How to calculate Keq from delta G (∆G)? – (Keq from Delta G)

The symbol K_eq denotes the equilibrium constant. K_eq helps in understanding the behavior of a chemical reaction at equilibrium.

∆G represents the Gibbs free energy change for a reaction. K_eq is related to ∆G (Delta G) by the formula ∆G = -RTln K_eq.

In this article, we will teach you how to find K_eq if the standard Gibbs free energy change (∆G°) of a reaction is given using ∆G = -RTln K_eq.

So, let’s start!

What is K_eq?

In chemistry, a reversible reaction is one that takes place in both forward as well as backward directions simultaneously.

aA + bB ⇌ cC + dD

Dynamic equilibrium is a state of balance achieved when the rate of forward reaction becomes equal to the rate of backward reaction.

K_eq helps in predicting quantitatively to what extent the reaction proceeds in the forward or reverse direction.

• If K_eq > 1: Equilibrium shifts forward
• If K_eq < 1: Equilibrium shifts backward
• If K_eq = 1: Dynamic equilibrium is maintained

What is ∆G (Delta G)?

Gibbs free energy change (∆G) refers to reversible work done on a system at constant temperature and pressure. In other words, ∆G quantifies the energy that is converted to useful work during a process.

It helps in determining the spontaneity of a chemical reaction and the direction in which the reaction progresses.

For the reversible reaction, aA + bB ⇌ cC + dD

• If ∆G < 0: The reaction is thermodynamically favored, thus spontaneous in the forward direction.
• If ∆G > 0: The reaction is non-spontaneous in the forward direction. Some sort of external interference, i.e., energy, is required to propel it forward.
• If ∆G = 0: The reaction stays at equilibrium, where the rate of forward reaction = rate of backward reaction.

How to find K_eq from ∆G (Delta G)?

K_eq is related to ∆G° by equation (i):

∆G° = -RT ln K_eq….Equation (i)

Where,

• ∆G° = Standard Gibbs free energy change for a chemical reaction (Units: J/mol or kJ/mol)
• R = Ideal gas constant (R= 8.314 J/K.mol)
• T= Absolute temperature (Unit: K)
• ln K_eq = Natural logarithm of the equilibrium constant (K) for the reaction (Unitless)

Under non-standard conditions, making K_eq the subject of the formula from equation (i) gives us:

ln K_eq = -∆G/RT

K_eq = e^–∆G/RT….Equation (ii)

Similarly, converting natural log (ln) to common log, equation (i) can be rewritten as:

∆G = – 2.303 RT log K_eq……Equation (iii)

Making K_eq the subject of the formula from equation (iii):

Log K_eq = -∆G/2.303RT

Taking antilog:

K_eq = 10^{–∆G/2.303RT}…equation (iv)

Thus, at a certain temperature, T, the K_eq of a reaction can be calculated from its ∆G (delta G) using any of the two equations, i.e., equations (ii) and (iv).  

Let’s see how to find K_eq from ∆G through the examples given below.

Solved examples for finding K_eq from ∆G? (K_eq from Delta G)

Example # 1: The Gibbs free energy change (∆G) for a reaction taking place at 300 K is -15.2 J/mol. Find the equilibrium constant (Keq) for the reaction.

Solution: As per the question statement: ∆G (Delta G) = -15.2 J/mol T= 300 K Keq =? We already know that R = 8.314 J/K.mol Applying the formula: ⇒ Keq = e-∆G/RT Substituting the known values: Keq = e-(-15.2)/(8.314 x 300) Keq = e0.006094 Finding the value of e0.006094 using a scientific calculator: ∴ Keq = 1.01 Result: The equilibrium constant (Keq) for the given reaction is 1.01. Further insight: ∆G < 0 indicates that the given reaction is spontaneous in the forward direction.  Keq > 1 further endorses that the rate of forward reaction is greater than the rate of backward reaction.

Example # 2: Calculate Keq for the reversible reaction of nitrogen with hydrogen at room temperature (298 K), given that the Gibbs free energy change (∆G) for the reaction is -725 J/mol. N2 (g) + 3 H2(g)  ⇌ 2 NH3(g)

Solution: As per the question statement: ∆G (Delta G) = -725 J/mol T= 298 K Keq =? We already know that R = 8.314 J/K.mol Applying the alternative formula: ⇒ Keq = 10-∆G/2.303RT Substituting the known values: Keq = 10-(-725)/(2.303 x 8.314 x 298) Keq = 100.127 Finding the value of 100.127 using a scientific calculator: ∴ Keq = 1.34 Result: The equilibrium constant (Keq) for the given reaction is 1.34.

Example # 3: The Gibbs free energy change (∆G) for a reaction is 9.56 kJ/mol at 313 K. Use ∆G (Delta G) to find Keq.

Solution: As per the question statement: ∆G = 9.56 kJ/mol = 9.56 x 1000 = 9560 J/mol (Converting kilojoules to joules to bring consistency in units) T= 313 K Keq =? We already know that R = 8.314 J/K.mol Applying the formula: ⇒ Keq = e-∆G/RT Substituting the known values: Keq = e-9560/(8.314 x 313) Keq = 10-3.67 Finding the value of 10-3.67 using a scientific calculator: ∴ Keq = 2.14 x 10-4 Result: The equilibrium constant (Keq) for the given reaction is 2.14 x 10-4 = 0.000214. Further insight: ∆G > 0 indicates that the given reaction is non-spontaneous in the forward direction.  Keq < 1 further endorses that the rate of backward reaction is greater than the rate of forward reaction.

FAQ

How do you calculate Keq from ∆G (Delta G)?

Keq can be calculated from ∆G using either one of the two formulae given below: Keq = e-∆G/RT  or Keq = 10-∆G/2.303 RT

How do you derive the equation for the relation between Gibbs free energy change and equilibrium constant?

∆G = ∆G° + RT lnQ Where, ∆G (Delta G) = Gibbs free energy change for a reaction ∆G°= Standard Gibbs free energy change for the reaction R = Ideal gas constant T = Absolute temperature Q = Reaction quotient At equilibrium, ∆G = 0 and Q = Keq (Equilibrium constant) Thus, the above equation is reduced to: 0 = ∆G°+ RT lnKeq Rearranging the above equation gives us the relation between Standard Gibbs free energy change and equilibrium constant. ∆G° = – RT lnKeq

What are the values of Keq and ∆G for a reaction that is at equilibrium?

At equilibrium, Keq = 1 and ∆G (Delta G) = 0.