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Wize University Chemistry Textbook > Thermodynamics Part 2: (2nd Law-entropy, 3rd Law, Gibbs Free Energy, Spontaneity)

Gibbs Free Energy + Spontaneity

3rd Law of ThermodynamicsPrevious Section
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Gibbs Free Energy and Spontaneity

  • It is not always convenient to calculate the entropy of the system and the surroundings to determine whether a process is spontaneous
  • Gibbs showed that at constant T and p another state function, ΔG , of the system can predict whether a reaction is spontaneous or not.
  • under constant T:
ΔG=ΔH−TΔS\Delta G=\Delta H-T \Delta S
  • under constant T and P:
  • If ∆𝐺 > 0 the reaction is
    non-spontaneous
    (
    spontaneous
    in the reverse direction)
  • If ∆𝐺 < 0 the reaction is
    spontaneous
    and reversible/irreversible
    irreversible
  • Just like enthalpy and entropy Gibbs free energy for a reaction can be calculated
ΔGproducts−ΔGreactants=ΔGrxn\Delta G_{products}-\Delta G_{reactants}=\Delta G_{rxn}

Note: we can use the same equation we saw for enthalpy of the reaction and entropy of the reaction. For Gibbs Free Energy just be aware that if the tables are given with values at 25C, this equation can only be used when reaction conditions are at 25C!
  • Gibbs free energy is a state function which allows us to set up the following relation
ΔGforward=−ΔGreserve\Delta G_{forward} =-\Delta G_{reserve}
  • Also recall how for deltaH0f of aqueous ions, it was not possible to separate the cation and anion contributions
  • deltaH0f=0 for H+
  • we can do the same for entropy and Gibbs free energy: deltaS0=0 for H+ and deltaG0f=0 for H+


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Spontaneity


Look at each row in the following table, use deltaG=deltaH -TdeltaS
Which are spontaneous, which are not?
Which are spontaneous only at a high/low temperature?
If we change the temperature, which could become non-spontaneous??
Which one is entropy driven to be spontaneous and which one is enthalpy driven to be spontaneous?

deltaG deltaH deltaS
- - +

- - -

- + +

+ + -'


We can be even more specific with our temperatures, rather than saying high or low temperatures to be spontaneous:

Write a < or > sign:

For Row 2: T deltaHdeltaS \frac{deltaH}{deltaS\ }

For Row 3: T deltaHdeltaS\frac{deltaH}{deltaS}

Now what if for row 2 we had placed the opposite sign?


What about for row 3 if we placed the opposite sign?
Using the following table, predict which of the following reactions would be spontaneous at 25°C:

If either of the reactions is non-spontaneous, at what temperature might it become spontaneous?

Consider reaction 1:
deltaG = deltaH - T(deltaS)
=9250J/mol - 298K(25.1J/molK)
=1770.2J/mol

Since deltaG is positive, it is non-spontaneous.

The temerature would need to be a lot higher to result in a negative delta G (spontaneous reaction).
Specifically the temperature would have to be greater than deltaH/deltaS.
deltaH/deltaS=9250J/mol / 25.1J/molK
=368.5K Therefore temp. would need to be higher than 368.5K for it to be spontaneous!

Want to check?
DeltaG= deltaH - TdeltaS
=9250J/mol - 369(25.1)
= - 11.9 aka the reaction is now spontaneous!!
Now consider reaction 2:
deltaG = deltaH - T(deltaS)
=2850J/mol - 298K(-125J/molK)
=2850 J/mol+ 37250J/mol
=40 100 J/mol

This reaction is also not spontaneous because the deltaG is a positive value.

For this reaction, since deltaH is positive and entropy is negative, both the deltaH part of the equation and the -T(deltaS) part of the equation end up positive no matter how you change the temperature in Kelvins. Therefore, temperature cannot be changed to make this reaction spontaneous and it will remain non-spontaneous!
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Predicting Spontaneity from Entropy

  • There are two main ways you may be asked to figure out the spontaneity of a process
  • If you are given the change in entropy, temperature, AND change in enthalpy (ΔH\Delta H)
  • If you are given the change in entropy of the system AND the change in entropy of the surroundings (from the 2nd law of thermodynamics)
  • Remember ΔSuniverse \Delta S_{universe}\ allows us to predict spontaneity and ΔSuniverse =ΔSsystem+ΔSsurr\Delta S_{universe\ }=\Delta S_{system}+\Delta S_{surr}
Exam Tip
This chart can be used to predict the spontaneity of a process if you're only given the change in entropy of the system and surroundings


