16.2 Free Energy

In the last section, we looked at ∆Suniverse and how it can help us predict the spontaneity of the reaction. This section will explore one more thermodynamic function, free energy, shown with the letter G that can also predict the spontaneity of the reaction. Free energy is also known as Gibbs Free Energy in honor of Josiah Willard Gibbs, a mathematical physics professor at the University of Yale who worked in many thermodynamics’ areas.
H(enthalpy), T(temperature), and S (entropy) are related to the Gibbs free energy (G) as per the formula below.
G = H – TS
In this equation, all three parameters G, H, and S are for the system.
For a reaction taking place at constant temperature T, change in Free energy can be expressed as follows.
∆G = ∆H – T∆S
∆G is the change in free energy, ∆H is the change in enthalpy and ∆S is the change in entropy of the system. T is the temperature in Kelvin. We can use this equation to predict the spontaneity of the reaction.
If we divide both sides of this equation with temperature T, we will get
∆G/T = ∆Hsystem/T – ∆Ssystem
In the last unit of this chapter, we established that -∆Hsystem/T = ∆Ssurrounding.
So, if we substitute -∆Hsystem/T with ∆Ssurrounding in the equation above, we will get
∆G/T = – ∆Ssurrounding – ∆S
Or – ∆G/T = ∆Ssystem + ∆Ssurrounding
In other words, we can say

• ∆G/T = ∆Suniverse (If temperature and pressure are kept constant)
  Recall from the last unit, if ∆Suniverse is positive, the reaction will be spontaneous.
  Based on – ∆G/T = ∆Suniverse equation, we can conclude, that if ∆G is negative for a reaction at constant temperature and pressure, we will have a positive value for ∆Suniverse and the reaction will be spontaneous.
  Since both enthalpy change, and entropy change for the system are thermodynamic quantities and can be calculated easily with the help of a thermodynamic data table; we often use this formula to predict the spontaneity of the reaction. If ∆G data for reactants and products is provided, you can also use that to get the ∆G reaction and predict the reaction’s spontaneity by using the formula below.
  ∆G reaction = ∑∆G products – ∑∆G reactants
  You can use this link below for the thermodynamic data table.
  https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic%20Values.pdf

Learning Check:
Question 4. Calculate ∆G for the following reaction and predict if the process will be spontaneous at 25 0C.
N2(g) + 3H2(g) 2NH3(g)
∆GN₂ 0.00 kJ/mol
∆G H₂ 0.00 kJ/mol
∆GNH₃ -16.48 kJ/mol

Since ∆G reaction = ∑∆G products – ∑∆G reactants
∆G reaction = [2(∆G (NH3)] – [∆G (N2) + 3 ∆G(H2)]
∆G reaction = [2(-16.48) ] – [1(0.00) + 3(0.00)]
∆G reaction = -32.96 kj
A negative value of ∆G shows that reaction will be spontaneous.

Question 5:
For the reaction given below, calculate ∆H and ∆S using the thermodynamic data table and then calculate ∆G and predict if the reaction will be spontaneous at 25 0C.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
∆H0 (kJ /mol) ∆S0 (J/mol K)
CH4 (g) -74.6 186.3
O2 (g) 0 205.2
CO2 (g) -393.51 213.8
H2O (l) -285.83 70.0

∆Hsystem = ∑∆H(products) – ∑∆H(reactants)
= ∑[∆H (CO2) + 2 ∆H(H2O)] – ∑[∆H (CH4) + 2 ∆H(O2)]
= [ -393.51 + (2 x -285.83)] – [-74.6 + (2 x 0)]
∆Hsystem = -965.26 + 74.6 = – 890.66 kJ
∆Ssystem = ∑∆S(products) – ∑∆S(reactants)
= ∑[∆S (CO2) + 2 ∆S(H2O)] – ∑[∆S (CH4) + 2 ∆S(O2)]
= [213.8 + 2(70)] – [186.3 + (2 x (205.2)]
∆Ssystem = -242.9 J = – 0.2429 kJ
∆G = ∆H – T∆S; T = 25 0C = (25 + 273.15) K = 298.15 K
∆G = – 890.66 – 298.15 x (-0.2429) = – -818.2 kJ
Since ∆G is negative, the reaction will be spontaneous at 25 0C.
Watch this YouTube video for another good example of this concept.

https://www.youtube.com/watch?v=XvuRJuXykyw

Learning Check:
Question 6: Calculate the value of ∆H and ∆S from the data given and based on that calculate ∆G for the reaction given below at 25 0C.
Question 7: Will this reaction be spontaneous at 25 0C?
C2H5OH (l) + 3O2 (g)2CO2 (g) + 3H2O (l)
∆H0 (kJ /mol) ∆S0 (J/mol K)
C2H5OH (l) -278 161
O2 (g) 0 205.2
CO2 (g) -393.51 213.8
H2O (l) -285.83 70.0

Answer: ∆H0 = -1367kJ; ∆S0 = -139 J/K; ∆G0 = -1326 kJ
Yes, it will be spontaneous because ∆G0 is negative.

