A reaction has K = 1.0. State the value of G^. [1 point]

A reaction at 298 K has G^ = -5.7\ kJ mol^-1. State whether K is greater or less than 1, and explain. [2 points]

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). A reaction at 298 K has G^ = -22.8\ kJ mol^-1 (R = 8.314\ J mol^-1K^-1). (a) Write the relationship between G^ and K. (b) Calculate K for the reaction. (c) State whether products or reactants are favored at equilibrium, and justify. (d) Explain what G^ = 0 would imply about K.

A 4-point quantitative FRQ on free energy and K. (a) Relationship (1 point): G^ = -RTln K. (b) Calculate K (1 point): ln K = - G^RT = --22800(8.314)(298) = 228002477 = 9.20; so K = e^9.20 = 9.9 × 10^3. (c) Favored (1 point): G^ 1, so products are favored at equilibrium. (d) G^ = 0 (1 point): if G^ = 0 then ln K = 0, so K = 1 and reactants and products are present in comparable amounts at equilibrium. Markers reward the relationship, the value of K, the product-favored conclusion, and the K = 1 interpretation for G^ = 0.

United StatesChemistrySyllabus dot point

How are the standard free energy change and the equilibrium constant related?

Topic 9.5 Free Energy and Equilibrium: relate the standard free energy change to the equilibrium constant using delta G standard equals minus RT ln K, and use delta G equals delta G standard plus RT ln Q for non-standard conditions.

A focused answer to AP Chemistry Topic 9.5, covering the relationship between the standard free energy change and the equilibrium constant, delta G standard equals minus RT ln K, the non-standard delta G equation, and how the sign of delta G standard relates to the size of K, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The standard free energy change and the equilibrium constant are linked by delta G standard equals minus RT ln K. This means a negative standard free energy change gives K greater than 1 (products favored), a positive value gives K less than 1 (reactants favored), and a value of zero gives K equal to 1. The more negative the standard free energy change, the larger K, and the further the reaction proceeds toward products. Under non-standard conditions the actual free energy change is delta G equals delta G standard plus RT ln Q, where Q is the reaction quotient; this shows that a reaction proceeds (delta G less than zero) until Q rises to K, at which point delta G equals zero and the system is at equilibrium. These equations connect the sign of delta G, the value of K, and the comparison of Q with K into one consistent picture.

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What this topic is asking
Standard free energy and K
The sign of delta G standard and the size of K
Non-standard conditions
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What this topic is asking

The College Board (Topic 9.5) wants you to relate the standard free energy change to the equilibrium constant using $\Delta G^\circ = -RT\ln K$ , and to use $\Delta G = \Delta G^\circ + RT\ln Q$ for non-standard conditions. This unites thermodynamics (Unit 9) with equilibrium (Unit 7): the sign and size of $\Delta G^\circ$ determine the value of $K$ .

Standard free energy and K

This equation translates a thermodynamic quantity ( $\Delta G^\circ$ ) into an equilibrium quantity ( $K$ ). Because of the minus sign and the logarithm, a more negative $\Delta G^\circ$ gives an exponentially larger $K$ , and a positive $\Delta G^\circ$ gives a $K$ below 1. The relationship is the bridge between the two big themes of the course.

The sign of delta G standard and the size of K

So a favorable reaction (negative $\Delta G^\circ$ ) has its equilibrium lying toward products, matching the qualitative reading of $K$ in Topic 7.5. The magnitude links to the magnitude of $K$ through the exponential, so even a modestly negative $\Delta G^\circ$ can give a large $K$ .

Non-standard conditions

Under conditions that are not standard, the actual free energy change uses the reaction quotient $Q$ :

\Delta G = \Delta G^\circ + RT\ln Q

This shows the driving force at any point in the reaction. When $Q < K$ , $\Delta G < 0$ and the reaction proceeds forward; as the reaction proceeds, $Q$ rises and $\Delta G$ becomes less negative; when $Q = K$ , $\Delta G = 0$ and the system is at equilibrium. This single equation contains both the standard relationship (when $Q = K$ , $\Delta G = 0$ and $\Delta G^\circ = -RT\ln K$ ) and the Q-versus-K direction rule of Unit 7.

Try this

Q1. A reaction has $K = 1.0$ . State the value of $\Delta G^\circ$ . [1 point]

Cue. $\Delta G^\circ = -RT\ln(1) = 0$ .

Q2. A reaction at $298$ K has $\Delta G^\circ = -5.7\ \text{kJ mol}^{-1}$ . State whether $K$ is greater or less than 1, and explain. [2 points]

Cue. Greater than 1; a negative $\Delta G^\circ$ makes $\ln K$ positive, so $K > 1$ and products are favored.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A reaction at

298

K has

\Delta G^\circ = -22.8\ \text{kJ mol}^{-1}

(

R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}

). (a) Write the relationship between

\Delta G^\circ

and

K

. (b) Calculate

K

for the reaction. (c) State whether products or reactants are favored at equilibrium, and justify. (d) Explain what

\Delta G^\circ = 0

would imply about

K

Show worked answer →

A 4-point quantitative FRQ on free energy and K.

(a) Relationship (1 point): $\Delta G^\circ = -RT\ln K$ .
(b) Calculate K (1 point): $\ln K = -\dfrac{\Delta G^\circ}{RT} = -\dfrac{-22800}{(8.314)(298)} = \dfrac{22800}{2477} = 9.20$ ; so $K = e^{9.20} = 9.9 \times 10^{3}$ .
(c) Favored (1 point): $\Delta G^\circ < 0$ gives $K > 1$ , so products are favored at equilibrium.
(d) $\Delta G^\circ = 0$ (1 point): if $\Delta G^\circ = 0$ then $\ln K = 0$ , so $K = 1$ and reactants and products are present in comparable amounts at equilibrium.

Markers reward the relationship, the value of $K$ , the product-favored conclusion, and the $K = 1$ interpretation for $\Delta G^\circ = 0$ .

AP 2021 (style)1 marksSection I (multiple choice). A reaction has

\Delta G^\circ > 0

. Its equilibrium constant

K

is (A) greater than 1 (B) equal to 1 (C) less than 1 (D) negative. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (C).

From $\Delta G^\circ = -RT\ln K$ , a positive $\Delta G^\circ$ makes $\ln K$ negative, so $K < 1$ (reactants favored). The trap is (D): an equilibrium constant cannot be negative; a positive $\Delta G^\circ$ gives a small positive $K$ .

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)