Extra Practice
Gibbs Free Energy: Spontaneity
If a reaction is nonspontaneous at 298 K with a negative ΔHo, then the reaction is:
Gibbs Free Energy: Spontaneity
For all of the following reactions both ΔGo and ΔHo are negative. For which reactions is ΔGo more negative than ΔHo? (You do not need thermochemical data to answer this question.)
1. SO2 (g) + SrO (s) → SrSO3 (s)
2. N2F4 (g) → 2 NF2 (g)
The Haber Bosch Process is an industrial process for “fixing” nitrogen, shown below. This process produces around 500 million tons of ammonia every year.
ΔfG0(NH3)=−16.4 kJ/mol\Delta_fG^0(NH_3)=-16.4\ kJ/mol
3/2 H2(g)+1/2 N2(g)→NH3(g)\rm 3/2\ H_{2(g)}\quad+\quad 1/2\ N_{2(g)}\quad\overrightarrow{\hspace{1.5cm}}\quad NH_{3(g)}
Gibbs Free Energy
Given the following reaction that occurs in the body, what is ∆Go?
AMP + H20(l) --> Adenosine + Pi + H+ ∆Go'=-13kJ/mol
∆Go'=∆Go + RTlnQ
Gibbs Free Energy: Spontaneity
Based on the following reactions, answer the following questions:
Reaction 1: CO2(g) + CaO(s) ⇌ CaCO3(s) ∆Go= -130.4 kJ/mol
Reaction 2: C(s) + O2(g) ⇌ CO2(g) ∆Go= -394.4 kJ/mol
For a reaction with a negative ΔGo value, which of the following statements is/are FALSE?
1. The equilibrium constant K is greater than one.
2. The reaction is spontaneous when all the reactants and products are in their standard states.
Gibbs Free energy: Spontaneity
For the reaction, A + B → C, ΔHo = +30 kJ; ΔSo=+50 J/K. Therefore the reaction is:
Gibbs Free Energy: Spontaneity
Which spontaneous process occurs because the enthalpy term dominates over the entropic term in the Gibbs free energy equation?
Gibbs Free Energy: Spontaneity
For all of the following reactions both ΔGo and ΔHo are negative. For which reactions is ΔGo more negative than ΔHo? (You do not need thermochemical data to answer this question.)
1. SO2 (g) + SrO (s) → SrSO3 (s)
2. N2F4 (g) → 2 NF2 (g)
A spontaneous process:
Gibbs Free Energy: Spontaneity
If a reaction is nonspontaneous at 298 K with a negative ΔHo, then the reaction is:
Gibbs Free Energy
Determine the value of ΔGo for these reactions.
A process cannot be spontaneous (product-favored) if __________________.
Gibbs Free Energy: Spontaneity
The oxidation of NO is shown below. Using the provided thermodynamic data answer the following questions
2NO(g)+O2(g)⇌2NO2(g)2NO(g)+O_2 (g)⇌2NO_2 (g)
∆H0=−57.1kJ/mol and ∆S0=−73J/K mol at 298K∆H^0=-57.1 kJ/mol \ and \ ∆S^0=-73 J/K\ mol\ at \ 298K
Gibbs Free Energy: Formation
The value of ΔGo for the following reaction is -2745 kJ. Use this value and the information on the Data Sheet to calculate the standard free energy of formation, ΔGfo of C4H10 (g) in kJ/mol.
C4H10 (g) + 13/2 O2 (g) →\rightarrow 4 CO2 (g) + 5 H2O (l)
ΔGfo (CO2(g)) = -394.4 kJ/mol ΔGfo (H2O(l)) = -237.13 kJ/mol
Gibbs Free Energy: Reaction quotient
The standard free energy change (ΔGo) for the reaction of ammonia with water is 29.05 kJ/mol at 25oC.
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
What is the value of ΔG (in kJ/mol) when the concentrations of NH3 (aq) = 0.10 M, NH4+ (aq) = 1.0x10-3 M, and OH- (aq) = 1.0x10-3 M?