Effect of Temperature on Gibbs Free Energy and Spontaneity of Reactions Chemistry Tutorial

Key Concepts

• Gibbs free energy (G) is also known as free energy or Gibbs energy.
• The sign of the change in Gibbs free energy (ΔG) for a chemical reaction tells us if the reaction is spontaneous or not:

  A reaction is spontaneous if ΔG_reaction < 0 (ΔG is negative)

  A reaction is nonspontaneous if ΔG_reaction > 0 (ΔG is positive)

  A reaction is at equilibrium if ΔG_reaction = 0

• The value of ΔG_reaction is dependent on temperature.⁽¹⁾
• The sign of ΔG_reaction does not change with changes in temperature if the sign of ΔH is the opposite of the sign of ΔS:

  (a) ΔH < 0 and ΔS > 0 then ΔG < 0 and reaction is spontaneous at any temperature.

  (b) ΔH > 0 and ΔS < 0 then ΔG > 0 and reaction is nonspontaneous at any temperature.

• The sign of ΔG_reaction does change with changes in temperature if the sign of ΔH is the same as the sign of ΔS:

  (a) ΔH < 0 and ΔS < 0 then the reaction is:

  ⚛ spontaneous at low temperature

  ⚛ nonspontaneous at high temperature

  (b) ΔH > 0 and ΔS > 0 then the reaction is:

  ⚛ nonspontaneous at low temperature

  ⚛ spontaneous at high temperature

• The temperature⁽²⁾ in Kelvin of a system at equilibrium, T_eq, can be calculated as follows:

  T_eq = ΔH ÷ ΔS

• A high temperature is defined as a temperature greater than T_eq

  high temperature: T > T_eq

• A low temperature is defined as a temperature less than T_eq

  low temperature: T < T_eq

• Use the table below to help you predict at what temperature a chemical reaction or a physical change will be spontaneous:

  	ΔHsystem	ΔSsystem	ΔGsystem (=ΔH - TΔS)		Change
  			ΔG sign	conditions
  Opposite sign	−	+	−	any temperature	Spontaneous
  	+	−	+	any temperature	Nonspontaneous
  Same sign	−	−	−	low temperature (T < Teq)	Spontaneous
  			+	high temperature (T > Teq)	Nonspontaneous
  	+	+	+	low temperature (T < Teq)	Nonspontaneous
  			−	high temperature (T > Teq)	Spontaneous

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Theory: How does a change in temperature effect Gibbs free energy and the spontaneity of a chemical reaction?

Recall that we can calculate the value of the Gibbs free energy change (ΔG_reaction) for a chemical reaction or a physical change at constant temperature and pressure using the equation given below:

ΔG_reaction = ΔH_reaction - TΔS_reaction

ΔH_reaction = enthalpy change for the reaction⁽³⁾ in kJ mol^-1

T = temperature of the system in kelvin (K)

ΔS_reaction = change in entropy for the reaction in kJ K^-1 mol^-1

Also recall that ΔG_reaction can be used to determine whether this chemical reaction or physical change will be spontaneous, nonspontaneous or at equilibrium at this particular temperature and pressure:

Putting all this information together, we can now say that a chemical reaction or a physical change will be:

• Spontaneous if ΔG_reaction < 0, that is, if ΔH_reaction - TΔS_reaction < 0
• At equilibrium if ΔG_reaction = 0, that is, if ΔH_reaction - TΔS_reaction = 0
• Nonspontaneous if ΔG_reaction > 0, that is, if ΔH_reaction - TΔS_reaction > 0

Let's think about a spontaneous chemical reaction first.
A reaction will ALWAYS be spontaneous⁽⁴⁾ at constant temperature and pressure if the following two conditions are met:

• it is exothermic (releases energy), that is, ΔH_reaction < 0
• the entropy of the chemical system increases, that is, ΔS_reaction > 0

and the temperature in kelvin for the chemical system will always be greater than 0 (T > 0).⁽⁵⁾
So, when we calculate ΔG

If you subtract a positive number from a negative number the result will always be negative, so, if ΔH_system is negative and ΔS_system is positive then ΔG_system is always negative and the reaction is always sponanteous.

