# Confusion in relation of Gibbs free energy and equilibrium constant

• Chemistry
• tbn032
In summary: O2(g) and you want to know the equilibrium constant for the reaction. You would use the equation Keq=Kp.
tbn032
Homework Statement
for the reaction
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
find Kp at 298 Kelvin?
Relevant Equations
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
SO2(g)+1/2O2(g)SO3(g);ΔHo=-98.32KJ/mole,ΔSo=-95J/(mole-K).
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp

Last edited:
Given the temperature 298 Kelvin, is this reaction at the 'standard state'? If so, the pressure is ambient or ~1 atm (100 kPa, or 1 bar).
https://en.wikipedia.org/wiki/Standard_state#Conventional_standard_states

Given the standard state, one should be able to determine the density of the gases. What does one's textbook indicate about STP, standard state and Gibbs free energy?

The problem statement one provides indicates gases.

Last edited:
my confusion is arising from the fact that in my book it is written that
Δ G = -RT ln Keq
Keq=KporKc as the standard state is referred in terms of partial pressure or concentration
Δ G = -RT ln Kp
Δ G = -RT ln Kk
if calculate the value of Keq from the equation Δ G = -RT ln Keq using R=8.3144598J⋅K-1⋅mol-1.the value of Keq is dimensionless
so I am not able to conclude if Keq=Kp or Keq=Kc
kp≠kc because Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).

Last edited:
The standard state for gases is 1 bar. So, in this example, the Keq is Kp. It is in terms of the partial pressures of the components in bars.

Incidentally, you omitted the superscript 0 in ##\Delta G^0##

Lord Jestocost
tbn032 said:
Homework Statement:: for the reaction
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
find Kp at 298 Kelvin?
Relevant Equations:: SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).

SO2(g)+1/2O2(g)SO3(g);ΔHo=-98.32KJ/mole,ΔSo=-95J/(mole-K).
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp
The equation is giving it up as you have SO2(g) you are dealing with gases

## 1. What is the difference between Gibbs free energy and equilibrium constant?

Gibbs free energy (G) is a thermodynamic quantity that measures the amount of energy available for a chemical reaction to occur at constant temperature and pressure. It takes into account both the enthalpy (H) and entropy (S) of the system. On the other hand, equilibrium constant (K) is a ratio of the concentrations of products and reactants at equilibrium. It is a measure of the extent to which a reaction proceeds towards products or reactants at a given temperature.

## 2. How are Gibbs free energy and equilibrium constant related?

There is a direct relationship between Gibbs free energy and equilibrium constant. The equilibrium constant (K) is related to the Gibbs free energy (G) through the equation: ΔG = -RTlnK, where R is the gas constant and T is the temperature in Kelvin. This means that the value of the equilibrium constant is directly related to the change in Gibbs free energy of a reaction. A more negative ΔG indicates a larger equilibrium constant and a more favorable reaction.

## 3. Can Gibbs free energy and equilibrium constant be used to predict the direction of a reaction?

Yes, both Gibbs free energy and equilibrium constant can be used to predict the direction of a reaction. If the value of ΔG is negative, the reaction will proceed in the forward direction and if it is positive, the reaction will proceed in the reverse direction. Similarly, if the equilibrium constant (K) is greater than 1, the reaction will favor the formation of products and if it is less than 1, the reaction will favor the formation of reactants.

## 4. How does temperature affect the relationship between Gibbs free energy and equilibrium constant?

Temperature has a significant effect on the relationship between Gibbs free energy and equilibrium constant. An increase in temperature leads to an increase in the value of the equilibrium constant, making the reaction more favorable. This is because an increase in temperature increases the randomness (entropy) of the system, resulting in a more negative ΔG value.

## 5. Can Gibbs free energy and equilibrium constant be used to determine the spontaneity of a reaction?

Yes, Gibbs free energy and equilibrium constant can be used to determine the spontaneity of a reaction. If the value of ΔG is negative, the reaction is spontaneous in the forward direction. However, if the value of ΔG is positive, the reaction is non-spontaneous in the forward direction and will only occur if the conditions are changed. A reaction with a negative ΔG and a large equilibrium constant (K) is highly spontaneous and will proceed almost completely to products.

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