Delta G = 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 = Natural logarithm of the equilibrium constant (K) for the reaction (Unitless)
Now let’s go through the above-mentioned components one by one.
Break down of the components of Delta G = -RTlnK
Gibbs free energy change (∆G) refers to reversible work done on a systemat 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 proceeds.
For a reversible reaction, A + B ⇌ C + D
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.
The equilibrium constant (K) is defined as the ratio of the molar concentration of products to the molar concentration of reactants raised to the number of moles of each in the balanced chemical equation.
For example, for the reaction aA + bB ⇌ cC + dD
The equilibrium constant (K) is given as:
K = [C]c[D]d/[A]a[B]b
If K > Q, the reaction proceeds forward
If K < Q, the reaction moves backward
If K = Q, the reaction maintains equilibrium
Where Q = reaction quotient
The equation Delta G = -RTlnK is derived from ∆G = ∆G° + RTln Q.
∆ G = ∆G° + RTln Q….equation (i)
Where ∆G°= Standard Gibbs energy change, i.e., the change in Gibbs free energy when all the substances are in their standard physical states.
At equilibrium, ∆G = 0, and Q = K, equation (i) is reduced to equation (ii):
0 = ∆ G° + RTln K
∆G° = -RTln K….equation (ii)
Where and how to use Delta G = -RTlnK? – Examples
The equation ∆G = -RTlnK can be used to find the Gibbs free energy change (Delta G) or equilibrium constant (K) at a constant temperature, depending upon which variable is unknown.
In either case, the value of R stays constant at 8.314 J/K.mol.
For example, The equilibrium constant for the reaction given below is 1.34 at 298K (room temperature).
N2(g) + 3 H2(g)⇌2 NH3(g)
Find the Gibbs free energy change (Delta G) using ∆G = -RTlnK.
Solution
As per the question statement:
Equilibrium constant (K) = 1.34
Temperature (T) = 298 K
Gibbs free energy change (∆G) =?
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Substituting the known values:
∆G = – (8.314) (298) (ln 1.34)
∴ ∆G = – 725.1 J/mol
Result: The Gibbs free energy change for the reaction is – 725 J/mol. A negative Delta G value implies that the formation of ammonia from hydrogen and nitrogen is spontaneous at room temperature.
Another example is – Find the equilibrium constant for a reaction taking place at 373 K, given that its Gibbs free energy change is -60 kJ/mol.
Consistency in units is very important; therefore, we have converted ∆G from kJ/mol to J/mol.
1 kJ = 1000 J
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Converting natural log into the common log:
∆G = -2.303 RT log K
Substituting the known values:
-60,000 = – 2.303 (8.314) (373) (log K)
60,000 = 7141.88 log K
log K = 60,000/7141.88
log K = 8.40
Find K by taking the antilog:
K = 108.40
∴ K = 2.52 x 108
Result: The equilibrium constant (K) for the reaction is 2.52 x 108.
More Examples involving Delta G = -RTlnK
Calculate ∆G for the reaction shown below at T = 423 K.
N2(g) + O2(g)⇌2 NO (g)
Given that, the equilibrium concentrations of N2, O2, and NO are 2.5 moles, 2.5 moles, and 0.10 moles, respectively.
The equilibrium constant expression for the above reaction is:
Now that we know the value of K, we can easily find ∆G by applying the formula ∆G = -RTlnK, substituting R = 8.314 J/K.mol and T = 423 K.
⇒ ∆G = -RTlnK
∆G = -(8.314)(423)[ln(1.6 x 10-3)]
∴ ∆G = 2.26 x 104 J/mol = 22.6 kJ/mol
Result: The Gibbs free energy change for the reaction is 22.6 kJ/mol, which hints at the non-spontaneity of the reaction.
Calculate ∆G for the ionization of acetic acid at T= 30°C. Given that, its equilibrium constant is 1.8 x 10-5.
CH3COOH(aq) + H2O (l)⇌CH3COO–(aq) + H3O+(aq)
As per the question statement:
T = 30°C
Converting °C into absolute scale.
1 K = 1°C + 273
T = 30 + 273 = 303 K
K = 1.8 x 10-5
Delta G =?
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Substituting the known values:
∆G = -(8.314)(303)[ln (1.8 x 10-5)]
∴ ∆G = 2.75 x 104 J/mol = 27.5 kJ/mol
Result: The Gibbs free energy change for the ionization of acetic acid is 27.5 kJ/mol.
FAQ
What does the equation Delta G = – RTlnK represent?