If ΔH_reaction < 0 and ΔS_reaction > 0 the reaction is spontaneous at any temperature.

Now, let's think about what would happen if you had an exothermic reaction (ΔH_system −) in which the entropy of the chemical system actually decreases (ΔS_system −).
The temperature of the system will still be positive as long as the tempertaure is expressed in kelvin.
So, let's try to arrive at a sign for ΔG_system the same way we did above:

IF the value of TΔS_system is smaller than the absolute value, or the magnitude, of ΔH_system (|ΔH_system|), then ΔG_system will be negative (ΔG_system < 0) and the reaction will be spontaneous.
BUT, if the value of of TΔS_system is larger than the absolute value, or the magnitude, of ΔH_system (|ΔH_system|), then ΔG_system will be positive (ΔG_system > 0) and the reaction will be nonspontaneous.
AND if the value of TΔS_system is the same as the absolute value, or the magnitude, of ΔH_system (|ΔH_system|), then ΔG_system will be zero (ΔG_system = 0) and the reaction will be at equilibrium.

The value of the TΔS_system term in the Gibbs free energy equation is critical in determining whether this system will be spontaneous or not.
Since we already said that ΔS_system is negative, that is, the entropy of the system decreases, the factor that must be deciding how large or small the value of the TΔS_system term is .... the temperature, T, in kelvin!

For a chemical system in which ΔH_system is negative and ΔS_system is also negative, then, if the temperature is

• very high then TΔS_system is very large, and ΔG will become positive, so reaction is nonsponanteous
• very low then TΔS_system is very small, and ΔG will become negative, so reaction is sponanteous

Consider now a chemical system which is endothermic (ΔH_system > 0) and in which there is a decrease in the entropy of the system (ΔS_system < 0) at constant temperature and pressure.
Let's see if we can arrive at the sign of ΔG_system and hence decide if such a reaction is spontaneous or not:

For a chemical system in which ΔH_system is positive and ΔS_system is negative, ΔG_system will ALWAYS be positive and therefore will ALWAYS be nonspontaneous.

But what if, for an endothermic reaction (ΔH_system > 0), the entropy change for the reaction is positive (ΔS_system > 0). Will this reaction be spontaneous or not?

Once again, whether nor not this reaction is spontaneous depends on the size of the TΔS_reaction term.

For a chemical reaction in which ΔH_system is positive and ΔS_system is negative, then, if the temperature is

• very high then TΔS_system is very large, and ΔG will become negative, so reaction is sponanteous
• very low then TΔS_system is very small, and ΔG will become positive, so reaction is nonsponanteous

We can summarise all this in one table as shown below:

We can use this information to help us predict the effect of a temperature change on the spontaneity of a chemical reaction or a physical change (at constant pressure).

Calculating the Temperature at Which a Chemical Reaction is Spontaneous

In the section above we found that if the sign of the enthalpy change and entropy change of a chemical reaction or physical change are the same, that is both positive or both negative, then whether the reaction is spontaneous or not depends on the temperature of this chemical system.
In other words, the sign of ΔG_reaction changes with changes in temperature if the sign of ΔH is the same as the sign of ΔS:

• ΔH and ΔS both negative (ΔH < 0 and ΔS < 0) then the reaction is

  (a) spontaneous at low temperature

  (b) nonspontaneous at high temperature

• ΔH and ΔS both positive (ΔH > 0 and ΔS > 0) then the reaction is

  (a) nonspontaneous at low temperature

  (b) spontaneous at high temperature

But what do we mean by "high temperature" and "low temperature"?

We are really refering to temperatures above the temperature at which the system is at equilibrium as "high temperature" and temperatures below the temperature at which the system is at equilibrium as "low temperature".
So we need to be able to calculate the temperature at which the system will be at equilibrium.