We know ∆G = ∆H – T∆S,
Since, T is the temperature in Kelvin, it will never be negative. Keeping that in mind, we can now summarize the significance of the sign of ∆H and ∆S, on the spontaneity of the reaction.
∆H ∆S ∆G

Dependency of ∆G on ∆H, ∆S and temperature can be summarized in the graph below.

Source: www.Openstax.org
Watch the video below on the same concept.

https://www.youtube.com/watch?v=8N1BxHsoOw&t=7s

If we apply this concept to the melting of ice or freezing of water , we can explain why increasing or decreasing the temperature is making the reaction spontaneous in one direction or the other.
H2O (s) H2O (l)
Let’s have a look at the melting of ice first. In melting of ice, solid phase is changing to liquid. This process will have a positive value of ∆S. We know that when ice melts, it absorbs energy, or we can say process will be endothermic, and ∆H will be positive for this reaction. Combining these two factors, now we can say a positive ∆S and a positive ∆H will make the reaction spontaneous at high temperature.
Now let’s see how freezing of water is different from melting of ice.
H2O (l) H2O (s)
In freezing of water, liquid water is changing to solid ice, so ∆S will be negative. Also, this process will be exothermic, as water releases energy when it freezes. So ∆H will be negative. As a result, a negative value of ∆S and negative value of ∆H will make the reaction spontaneous at low temperature. These results are consistent with our observation, water freezes spontaneously to solid ice at temperature lower than 0 0C, and ice melts spontaneously at a temperature greater than 0 0C,
This concept only helps us predict the spontaneity of the reaction; it does not talk about the rate of reaction. In some cases, we see that based on thermodynamic data, the reaction should occur spontaneously, but the reaction rate is so slow that we may never observe the reaction happening. Some reactions are thermodynamically favored but kinetically not favored to occur spontaneously.
The conversion of Diamond to graphite is an excellent example of such type of reaction.
C(Diamond) C(Graphite)
∆Gsystem = ∑∆G(Graphite) – ∑∆G(Diamond)
∆G0 (kJ /mol)
C(Diamond) 2.90
C(Graphite) 0

Based on the thermodynamic data table value
∆Gsystem = 0 – 2.90 = -2.90 Kj
Since ∆Gsystem is negative, the conversion of Diamond to Graphite should be spontaneous at room temperature. However, we do not see a diamond changing to graphite on its own because the reaction is extremely slow. You can say this reaction is thermodynamically favorable but kinetically not favorable. Hence the saying “diamonds are forever” is still true.
Based on this concept, we now know that a reaction is spontaneous if ∆Gsystem is negative and non-spontaneous if ∆Gsystem is positive. Extending this concept further, we can say, when ∆Gsystem = 0 the reaction will be at equilibrium.
Using the relation ∆G = ∆H – T∆S , this concept can also help us find the temperature at which a reaction will reach equilibrium. This formula can also be used to calculate the equilibrium constant for a reaction.
Question 8: Using the thermodynamic data table, find the boiling point of water based on the reaction below.
H2O (l) H2O (g),
∆H (kj/mol) ∆S (J/mol.K)
H2O (l) – 285.83 70.0
H2O (g), – 241.82 188.8

At boiling point, the liquid and gas phases are in equilibrium.
Hence ∆Gsystem = 0
Since ∆G = ∆H – T∆S
Substituting the value of ∆Gsystem = 0, we can say
∆H – T∆S = 0
Or ∆H = T∆S
We can use a thermodynamic data table to calculate the value of ∆Hsystem and ∆Ssystem
∆Hsystem = ∑∆HH2O(g) – ∑∆ HH2O(l)
= – 241.82 – – 285.83 = 44.01 kJ
∆Hsystem = 44.01 kJ
∆Ssystem = ∑∆S(products) – ∑∆S(reactants)
∆Ssystem = ∑∆SH2O(g) – ∑∆ S H2O(l)
∆Ssystem = 188.8 – 70.0 = 118.8 J/K
∆Ssystem = 0.1188 kJ/K
Since ∆H = T∆S
44.01 = T (0.1188)
T = 44.01 kj / (0.1188 kj/K ) = 370.5 K
Boiling point = 370.5 K
= (370.5 – 273.15) 0C = 97.3 0C

Learning Check:
Question 9: Based on thermodynamic data, find the equilibrium constant for the following reaction under standard conditions.
N2 (g) + 3H2(g) 2NH3(g)
∆G0 KJ/mol
N2 (g) 0
H2(g) 0
NH3(g) -16.5

Solution:
∆G0reaction = ∑∆ G0 (products) – ∑∆ G0 (reactants)
= [2∆G0 (NH3)] – [∆G0 (N2) + 3∆G0 (H2)]
[ 2 (-16.5)] – [0 + 3(0)] = -33 Kj
= -33000 J
∆G0 = – RT ln K
Under standard conditions, T = 25 0C = 25 + 273.15 = 298.15 K
-33000 = -8.314 x 298.15 x ln K
ln K = -33000 / (-8.314 x 298.15)
ln K = 13.3
K = e^13.3
K = 6.05 x 105