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📝 Equilibrium constant and Gibbs free energy (13 MCQs)

📖 From Principles of Biochemistry • 1. The Foundations of Biochemistry • 13 questions available

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What is Equilibrium constant and Gibbs free energy?

Definition:
The equilibrium constant ( $K_{eq}$ ) and Gibbs free energy ( $\Delta G$ ) are thermodynamically linked, where the standard free energy change $\Delta G^\circ$ is related to $K_{eq}$ by the equation $\Delta G^\circ = -RT \ln K_{eq}$ , where $R$ is the gas constant and $T$ is the absolute temperature, and this relationship indicates that a large $K_{eq}$ (favoring products) corresponds to a negative $\Delta G^\circ$ (spontaneous reaction), providing a quantitative basis for predicting the direction and extent of biochemical reactions.

Working:
This relationship works by allowing scientists to calculate the equilibrium position of a reaction from $\Delta G^\circ$ , and vice versa; for example, if $K_{eq} > 1$ , then $\Delta G^\circ < 0$ , meaning the reaction proceeds spontaneously toward products, and if $K_{eq} < 1$ , then $\Delta G^\circ > 0$ , meaning it is unfavorable; in cells, reactions are often far from equilibrium, and the actual $\Delta G$ depends on concentrations, described by $\Delta G = \Delta G^\circ + RT \ln Q$ , where $Q$ is the reaction quotient, allowing control over metabolic fluxes.

Example:
A simple example is the hydrolysis of ATP, which has a $K_{eq}$ very large (about $10^5$ ) under standard conditions, giving a $\Delta G^\circ = -30.5 \text{ kJ/mol}$ , indicating that the reaction strongly favors products, and this thermodynamic favorability is why ATP is a good energy carrier, and the actual $\Delta G$ in cells is even more negative due to the high ATP/ADP ratio.

Reason:
Understanding the equilibrium constant and Gibbs free energy is essential for predicting reaction spontaneity, designing metabolic engineering strategies, and interpreting enzyme kinetics, making it a cornerstone of biochemistry and cellular bioenergetics.

Easy

Medium

Hard

📝 All Equilibrium constant and Gibbs free energy MCQs

Q1. For a reaction at a specified temperature, which relationship correctly connects the equilibrium constant $K_{eq}$ with the standard free-energy change $\Delta G^\circ$ ?

A.A larger positive

\Delta G^\circ

corresponds to a larger

K_{eq}

B.A negative

\Delta G^\circ

corresponds to

K_{eq} > 1

✅

K_{eq}

is independent of

\Delta G^\circ

D.A reaction with

K_{eq} = 1

must have a large negative

\Delta G^\circ

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The relationship is $\Delta G^\circ = -RT \ln K_{eq}$ . Therefore, when $\Delta G^\circ$ is negative, $\ln K_{eq}$ is positive and $K_{eq} > 1$ . A larger equilibrium constant indicates greater product favorability under standard conditions.

Q2. A biochemical reaction has $K_{eq} = 1$ at a particular temperature. What can be concluded about its standard free-energy change?

A.It is strongly negative

B.It is strongly positive

C.It is approximately zero ✅

D.It cannot be related to

K_{eq}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Because $\Delta G^\circ = -RT \ln K_{eq}$ , setting $K_{eq} = 1$ makes $\ln(1) = 0$ . Thus the standard free-energy change is zero, although this does not mean the reaction cannot proceed in either direction under nonstandard conditions.

Q3. Two reactions occur at the same temperature. Reaction X has $K_{eq} = 100$ , while reaction Y has $K_{eq} = 0.01$ . Which interpretation is most accurate?

A.Both reactions have identical standard free-energy changes

B.X has a more negative

\Delta G^\circ

and is more product-favored under standard conditions ✅

C.Y has a more negative

\Delta G^\circ

because its

K_{eq}

is smaller

D.Neither reaction has a predictable relationship between

K_{eq}

and

\Delta G^\circ

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A larger $K_{eq}$ means a larger positive value of $\ln K_{eq}$ , which makes $\Delta G^\circ$ more negative. Therefore, X is thermodynamically more product-favored under standard conditions than Y.

Q4. A student claims that a reaction with $K_{eq} = 500$ must proceed rapidly because its equilibrium strongly favors products. Which response best evaluates the claim?

A.Correct, because thermodynamic favorability determines reaction speed

B.Correct, because a large

K_{eq}

means a small activation energy

C.Incorrect, because

K_{eq}

describes equilibrium composition rather than the rate of reaching equilibrium ✅

D.Incorrect, because

K_{eq}

has no connection with free energy

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A large $K_{eq}$ indicates that products are favored at equilibrium and corresponds to a negative $\Delta G^\circ$ . However, reaction rate depends on kinetic factors such as activation energy, so thermodynamic favorability does not guarantee rapid reaction.

Q5. A metabolic reaction has $K_{eq} = 0.001$ . A researcher concludes that the reaction is impossible in a living cell. Which reasoning best identifies the problem?

A.A small

K_{eq}

means the reaction is thermodynamically impossible

B.A small

K_{eq}

means the reaction favors reactants under standard conditions, but cellular concentrations can alter the actual

\Delta G

✅

C.A small

K_{eq}

proves that ATP must always be consumed

D.A small

K_{eq}

means the reaction must have a high activation energy

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A small equilibrium constant indicates product formation is unfavorable under standard conditions, but cellular concentrations may differ greatly from standard conditions. The actual free-energy change depends on the reaction quotient, so the reaction can still proceed in cells.

