AP Chemistry Practice Quiz: Free Energy and Equilibrium

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 11 questions to check your progress.

Question 1 of 11

A process is described as "thermodynamically favored" under standard conditions. What does this imply about the value of its equilibrium constant, K?

A)K > 1B)K < 1C)K = 1D)K = 0

All Questions (11)

A process is described as "thermodynamically favored" under standard conditions. What does this imply about the value of its equilibrium constant, K?

A) K > 1

B) K < 1

C) K = 1

D) K = 0

Correct Answer: A

The provided content states that the phrase 'thermodynamically favored' (ΔG° < 0) means that the products are favored at equilibrium, which corresponds to an equilibrium constant K > 1.

For a chemical reaction at equilibrium under standard conditions, the equilibrium constant K is found to be 0.01. Which statement about the standard free energy change, ΔG°, is correct?

A) ΔG° is negative.

B) ΔG° is positive.

C) ΔG° is zero.

D) The sign of ΔG° cannot be determined from K.

Correct Answer: B

According to the provided content, processes with ΔG° > 0 favor reactants, which corresponds to an equilibrium constant K < 1. Since K = 0.01 is less than 1, ΔG° must be positive.

Consider the relationship ΔG° = −RT ln K. If a reaction has a large positive ΔG°, what must be true about the value of ln K and K?

A) ln K must be positive, so K > 1.

B) ln K must be negative, so K < 1.

C) ln K must be zero, so K = 1.

D) ln K must be positive, so K < 1.

Correct Answer: B

For ΔG° to be positive in the equation ΔG° = −RT ln K, the term (−RT ln K) must be positive. Since R and T are always positive, ln K must be negative for the entire term to be positive. If ln K is negative, K must be a value between 0 and 1 (K < 1).

A reaction has an equilibrium constant K = 5.0 x 10⁻⁴ at 298 K. Which statement correctly describes the standard Gibbs free energy change (ΔG°) and the favorability of the process?

A) ΔG° < 0, and the process is thermodynamically favored.

B) ΔG° > 0, and the process is not thermodynamically favored.

C) ΔG° = 0, and the system is at equilibrium under standard conditions.

D) ΔG° > 0, and the process is thermodynamically favored.

Correct Answer: B

The equilibrium constant K is much less than 1. According to the content, processes with K < 1 favor reactants and have a positive standard free energy change (ΔG° > 0). A process with ΔG° > 0 is not thermodynamically favored under standard conditions.

If the standard free energy change (ΔG°) for a reaction is very close to zero, what is the approximate value of the equilibrium constant (K)?

A) K is very large (K >> 1).

B) K is very small (K << 1).

C) K is close to 1.

D) K is close to -1.

Correct Answer: C

The content explicitly states, 'When ΔG° is near zero, K is close to 1.' This is because if ΔG° is zero in the equation ΔG° = −RT ln K, then ln K must be zero, which means K = 1.

Which of the following sets of conditions correctly describes a reaction that is thermodynamically favored under standard conditions?

A) ΔG° > 0 and K > 1

B) ΔG° < 0 and K > 1

C) ΔG° > 0 and K < 1

D) ΔG° < 0 and K < 1

Correct Answer: B

The definition of a thermodynamically favored process is one with a negative standard free energy change (ΔG° < 0). The content explicitly links this condition to an equilibrium constant greater than 1 (K > 1), where products are favored at equilibrium.

Using the equation K = e^−ΔG°/RT, if ΔG° for a reaction is a large negative value, what does this imply about the exponent and the resulting value of K?

A) The exponent (−ΔG°/RT) is a large negative number, leading to K < 1.

B) The exponent (−ΔG°/RT) is a large positive number, leading to K > 1.

C) The exponent (−ΔG°/RT) is close to zero, leading to K ≈ 1.

D) The exponent (−ΔG°/RT) is a large positive number, leading to K < 1.

Correct Answer: B

If ΔG° is a large negative number, the term −ΔG° becomes a large positive number. Since R and T are positive, the entire exponent (−ΔG°/RT) is a large positive number. Raising the base e to a large positive power results in a very large number, so K will be much greater than 1.

Reaction X has a standard free energy change, ΔG°x = -25 kJ/mol. Reaction Y has a standard free energy change, ΔG°y = +25 kJ/mol. Which statement accurately compares their equilibrium constants, Kx and Ky, at the same temperature?

A) Kx > 1 and Ky < 1

B) Kx < 1 and Ky > 1

C) Both Kx and Ky are greater than 1.

D) Both Kx and Ky are less than 1.

Correct Answer: A

For Reaction X, ΔG°x is negative, which means it is thermodynamically favored and its equilibrium constant Kx must be greater than 1. For Reaction Y, ΔG°y is positive, which means it is not thermodynamically favored (reactants are favored) and its equilibrium constant Ky must be less than 1.

A chemical reaction is found to have an equilibrium constant, K, of 150 at a certain temperature. What must be the sign of the standard Gibbs free energy change, ΔG°, for this reaction?

A) Positive

B) Negative

C) Zero

D) It depends on the temperature.

Correct Answer: B

The provided content states that processes with ΔG° < 0 favor products, which corresponds to K > 1. Since K = 150 is greater than 1, the standard Gibbs free energy change, ΔG°, must be negative.

Under which condition is the equilibrium constant, K, expected to deviate most strongly from 1?

A) When ΔG° is very close to zero.

B) When the absolute value of ΔG° is much larger than the value of RT.

C) When ΔG° has the same value as RT.

D) When the temperature, T, is very low, regardless of ΔG°.

Correct Answer: B

The provided content directly states, 'When ΔG° is much larger or smaller than RT, K deviates strongly from 1.' This means a large magnitude of ΔG° relative to RT leads to K values that are either very large (K >> 1) or very small (K << 1).

A chemist observes that a particular reaction at standard conditions has an equilibrium mixture consisting almost entirely of reactants. Which of the following provides the best thermodynamic explanation for this observation?

A) ΔG° is large and negative, so K > 1.

B) ΔG° is near zero, so K ≈ 1.

C) ΔG° is large and positive, so K < 1.

D) ΔG° is large and negative, so K < 1.

Correct Answer: C

If the equilibrium mixture consists almost entirely of reactants, it means the reaction favors reactants. According to the content, processes that favor reactants have an equilibrium constant K < 1. This condition corresponds to a positive standard free energy change (ΔG° > 0). A mixture of 'almost entirely reactants' implies K is much less than 1, which corresponds to a large positive ΔG°.