First we recall the relationship between the change in Gibbs free energy (ΔG_reaction), enthalpy change (ΔH_reaction), entropy change (ΔS_reaction) and temperature of the system in kelvin (T):

ΔG_reaction = ΔH_reaction - TΔS_reaction

and the condition for the chemical reaction or physical change to be at equilibrium, that is:

ΔG_reaction = 0

By combining these two expressions we arrive at an expression describing the system at equilibrium (at constant pressure):

ΔH_reaction - T_eqΔS_reaction = 0

Where T_eq is the temperature at which the system is at equilibrium (at constant pressure).

We can now re-arrange this relationship as follows:

So high temperature means a temperature (T) greater than T_eq, and, low temperature means a temperature (T) less than T_eq:

• high temperature: T > T_eq
• low temperature: T < T_eq

Let's consider a physical change we are familiar with: solid water (H₂O_(s)) melting to form liquid water (H₂O_(l)) according to the balanced chemical equation given below:

H₂O_(s) → H₂O_(l)

For this physical change, ΔH = 6.008 kJ mol^-1 and ΔS = 21.99 J K^-1 mol^-1.
We note that ΔH and ΔS are both positive, so, using the table in the previous section, we predict that this reaction will be spontaneous at high temperature (T > T_eq) and nonspontaneous at low temperature (T < T_eq).
Now, we can calculate the temperature (T_eq) at which this chemical system is at equilibrium:

ΔH_reaction = 6.008 kJ mol^-1

ΔS_reaction = 21.99 J K^-1 mol^-1 = 21.99 ÷ 1000 = 0.02199 kJ K^-1 mol^-1

Substituting these values into the expression for temperature (T_eq):

This system is at equilibrium at a temperature of 273.2 K (and constant pressure).

This physical change will be spontaneous at temperatures (T) greater than T_eq, that is, when T > 273.2 K (and constant pressure).
In other words, at temperatures above 273.2 K solid water will melt spontaneously to form liquid water (at constant pressure).
Ofcourse, we can convert this to a temperature in °C by subtracting 273.2 ⁽⁶⁾

T_eq = 273.2 K - 273.2 = 0°C

and we find that solid water, ice, melts to liquid water at temperatures above 0°C which is confirmed by our daily experience at atmospheric pressure.

Now let's consider the reverse of this process, liquid water (H₂O_(l)) freezing to produce solid water (H₂O_(s)) according to the following balanced chemical reaction:

H₂O_(l) → H₂O_(s)

For this physical change ΔH = −6.008 kJ mol^-1 and ΔS = −21.99 J K^-1 mol^-1 = −0.02199 kJ K^-1 mol^-1
Note that ΔH and ΔS are both negative, so whether or not this change is spontaneous or not is dependent on temperature; spontaneous at low temperature (T < T_eq) and nonspontaneous at high temperature (T > T_eq).
Let's calculate the temperature at which this system is at equilibrium:

This physical change is therefore spontaneous at low tempertures, temperatures less than 273.2 K (T < 273.2 K or T < 0°C)
Once again this is confirmed by our everyday experience of needing to put liquid water in the freezer in order to freeze it and turn it into iceblocks!

Worked Example : Using Gibbs Free Energy to Determine the Temperature at which a Reaction is Spontaneous

Question:

The commercial production of ammonia is extremely important because ammonia is used to produce fertiliser and without these fertilisers the world's farm lands would quickly become depleted in nitrogen and unable to sustain the crops that feed us.
Ammonia gas (NH_3(g)) can be produced from nitrogen gas (N_2(g)) and hydrogen gas (H_2(g)) according to the following balanced chemical equation:

N_2(g) + 3H_2(g) → 2NH_3(g)

For this chemical reaction at 1 atmosphere pressure (101.3 kPa), ΔH = -92.2 kJ mol^-1 and ΔS = -198.7 J K^-1 mol^-1.
At what temperatures in °C will this reaction be spontaneous at 1 atmosphere pressure?

Solution:

(Based on the StoPGoPS approach to problem solving.)

Footnotes