Please do not block ads on this website.
No ads = no money for us = no free stuff for you!
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:
Spontaneity of Reaction at Constant Temperature and Pressure 
ΔG_{reaction} 
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 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
Gibbs Free Energy equation: 
ΔG_{system} 
= 
ΔH_{system} 
− 
(T 
× 
ΔS_{system}) 
Sign of each term : 
ΔG_{system} 
= 
− 
− 
(+ 
× 
+) 


= 
− 
− 
(+) 


Result : 
ΔG_{system} 
= 
− 




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:
Gibbs Free Energy equation: 
ΔG_{system} 
= 
ΔH_{system} 
− 
(T 
× 
ΔS_{system}) 
Sign of each term : 
ΔG_{system} 
= 
− 
− 
(+ 
× 
−) 


= 
− 
− 
(−) 


Result : 
ΔG_{system} 
= 
− 
+ 
? 


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:
Gibbs Free Energy equation: 
ΔG_{system} 
= 
ΔH_{system} 
− 
(T 
× 
ΔS_{system}) 
Sign of each term : 
ΔG_{system} 
= 
+ 
− 
(+ 
× 
−) 


= 
+ 
− 
(−) 


Result : 
ΔG_{system} 
= 
+ 
+ 



Result : 
ΔG_{system} 
= 
+ 




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?
Gibbs Free Energy equation: 
ΔG_{system} 
= 
ΔH_{system} 
− 
(T 
× 
ΔS_{system}) 
Sign of each term : 
ΔG_{system} 
= 
+ 
− 
(+ 
× 
+) 
Result : 
ΔG_{system} 
= 
+ 
− 
(+) 


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:

ΔH_{system} 
ΔS_{system} 
ΔG_{system} (=ΔH  TΔS) 
Change 
ΔG sign 
conditions 
Opposite sign 
− 
+ 
− 
any temperature 
Spontaneous 
+ 
− 
+ 
any temperature 
Nonspontaneous 
Same sign 
− 
− 
− 
low temperature (T < T_{eq}) 
Spontaneous 
+ 
high temperature (T > T_{eq}) 
Nonspontaneous 
+ 
+ 
+ 
low temperature (T < T_{eq}) 
Nonspontaneous 
− 
high temperature (T > T_{eq}) 
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).
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_{eq}ΔS_{reaction} = 0
Where T_{eq} is the temperature at which the system is at equilibrium (at constant pressure).
We can now rearrange this relationship as follows:
Add T_{eq}ΔS_{reaction} to both sides of the equation 
ΔH_{reaction}  T_{eq}ΔS_{reaction} + T_{eq}ΔS_{reaction} 
= 
0 + T_{eq}ΔS_{reaction} 
ΔH_{reaction} 
= 
T_{eq}ΔS_{reaction} 
Divide both sides of the equation by ΔS_{reaction} 
ΔH_{reaction} ΔS_{reaction} 
= 
T_{eq}ΔS_{reaction}
ΔS_{reaction} 
ΔH_{reaction} ΔS_{reaction} 
= 
T_{eq} 
Which we can also write as 
T_{eq} 
= 
ΔH_{reaction} ΔS_{reaction} 
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 (H_{2}O_{(s)}) melting to form liquid water (H_{2}O_{(l)}) according to the balanced chemical equation given below:
H_{2}O_{(s)} → H_{2}O_{(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:
T_{eq} 
= 
ΔH_{reaction} ΔS_{reaction} 
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}):
T_{eq} 
= 
ΔH_{reaction} ΔS_{reaction} 
T_{eq} 
= 
6.008 kJ mol^{1} 0.02199 kJ K^{1} mol^{1} 
T_{eq} 
= 
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 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 (H_{2}O_{(l)}) freezing to produce solid water (H_{2}O_{(s)}) according to the following balanced chemical reaction:
H_{2}O_{(l)} → H_{2}O_{(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:
T_{eq} 
= 
ΔH_{reaction} ΔS_{reaction} 
T_{eq} 
= 
6.008 kJ mol^{1} 0.02199 kJ K^{1} mol^{1} 
T_{eq} 
= 
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!
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!