Gibbs Free Energy Problems

Gibbs free energy is a thermodynamic potential that is used to calculate the amount of work done by a system at constant temperature and pressure. This maximum amount of work can be done only in a completely reversible process. A change in this energy is the difference between the heat released during a process and the heat released for the same process. This Gibbs free energy change becomes zero when equilibrium is attained. When a spontaneous reaction occurs, this energy becomes negative and when they are non-spontaneous, this energy is positive. 

Relationship between Gibbs Free Energy and Equilibrium Constant

The rate of forward reactions and backward reactions become equal at equilibrium. The driving force in each direction is equal because the free energy of reactants and products in equilibrium is equivalent (∆G=0). 

At equilibrium, the reaction quotient is Q= Keq. 

So from both of these equations, a reaction that has reached equilibrium ∆G is 

∆G= ∆G° + RTlnQ

0= ∆G° + RTlnQ

∆G° = -RTlnQ

This equation can be used to find ∆G from the equilibrium constant and vice versa. Knowing either ∆G or Keq helps us to determine whether the reactants and products are favoured at equilibrium. 

Gibbs free energy problems questions

You can solve Gibbs free energy problems if the value of the equilibrium constant is given and vice versa. 

Question: 

Consider that the standard energy of formation (∆Gf) for ammonia is -16.6 kJ/mol. Determine the value of the equilibrium constant at 25 °C (298 K) for the following reaction.  

N2(g) + 3H2(g) ⇄ 2NH3(g)

 Solution:  

We know the value of ∆Gf , so we can calculate the value of ∆G from the given equation 

ΔG°rxn = ΣnΔG°f (products) – ΣnΔG°f(reactants) 

And because both nitrogen and hydrogen are in their elemental states, the ∆Gf is zero for both of them. 

 ΔG°rxn = ΣnΔG°f(products) – ΣnΔG°f(reactants) 

ΔG°rxn = 2(-16.6 kJ/mol) – [(0 kJ/mol) + 3(0 kJ/mol)]

ΔG°rxn = -33.2 kJ/mol

After Solving for ∆G, solve for Keq ΔG° = – RT ln Keq 

ΔG∘=−RT ln Keq

ln Keq=−ΔG∘RT

Keq= e−ΔG∘RT

Before substituting values into the equation, remember that we are using R= 8.314 J/K mol. 

So we have to convert ∆G= -33.2 kJ/ mol into ∆G= -33,200 J/mol. 

Then to calculate Keq 

Keq=e−ΔG∘RT

Keq=e−33,200 J/mol (8.314 J/K⋅mol)(298 K)

Keq=e13.4

Keq= 6.6×105

Keq is very large, indicating that the products are heavily favoured at 25°C. 

Question: 

The equilibrium constant for the given reaction is 16 X 10-12 at 25°C. 

AgCl(s) ⇄ Ag+(aq) + Cl-(aq)

Calculate ∆G° 

Solution: 

Substitute the given values in the equation of ∆G°

ΔG∘=−RT ln Keq

ΔG∘=−(8.314 J/K⋅mol)(298 K) ln (1.6×10−10)

ΔG∘=−55,884 J/mol

Since ∆G° is big, it favours the formation of reactants over products. 

Gibbs Free Energy And Emf of a cell

When we pass a charge through a galvanic cell in a reverse pathway, a maximum amount of work is done. The amount of reversible work done by the galvanic cell is responsible for the decrease in Gibbs free energy in the reaction. 

So work done is 

W = nFE (cell)

Here,

W represents work done

nF represents total charge passed and;

E(cell) = emf of the cell

Δr G = −W

Δr G = −nFE(cell)

This equation is used to calculate standard cell potential. 

Δr G° = −nFE°(cell)

This reaction represents Gibbs free energy of a reaction is known to be Extensive Thermodynamic Property which indicates that value depends on n. 

Gibbs free energy problems questions for the relationship between Gibbs free energy and emf

You can find the meaning of Gibbs free energy problems by solving these Gibbs free energy problems. 

Question:

Determine the emf and ∆G of the cell reaction at 298 K for the given cell. 

Mg(s) ∣ Mg2+ (0.01M) || Ag+(0.0001M) | Ag(s) 

Given : [E°Mg2+ /Mg =−2.37V,

E°Ag+/Ag  =+0.80V]

Solution:

At anode:       Mg →  Mg2+ +2e 

At Cathode:.   [ Ag+ +e →Ag]×2 

____________________________

       Mg+ 2Ag+→Mg2++2Ag; n=2 

E cell =E cathode −Eanode

 =E Ag+ /Ag− E Mg 2+ /Mg

=0.80V−(−2.37V)=3.17V

Substitute above value in Nernst Equation

Ecell= E°cell – 0.059/ n log Mg2+/ [Ag+]2

We get,

Ecell= 3.17- 0.059/2 log 10-2/ (10-4)2

Ecell= 3.17 – 0.0295 log 106

Ecell= 3.17- 0.117 V = 2.993 V

Now substitute the value of Ecell in the below equation

∆G= -nFEcell

      = 2×96500C mol−1 ×2.993V 

ΔG=−577649 Jmol−1

 =−577.649 kJmol−1

Conclusion

Gibbs free energy is used to measure the maximum work done by a thermodynamic system if the temperature and pressure of the system are constant. This maximum amount of work can be done only if we follow a reversible process. The SI unit of Gibbs free energy is Joules. We have mentioned some Gibbs free energy problems in this equation and their solutions.