The equation Delta G or (∆G) = – RTlnK represents the relationship between the Gibbs free energy change (∆G) and equilibrium constant (K) of a reversible chemical reaction.
In ∆G = -RTlnK:
Delta G = Gibbs free energy change
R = Ideal or molar gas constant (R= 8.314 J/K.mol)
T = Absolute temperature
ln = Natural log (loge x = ln x)
K = Equilibrium constant
How to find the equilibrium constant (K) using the equation Delta G = -RTlnK?
K can be determined using any of the two formulae:
K = e–∆G/RT or K = 10–∆G/2.303 RT
Which of the following formulae relate the free energy change to the equilibrium constant?
A) – ∆G = RT log K
B) – ∆G = (RT log K)/2.303
C) – ∆G = 2.303 RT log K
D) ∆G = RT ln K
Option C is the correct answer. The formula ∆G = -RTln K can be rewritten as ∆G = -2.303 RTlog K or – ∆G = 2.303 RT log K.
Which of the following options gives the correct formula for finding Gibbs free energy change (Delta G) of a reaction?
A) ∆G = ∆H – T∆S
B) ∆G = -RTlnK
C) ∆G = Σ∆Gf (products) – Σ∆Gf (reactants)
D) ∆G = -nFEcell
E) All of the above
Option E is the correct answer. All the above-mentioned formulae can be used to calculate ∆G under different circumstances.
∆G = ∆H – T∆S is used to find ∆G of a thermodynamic reaction, given its temperature (T), enthalpy change (∆H), and entropy change (∆S).
Delta G = -RTlnK relates ∆G to the equilibrium constant (K) of a reversible gaseous reaction.
∆G = Σ∆Gf (products) – Σ∆Gf (reactants) is used to find ∆G given the Gibbs free energy change of formation of the different products and reactants involved in a balanced chemical equation.
∆G = -nFEcell is applied to find ∆G of an electrochemical cell supported by a redox reaction. It directly relates the Gibbs free energy change with the electrical potential difference (Ecell).
The equation ∆G = -RTlnK is very important in understanding the relationship between the energetics of a chemical reaction and its equilibrium states.
As per the equation ∆G = -RTlnK, the Gibbs free energy change (Delta G) of a reaction is directly linked to its equilibrium constant (K).
In this article, you will find plenty of examples of using Delta G = -RTlnK, but before that, let us start from the basics.
Delta G = 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 = Natural logarithm of the equilibrium constant (K) for the reaction (Unitless)
Now let’s go through the above-mentioned components one by one.
Break down of the components of Delta G = -RTlnK
Gibbs free energy change (∆G) refers to reversible work done on a systemat 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 proceeds.
For a reversible reaction, A + B ⇌ C + D
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.
The equilibrium constant (K) is defined as the ratio of the molar concentration of products to the molar concentration of reactants raised to the number of moles of each in the balanced chemical equation.
For example, for the reaction aA + bB ⇌ cC + dD
The equilibrium constant (K) is given as:
K = [C]c[D]d/[A]a[B]b
If K > Q, the reaction proceeds forward
If K < Q, the reaction moves backward
If K = Q, the reaction maintains equilibrium
Where Q = reaction quotient
The equation Delta G = -RTlnK is derived from ∆G = ∆G° + RTln Q.
∆ G = ∆G° + RTln Q….equation (i)
Where ∆G°= Standard Gibbs energy change, i.e., the change in Gibbs free energy when all the substances are in their standard physical states.
At equilibrium, ∆G = 0, and Q = K, equation (i) is reduced to equation (ii):
0 = ∆ G° + RTln K
∆G° = -RTln K….equation (ii)
Where and how to use Delta G = -RTlnK? – Examples
The equation ∆G = -RTlnK can be used to find the Gibbs free energy change (Delta G) or equilibrium constant (K) at a constant temperature, depending upon which variable is unknown.
In either case, the value of R stays constant at 8.314 J/K.mol.
For example, The equilibrium constant for the reaction given below is 1.34 at 298K (room temperature).
N2(g) + 3 H2(g)⇌2 NH3(g)
Find the Gibbs free energy change (Delta G) using ∆G = -RTlnK.
Solution
As per the question statement:
Equilibrium constant (K) = 1.34
Temperature (T) = 298 K
Gibbs free energy change (∆G) =?
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Substituting the known values:
∆G = – (8.314) (298) (ln 1.34)
∴ ∆G = – 725.1 J/mol
Result: The Gibbs free energy change for the reaction is – 725 J/mol. A negative Delta G value implies that the formation of ammonia from hydrogen and nitrogen is spontaneous at room temperature.
