Thermodynamic Favorability (College Board AP® Chemistry) : Study Guide

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Updated on

Thermodynamic Favorability

  • Gibbs free energy provides an effective way of focusing on a reaction system at constant temperature and pressure to determine whether the reaction is thermodynamically favorable 

  • A thermodynamically favorable reaction is one that occurs without the need for additional energy input after the reactants are mixed

    • A thermodynamically favored reaction will have a negative ΔGo value 

  • For a reaction to be spontaneous, Gibbs free energy must be a negative value (ΔGo ≤ 0)

Examiner Tips and Tricks

Historically, a thermodynamically favorable reaction has been referred to as a spontaneous reaction.

Thermodynamically favored is the preferred term as it avoids the common misunderstanding that the reaction happens suddenly or without cause.

Thermodynamic Favorability & Temperature Changes

  • The temperature conditions for a process to be thermodynamically favored (whereby ΔGo is negative and less than 0) can be predicted from the signs of ΔHo and ΔSo

Entropy - Gibbs Equation, downloadable AS & A Level Chemistry revision notes
  • If ΔHis negative (exothermic reaction) and ΔSo is positive:

    • This will result in a negative  ΔG 

    • The reaction is always thermodynamically favored

    • No calculation of ΔG would be necessary to determine this

  • If ΔHis negative (exothermic reaction) and ΔSo is negative:

    • At high temperatures -TΔSwill be very large and positive and will overcome ΔHo

    • At high temperatures ΔGo is positive

    • The reaction is thermodynamically favored at low temperatures

ΔGand Exothermic Reactions

Entropy - Feasibility of Exothermic Reactions, downloadable AS & A Level Chemistry revision notes

If ΔHis negative and If ΔSo is negative the reaction will only be thermodynamically favored at low temperatures

  • If ΔHis positive (endothermic reaction) and ΔSo is negative:

    • Both ΔHo and -TΔSo will be positive

    • This results in a positive ΔGo 

    • Therefore, regardless of the temperature, endothermic with a negative ΔSo is not thermodynamically favored

    • No calculation of ΔGo is necessary to determine that the process is thermodynamically unfavored

  • If ΔHis positive (endothermic reaction) and ΔSo is positive

    •  ΔHo is positive and -TΔSo will be negative

    • At low temperatures -TΔSo cannot overcome the larger ΔHo

    • At low temperatures, ΔGo is positive and the reaction is thermodynamically favored 

    • At higher temperatures, the second term will become negative enough to overcome the ΔH resulting in a negative ΔGo

ΔGand Endothermic Reactions

Entropy - Feasibility of Endothermic Reactions, downloadable AS & A Level Chemistry revision notes

If ΔHis positive and If ΔSo is positive the reaction will only be thermodynamically favored at a high temperatures

Summary table showing the relationship between ΔH, ΔSand ΔGo

ΔH

ΔS

ΔG  < 0 favoured at:

 Reaction is:

 < 0

 > 0 

all temperatures

thermodynamically favored

 > 0

< 0

no temperatures

thermodynamically unfavored 

 > 0

 > 0

high temperatures

thermodynamically favoured as temperature increases

 < 0

< 0

low temperatures

thermodynamically unfavored as temperature increases

Worked Example

For a reaction with the following enthalpy and entropy changes:

  • ΔS= -75.8 kJ mol-1

  • ΔH= -13.65 kJ mol-1

a) Calculate ΔGo at 298 K. 

b) Calculate the temperature the reaction will become thermodynamically favored.

Answer:

a) At 298 K:

  • Convert ΔSo from J K-1 mol-1 into kJ K-1 mol-1

    • = fraction numerator negative 75.8 space over denominator 1000 end fraction

    • = – 0.0758 kJ K-1 mol-1

  • ΔGo= ΔHo – TΔSo

    • ΔGo = -13.65 – 298 x (–0.0758) 

    • ΔGo= + 8.94 kJ mol-1

b) The temperature at which the reaction will become thermodynamically favored:

  • Rearrange ΔGo= ΔHo – TΔSo to find T

    • When ΔGo= 0, then straight T equals space fraction numerator straight capital delta H to the power of straight o over denominator straight capital delta S to the power of straight o end fraction

    • T= fraction numerator negative 13.65 over denominator negative 0.0758 end fraction equals space 180 space straight K

  • When ΔHand ΔSare both negative, decreasing the temperature causes reactions to be thermodynamically favored 

  • Therefore below 180 K, the reaction is thermodynamically favored, and above 180 K the reaction is not thermodynamically favored 

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Alexandra Brennan

Author: Alexandra Brennan

Expertise: Chemistry Content Creator

Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.

Stewart Hird

Reviewer: Stewart Hird

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

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