΔH, and entropy change, ΔS

Worked example

Determining if a reaction is feasible / spontaneous

  1. Calculate the Gibbs free energy change for the following reaction at 298 K
  2. Determine whether the reaction is feasible.

2Ca (s) + O(g) → 2CaO (s)         ΔH = -635.5 kJ mol-1

  • S[Ca(s)] = 41.00 J K-1 mol-1
  • S[O2(g)] = 205.0 J K-1 mol-1
  • S[CaO(s)] = 40.00 J K-1 mol-1

 

Answer 1:

Step1: Calculate ΔSsystem

    • ΔSsystemꝋ = ΣΔSproducts – ΣΔSreactants

    • ΔSsystemꝋ = (2 x ΔS [CaO(s)]) –  (2 x ΔS [Ca(s)] + ΔSꝋ [O2(g)])
      • = (2 x 40.00) – (2 x 41.00 + 205.0)
      • = -207.0 J K-1 mol-1

Step 2: Convert ΔSꝋ to kJ K-1 mol-1

    • ΔSsystemꝋ = fraction numerator negative 207.0 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent over denominator 1000 end fraction -0.207 kJ mol-1

Step 3: Calculate ΔG

    • ΔG = ΔHreaction – TΔSsystem

ΔG  = -635.5 – (298 x -0.207)

          = –573.8 kJ mol-1

 

Answer 2: 

    • Since ΔG is negative, the reaction is feasible

Factors affecting ΔG and the spontaneity / feasibility of a reaction

  • We can also look at the values for ΔH and ΔS to determine whether the reaction is spontaneous / feasible at a given temperature (T)
  • The Gibbs equation will be used to explain what will affect the spontaneity / feasibility of a reaction for exothermic and endothermic reactions

Gibbs free energy equation

Gibbs Free Energy Equation

Exothermic reactions

  • In exothermic reactions, ΔHreaction is negative
  • If the ΔSsystem is positive:
    • Both the first and second terms will be negative
    • Resulting in a negative ΔG so the reaction is feasible
    • Therefore, regardless of the temperature, an exothermic reaction with a positive ΔSsystem will always be feasible

  • If the ΔSsystem is negative:
    • The first term is negative and the second term is positive
    • At very high temperatures, the –TΔSsystemꝋ will be very large and positive and will overcome ΔHreaction
    • Therefore, at high temperatures ΔGꝋ is positive and the reaction is not feasible

  • Since the relative size of an entropy change is much smaller than an enthalpy change, it is unlikely that TΔS > Δas temperature increases
  • These reactions are therefore usually spontaneous under normal conditions

Flow chart to determine the feasibility of exothermic reactions

Feasibility summary of an exothermic reaction

The diagram shows under which conditions exothermic reactions are feasible

Endothermic reactions

  • In endothermic reactions, ΔHreaction is positive
  • If the ΔSsystem is negative:
    • Both the first and second term will be positive
    • Resulting in a positive ΔG so the reaction is not feasible
    • Therefore, regardless of the temperature, endothermic with a negative ΔSsystem will never be feasible

  • If the ΔSsystem is positive:
    • The first term is positive and the second term is negative
    • At low temperatures, the –TΔSsystemꝋ will be small and negative and will not overcome the larger ΔHreaction
    • Therefore, at low temperatures ΔGꝋ is positive and the reaction is not feasible
    • The reaction is more feasible at high temperatures as the second term will become negative enough to overcome the ΔHreaction resulting in a negative ΔG

  • This tells us that for certain reactions which are not feasible at room temperature, they can become feasible at higher temperatures
    • An example of this is found in metal extractions, such as the extraction if iron in the blast furnace, which will be unsuccessful at low temperatures but can occur at higher temperatures (~1500 oC in the case of iron)

Flow chart to determine the feasibility of endothermic reactions

Feasibility summary of an endothermic reaction

The diagram shows under which conditions endothermic reactions are feasible

Summary of factors affecting Gibbs free energy 

If ΔH .... And if ΔS .... Then ΔG is Spontaneous?  Because

is negative

< 0

exothermic

is positive

> 0

more disorder

always negative

< 0

Always Forward reaction spontaneous at any T

is positive

> 0

endothermic

is negative

< 0

more order

always positive

> 0

Never Reverse reaction spontaneous at any T

is negative

< 0

exothermic

is negative

< 0

more order

negative at low T

positive high T

Dependent on T

Spontaneous only at low T

TΔS < H

is positive

> 0

endothermic

is positive

> 0

more disorder

negative at high T

positive low T

Dependent on T

Spontaneous only at high T

TΔS > H

Temperature & Spontaneity

  • Rearranging the Gibbs equation allows you to determine the temperature at which a non-spontaneous reaction become feasible

ΔG = ΔHreaction - TΔSsystem

  • Remember, for a reaction to be feasible ΔGΘꝋmust be zero or negative
    • 0 = ΔH - TΔS
    • ΔH = TΔS
    • T fraction numerator increment H to the power of theta over denominator increment S to the power of theta end fraction

Worked example

At what temperature will the reduction of aluminium oxide with carbon become spontaneous?

Al2O3 (s) + 3C (s)→ 2Al (s) + 3CO (g)             

ΔH = +1336 kJ mol-1

ΔS = +581 J K-1 mol-1

Answer:

  • If ΔG = 0 then T = fraction numerator increment H to the power of theta over denominator increment S to the power of theta end fraction
  • Covert ΔS to kJ K-1 mol-1 by dividing by 1000
  • T = fraction numerator 1336 over denominator open parentheses 581 over 1000 close parentheses end fraction = 2299 K

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Philippa

Author: Philippa

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.