Q6. A biochemical reaction has $\Delta G^\circ = +12$ kJ/mol. Under cellular conditions, the concentrations of products are greatly reduced relative to reactants. What is the most reasonable prediction?

A.The reaction must remain unfavorable because

\Delta G^\circ

never changes

B.The reaction can become favorable because the reaction quotient can make the actual

\Delta G

negative ✅

C.The reaction becomes favorable only if

K_{eq}

becomes negative

D.The reaction cannot proceed unless its activation energy becomes zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Standard free energy describes a reference state, whereas actual free energy depends on concentrations through the reaction quotient. Removing products can lower the reaction quotient sufficiently to make actual $\Delta G$ negative despite positive $\Delta G^\circ$ .

Q7. A reaction has $K_{eq} = 10^6$ , but an enzyme-catalyzed version reaches equilibrium much faster than the uncatalyzed reaction. Which conclusion follows?

A.The enzyme increases

K_{eq}

by stabilizing products

B.The enzyme makes

\Delta G^\circ

more negative

C.The enzyme changes the equilibrium position but not the reaction tendency

D.The enzyme changes the rate without changing

K_{eq}

\Delta G^\circ

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: An enzyme lowers the activation energy for reaching equilibrium but does not alter the thermodynamic difference between reactants and products. Therefore, $K_{eq}$ and $\Delta G^\circ$ remain unchanged while the reaction proceeds faster.

Q8. A student calculates $\Delta G^\circ = -RT \ln(0.2)$ and concludes that the reaction is strongly product-favored because the result is negative. What is the error?

A.The logarithm of

0.2

is positive

B.The negative sign was incorrectly interpreted;

\ln(0.2)

is negative, making

\Delta G^\circ

positive ✅

R

must be negative

K_{eq}

cannot be used to calculate

\Delta G^\circ

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since $0.2 < 1$ , its natural logarithm is negative. Multiplying by the leading negative sign makes $\Delta G^\circ$ positive. Thus the reaction is reactant-favored under standard conditions rather than product-favored.

Q9. A graph plots $\Delta G^\circ$ on the vertical axis against $\ln K_{eq}$ on the horizontal axis for several reactions at the same temperature. What pattern should the data approximately follow?

A.A straight line with positive slope

B.A horizontal line above zero

C.A straight line with negative slope ✅

D.A curved line that always remains positive

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The relationship $\Delta G^\circ = -RT \ln K_{eq}$ predicts a linear relationship at constant temperature. The coefficient $-RT$ is negative, so increasing $\ln K_{eq}$ produces increasingly negative values of $\Delta G^\circ$ .

Q10. Reaction A has $\Delta G^\circ = -20$ kJ/mol, while reaction B has $\Delta G^\circ = -5$ kJ/mol at the same temperature. Which prediction is best supported?

A.Reaction A has a smaller

K_{eq}

than reaction B

B.Reaction A has a larger

K_{eq}

than reaction B ✅

C.Both reactions must have

K_{eq} = 1

D.The values of

K_{eq}

cannot be compared from

\Delta G^\circ

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At the same temperature, more negative standard free energy corresponds to a larger equilibrium constant. Therefore, reaction A has a larger $K_{eq}$ and is more strongly product-favored under standard conditions than reaction B.

Q11. A pathway contains two reactions. Reaction 1 has $\Delta G^\circ = -30$ kJ/mol and reaction 2 has $\Delta G^\circ = +10$ kJ/mol. If the reactions occur sequentially, what is the combined standard free-energy change?

A.+40 kJ/mol

B.-40 kJ/mol

C.-20 kJ/mol ✅

D.+20 kJ/mol

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Standard free-energy changes are additive for sequential reactions. Adding $-30$ kJ/mol and $+10$ kJ/mol gives $-20$ kJ/mol, indicating that the overall pathway is product-favored under standard conditions.

Q12. A reaction has $K_{eq} = 10$ , and another independent reaction has $K_{eq} = 100$ . They are combined to form an overall process whose equilibrium constant is the product of the individual constants. What is the resulting $K_{eq}$ ?

A.10

B.90

C.1000 ✅

D.0.001

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For reactions that are added together, their equilibrium constants multiply. Therefore, the combined equilibrium constant is $10 \times 100 = 1000$ . This large value corresponds to a negative overall standard free-energy change.

Q13. At a fixed temperature, reaction P has $K_{eq} = 10^4$ , while reaction Q has $K_{eq} = 10^{-4}$ . A student says their standard free-energy changes differ only slightly because the numerical $K_{eq}$ values are both four powers of ten from 1. Which evaluation is best?

A.Correct, because both constants are equally close to 1

B.Correct, because the sign of

K_{eq}

determines free energy

C.Incorrect, because their logarithms have opposite signs, so their standard free-energy changes have opposite signs ✅

D.Incorrect, because equilibrium constants cannot be compared using logarithms

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Taking logarithms reveals the key difference: $\ln(10^4)$ is positive whereas $\ln(10^{-4})$ is negative. Consequently, their standard free-energy changes have opposite signs and equal magnitudes at the same temperature.

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