 # 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 ΔGreaction < 0 (ΔG is negative)

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

A reaction is at equilibrium if ΔGreaction = 0

• The value of ΔGreaction is dependent on temperature.(1)
• The sign of ΔGreaction 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 ΔGreaction 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, Teq, can be calculated as follows:

Teq = ΔH ÷ ΔS

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

high temperature: T > Teq

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

low temperature: T < Teq

• 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 − + − any temperature Spontaneous + − + any temperature Nonspontaneous − − − 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 (ΔGreaction) for a chemical reaction or a physical change at constant temperature and pressure using the equation given below:

ΔGreaction = ΔHreaction - TΔSreaction

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

T = temperature of the system in kelvin (K)

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

Also recall that ΔGreaction 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:

Spontaneity of Reaction at Constant Temperature and Pressure
ΔGreaction Change
ΔG < 0 Spontaneous
ΔG = 0 Equilibrium
ΔG > 0 Nonspontaneous

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

• Spontaneous if ΔGreaction < 0, that is, if ΔHreaction - TΔSreaction < 0
• At equilibrium if ΔGreaction = 0, that is, if ΔHreaction - TΔSreaction = 0
• Nonspontaneous if ΔGreaction > 0, that is, if ΔHreaction - TΔSreaction > 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, ΔHreaction < 0
• the entropy of the chemical system increases, that is, ΔSreaction > 0

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

 Gibbs Free Energy equation: Sign of each term :   Result : ΔGsystem = ΔHsystem − (T × ΔSsystem) ΔGsystem = − − (+ × +) = − − (+) ΔGsystem = −

If you subtract a positive number from a negative number the result will always be negative, so, if ΔHsystem is negative and ΔSsystem is positive then ΔGsystem is always negative and the reaction is always sponanteous.

If ΔHreaction < 0 and ΔSreaction > 0 the reaction is spontaneous at any temperature.

Now, let's think about what would happen if you had an exothermic reaction (ΔHsystem −) in which the entropy of the chemical system actually decreases (ΔSsystem −).
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 ΔGsystem the same way we did above:

 Gibbs Free Energy equation: Sign of each term :   Result : ΔGsystem = ΔHsystem − (T × ΔSsystem) ΔGsystem = − − (+ × −) = − − (−) ΔGsystem = − + ?

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

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

For a chemical system in which ΔHsystem is negative and ΔSsystem is also negative, then, if the temperature is

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

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

 Gibbs Free Energy equation: Sign of each term :   Result : Result : ΔGsystem = ΔHsystem − (T × ΔSsystem) ΔGsystem = + − (+ × −) = + − (−) ΔGsystem = + + ΔGsystem = +

For a chemical system in which ΔHsystem is positive and ΔSsystem is negative, ΔGsystem will ALWAYS be positive and therefore will ALWAYS be nonspontaneous.

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

 Gibbs Free Energy equation: Sign of each term : Result : ΔGsystem = ΔHsystem − (T × ΔSsystem) ΔGsystem = + − (+ × +) ΔGsystem = + − (+)

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

For a chemical reaction in which ΔHsystem is positive and ΔSsystem is negative, then, if the temperature is

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

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

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

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).

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## 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 ΔGreaction 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 (ΔGreaction), enthalpy change (ΔHreaction), entropy change (ΔSreaction) and temperature of the system in kelvin (T):

ΔGreaction = ΔHreaction - TΔSreaction

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

ΔGreaction = 0

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

ΔHreaction - TeqΔSreaction = 0

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

We can now re-arrange this relationship as follows:

 Add TeqΔSreaction to both sides of the equation ΔHreaction - TeqΔSreaction + TeqΔSreaction = 0 + TeqΔSreaction ΔHreaction = TeqΔSreaction Divide both sides of the equation by ΔSreaction ΔHreactionΔSreaction = TeqΔSreactionΔSreaction ΔHreactionΔSreaction = Teq Which we can also write as Teq = ΔHreactionΔSreaction

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

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

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 > Teq) and nonspontaneous at low temperature (T < Teq).
Now, we can calculate the temperature (Teq) at which this chemical system is at equilibrium:

 Teq = ΔHreactionΔSreaction

ΔHreaction = 6.008 kJ mol-1

ΔSreaction = 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 (Teq):

 Teq = ΔHreactionΔSreaction Teq = 6.008 kJ mol-1 0.02199 kJ K-1 mol-1 Teq = 273.2 K

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 Teq, 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)

Teq = 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 < Teq) and nonspontaneous at high temperature (T > Teq).
Let's calculate the temperature at which this system is at equilibrium:

 Teq = ΔHreactionΔSreaction Teq = 6.008 kJ mol-1 0.02199 kJ K-1 mol-1 Teq = 273.2 K

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!

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## 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 (NH3(g)) can be produced from nitrogen gas (N2(g)) and hydrogen gas (H2(g)) according to the following balanced chemical equation:

N2(g) + 3H2(g) → 2NH3(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.)

STOP STOP! State the Question.
What is the question asking you to do?
Calculate the temperature in °C at which the reaction is spontaneous.

T = ? °C

PAUSE PAUSE to Prepare a Game Plan
(1) What information (data) have you been given in the question?