Another example is – Find the equilibrium constant for a reaction taking place at 373 K, given that its Gibbs free energy change is -60 kJ/mol.
Consistency in units is very important; therefore, we have converted ∆G from kJ/mol to J/mol.
1 kJ = 1000 J
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Converting natural log into the common log:
∆G = -2.303 RT log K
Substituting the known values:
-60,000 = – 2.303 (8.314) (373) (log K)
60,000 = 7141.88 log K
log K = 60,000/7141.88
log K = 8.40
Find K by taking the antilog:
K = 108.40
∴ K = 2.52 x 108
Result: The equilibrium constant (K) for the reaction is 2.52 x 108.
More Examples involving Delta G = -RTlnK
Calculate ∆G for the reaction shown below at T = 423 K.
N2(g) + O2(g)⇌2 NO (g)
Given that, the equilibrium concentrations of N2, O2, and NO are 2.5 moles, 2.5 moles, and 0.10 moles, respectively.
The equilibrium constant expression for the above reaction is:
Now that we know the value of K, we can easily find ∆G by applying the formula ∆G = -RTlnK, substituting R = 8.314 J/K.mol and T = 423 K.
⇒ ∆G = -RTlnK
∆G = -(8.314)(423)[ln(1.6 x 10-3)]
∴ ∆G = 2.26 x 104 J/mol = 22.6 kJ/mol
Result: The Gibbs free energy change for the reaction is 22.6 kJ/mol, which hints at the non-spontaneity of the reaction.
Calculate ∆G for the ionization of acetic acid at T= 30°C. Given that, its equilibrium constant is 1.8 x 10-5.
CH3COOH(aq) + H2O (l)⇌CH3COO–(aq) + H3O+(aq)
As per the question statement:
T = 30°C
Converting °C into absolute scale.
1 K = 1°C + 273
T = 30 + 273 = 303 K
K = 1.8 x 10-5
Delta G =?
We already know that R = 8.314 J/K.mol
Applying the formula:
⇒ ∆G = -RTlnK
Substituting the known values:
∆G = -(8.314)(303)[ln (1.8 x 10-5)]
∴ ∆G = 2.75 x 104 J/mol = 27.5 kJ/mol
Result: The Gibbs free energy change for the ionization of acetic acid is 27.5 kJ/mol.
FAQ
What does the equation Delta G = – RTlnK represent?
The equation Delta G or (∆G) = – RTlnK represents the relationship between the Gibbs free energy change (∆G) and equilibrium constant (K) of a reversible chemical reaction.
In ∆G = -RTlnK:
Delta G = Gibbs free energy change
R = Ideal or molar gas constant (R= 8.314 J/K.mol)
T = Absolute temperature
ln = Natural log (loge x = ln x)
K = Equilibrium constant
How to find the equilibrium constant (K) using the equation Delta G = -RTlnK?
K can be determined using any of the two formulae:
K = e–∆G/RT or K = 10–∆G/2.303 RT
Which of the following formulae relate the free energy change to the equilibrium constant?
A) – ∆G = RT log K
B) – ∆G = (RT log K)/2.303
C) – ∆G = 2.303 RT log K
D) ∆G = RT ln K
Option C is the correct answer. The formula ∆G = -RTln K can be rewritten as ∆G = -2.303 RTlog K or – ∆G = 2.303 RT log K.
Which of the following options gives the correct formula for finding Gibbs free energy change (Delta G) of a reaction?
A) ∆G = ∆H – T∆S
B) ∆G = -RTlnK
C) ∆G = Σ∆Gf (products) – Σ∆Gf (reactants)
D) ∆G = -nFEcell
E) All of the above
Option E is the correct answer. All the above-mentioned formulae can be used to calculate ∆G under different circumstances.
∆G = ∆H – T∆S is used to find ∆G of a thermodynamic reaction, given its temperature (T), enthalpy change (∆H), and entropy change (∆S).
Delta G = -RTlnK relates ∆G to the equilibrium constant (K) of a reversible gaseous reaction.
∆G = Σ∆Gf (products) – Σ∆Gf (reactants) is used to find ∆G given the Gibbs free energy change of formation of the different products and reactants involved in a balanced chemical equation.
∆G = -nFEcell is applied to find ∆G of an electrochemical cell supported by a redox reaction. It directly relates the Gibbs free energy change with the electrical potential difference (Ecell).
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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