Balanced chemical equation: N2(g) + 3H2(g) → 2NH3(g)

Conditions: constant pressure of 1 atmosphere (101.3 kPa)

ΔH = -92.2 kJ mol-1

ΔS = -198.7 J K-1 mol-1

(2) What is the relationship between what you know and what you need to find out?

(a) Equilibrium Temperature in kelvin: Teq = ΔH ÷ ΔS

(b) Convert temperature in kelvin to temperature in °C:

T(°C) = T(K) - 273.2

(c) Refer to the table below:

ΔHsystem ΔSsystem ΔGsystem
(=ΔH - TΔS)
Change
+ Spontaneous
+ + Nonspontaneous
− (T < Teq) Spontaneous
+ (T > Teq) Nonspontaneous
+ + + (T < Teq) Nonspontaneous
− (T > Teq) Spontaneous

GO GO with the Game Plan

(a) Equilibrium Temperature in kelvin: Teq = ΔH ÷ ΔS

ΔH = −92.2 kJ mol-1

ΔS = −198.7 J K-1 mol-1 = −198.7 J K-1 mol-1 ÷ 1000 J kJ-1 = −0.1987 kJ K-1 mol-1

Teq = ΔH ÷ ΔS = −92.2 kJ mol-1 ÷ −0.1987 kJ K-1 mol-1 = 464.0 K

(b) Convert temperature in kelvin to temperature in °C:

T(°C) = T(K) - 273.2 = 464.0 - 273.2 = 190.8° C

(c) Refer to the table below (ΔH < 0 and ΔS < 0):

ΔHsystem ΔSsystem ΔGsystem
(=ΔH - TΔS)
Change
+ Spontaneous
+ + Nonspontaneous

(−92.2)

(−0.1987)

(T < Teq)
(T < 190.8°C)
Spontaneous
+
(T > Teq)
(T > 190.8°C )
Nonspontaneous
+ + + (T < Teq) Nonspontaneous
− (T > Teq) Spontaneous

This chemical reaction will be spontaneous at temperatures less than 190.8°C

PAUSE PAUSE to Ponder Plausibility

Yes, we have calculated the temperature in °C below which the reaction will be spontaneous at 1 atmosphere pressure.

The reaction is exothermic: N2(g) + 3H2(g) → 2NH3(g) + 92.2 kJ mol-1
Consider the system at equilibrium, if you perturb (or disturb) that equilibrium by raising the temperature then by le Chatelier's Principle the system will react to minimise the effect of the change, that is, use some of the heat to drive the reaction in the reverse direction consuming ammonia and producing more nitrogen gas and hydrogen gas. That is, "high" temperatures will favour the reverse reaction. The forward reaction will be favoured at "lower" tempertaures.
This seems to support the idea that the reaction would be spontaneous at a lower temperature than the temperature at which the system was at equilibrium.

Use ΔH = -92.2 kJ mol-1, ΔS = -0.1987 kJ K-1 mol-1, and a temperature a bit less than 190.8°C, say 190°C (190 + 273.2 = 463.2 K) to calculate ΔG and make sure it is negative:
ΔG = -92.2 - (463.2 × -0.1987) = -0.16 (ΔG < 0 therefore spontaneous).

Similarly, use a temperature slightly more than 190.8°C (say 191°C = 191+ 273.2 = 464.2 K) to calculate ΔG and make sure it is now positive:
ΔG = -92.2 - (464.2 × -0.1987) = 0.037 (ΔG > 0 therefore nonspontaneous).

We are therefore reasonably confident that our answer is plausible. (7)

STOP STOP! State the Solution
This chemical reaction will be spontaneous at temperatures less than 190.8°C (at constant 1 atmosphere pressure)

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Footnotes

(1) The value of G is also dependent on pressure but the following discussion will assume constant atmospheric pressure.

(2) Kelvin (K) is the S.I. unit of temperature. Non S.I. units of temperature include °C and °F. You may need to learn how to convert these are other units of temperature to kelvin (K).

(3) The enthalpy change for a reaction is also referred to as the heat of reaction.

(4) Remember that the sign of ΔG tells us whether the reaction is spontaneous or not but it DOES NOT tell us whether the reaction is fast or slow. Some spontaneous reactions occur very rapidly, some occur very, very slowly. The sign of ΔG DOES NOT tell us anything about the reaction rate.

(5) The third law of thermodynamics can be generalised as saying that it is impossible to reach absolute zero, 0 K, in a finite number of steps. It is therefore impossible for a temperature to be below 0 K, that is, negative temperatures in units of kelvin are impossible.
Therefore the temperature of any chemical system must be greater than 0 K (temperature in kelvin is always positive)
Note that 0 K = −273.15°C
So it is very important that we express temperatures in kelvin (K)!

(6) We usually give the value of absolute zero (0 K) as -273.15 °C, however, in these calculations we were only justified in using 4 significant figures, so when we do the subtraction to convert K to °C we are only justified in 1 decimal place. Therefore, we round 273.15 up to 273.2 and subtract 273.2 from the temperature in kelvin.

(7) One of the reasons that this chemical reaction was so difficult to commercialise is that it occurs spontaneously but slowly at room temperature and pressures, and, the yield of ammonia is very, very, small, that is, the equilibrium position lies well to the left!