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📝 Gibbs free energy equation ΔG=ΔH-TΔS (12 MCQs)

📖 From Principles of Biochemistry • 1. The Foundations of Biochemistry • 12 questions available

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What is Gibbs free energy equation ΔG=ΔH-TΔS?

Definition:
The Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$ is a fundamental thermodynamic equation that determines the spontaneity of a reaction, where $\Delta G$ is the change in free energy, $\Delta H$ is the change in enthalpy (heat content), $T$ is the absolute temperature, and $\Delta S$ is the change in entropy; a negative $\Delta G$ means the reaction is exergonic (spontaneous), while a positive $\Delta G$ means it is endergonic (non-spontaneous) and requires energy input.

Working:
This equation works by balancing the enthalpy (energy stored in bonds) and entropy (disorder) contributions to determine if a reaction can proceed without external energy; at constant temperature and pressure, the sign of $\Delta G$ predicts the direction of a reaction, and in cells, endergonic reactions are coupled to exergonic ones (e.g., ATP hydrolysis with $\Delta G \approx -30.5 \text{ kJ/mol}$ ) to drive biosynthesis, and the equilibrium constant $K_{eq}$ is related by $\Delta G^\circ = -RT \ln K_{eq}$ .

Example:
A simple example is the conversion of glucose to glucose-6-phosphate, which has $\Delta G^\circ = +13.8 \text{ kJ/mol}$ (endergonic), but is coupled to ATP hydrolysis ( $\Delta G^\circ = -30.5 \text{ kJ/mol}$ ), making the overall reaction exergonic and feasible, illustrating how cells use coupled reactions to drive thermodynamically unfavorable processes.

Reason:
The Gibbs free energy equation is central to biochemistry because it allows prediction of reaction feasibility, energy coupling in metabolism, and the direction of biochemical pathways, and it is essential for understanding bioenergetics, enzyme function, and drug action, making it a cornerstone of biochemical thermodynamics.

Easy

Medium

Hard

📝 All Gibbs free energy equation ΔG=ΔH-TΔS MCQs

Q1. For a process at constant temperature and pressure, which expression correctly predicts whether the process is thermodynamically favorable from the signs and magnitudes of $\Delta H$ and $\Delta S$ ?

\Delta G=\Delta H+T\Delta S

\Delta G=\Delta H-T\Delta S

✅

\Delta G=T\Delta H-\Delta S

\Delta G=\Delta H/T-\Delta S

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The correct relationship is $\Delta G=\Delta H-T\Delta S$ . The enthalpy change contributes directly, while the entropy term is multiplied by absolute temperature. A negative $\Delta G$ indicates thermodynamic favorability under the specified conditions.

Q2. A reaction has $\Delta H<0$ and $\Delta S>0$ . Without knowing their numerical magnitudes, what can be concluded about $\Delta G$ as temperature changes?

A.It must always be positive

B.It must always equal zero

C.It must always be negative ✅

D.Its sign depends only on pressure

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When $\Delta H<0$ and $\Delta S>0$ , both terms make $\Delta G$ more negative because $\Delta H$ is negative and $-T\Delta S$ is also negative. Therefore, $\Delta G$ is negative at every positive absolute temperature.

Q3. Two processes have the following properties: Process X has $\Delta H=-10$ kJ/mol and $\Delta S=-20$ J/(mol·K), while Process Y has $\Delta H=+5$ kJ/mol and $\Delta S=+40$ J/(mol·K). At 300 K, which process has the more favorable $\Delta G$ ?

A.Process X, because its enthalpy is negative

B.Process Y, because its entropy contribution is sufficiently favorable ✅

C.Both have the same

\Delta G

D.Neither can have a negative

\Delta G

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For X, $\Delta G=-10-(300)(-0.020)=-4$ kJ/mol. For Y, $\Delta G=5-(300)(0.040)=-7$ kJ/mol. Thus Y is more favorable despite having positive enthalpy because its favorable entropy contribution is larger.

Q4. A student argues that any process with positive $\Delta H$ cannot occur spontaneously. A measured process has $\Delta H=+12$ kJ/mol and $\Delta S=+60$ J/(mol·K). At 300 K, how should the student's reasoning be evaluated?

A.Correct, because positive enthalpy always prevents spontaneity

B.Incorrect, because

\Delta S

makes

\Delta G=-6

kJ/mol ✅

C.Correct, because entropy is irrelevant when

\Delta H>0

D.Incorrect, because positive

\Delta H

always makes

\Delta G

zero

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student's reasoning ignores the entropy term. At 300 K, $\Delta G=12-(300)(0.060)=12-18=-6$ kJ/mol. Therefore, despite being endothermic, the process has a negative free-energy change under these conditions.

Q5. A researcher observes that a molecular assembly becomes more ordered while releasing heat. Which combination most strongly explains why the assembly can still have a negative $\Delta G$ ?

A.Positive

\Delta H

and negative

\Delta S

B.Negative

\Delta H

and negative

\Delta S

✅

C.Positive

\Delta H

and positive

\Delta S

D.Zero

\Delta H

and negative

\Delta S

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Greater molecular order commonly corresponds to a negative entropy change, which opposes favorable free energy. However, if the process releases enough heat so that the negative $\Delta H$ outweighs the positive $-T\Delta S$ contribution, the overall $\Delta G$ can still be negative.

Q6. An enzyme-catalyzed reaction has $\Delta H=-8$ kJ/mol and $\Delta S=-10$ J/(mol·K). A scientist claims increasing temperature must make the reaction more favorable because temperature always increases molecular motion. What happens to $\Delta G$ when temperature rises from 300 K to 500 K?

A.It changes from

-5

-3

kJ/mol, becoming less favorable ✅

B.It changes from

-11

-13

kJ/mol, becoming more favorable

C.It remains exactly

-8

kJ/mol

D.It becomes positive at 500 K

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At 300 K, $\Delta G=-8-(300)(-0.010)=-5$ kJ/mol. At 500 K, $\Delta G=-8+5=-3$ kJ/mol. Because $\Delta S$ is negative, increasing temperature makes the $-T\Delta S$ contribution increasingly positive.

Q7. A proposed metabolic transformation has $\Delta H=+30$ kJ/mol and $\Delta S=+100$ J/(mol·K). A coupled energy-releasing reaction contributes $\Delta G=-40$ kJ/mol. At 298 K, what is the approximate net $\Delta G$ for the coupled system?

+19.8

kJ/mol

+10.0

kJ/mol

-19.8

kJ/mol ✅

-70.0

kJ/mol

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For the metabolic transformation, $\Delta G=30-(298)(0.100)=0.2$ kJ/mol. Adding the coupling reaction gives $0.2-40=-39.8$ kJ/mol, approximately $-40$ kJ/mol. Thus coupling can drive an otherwise unfavorable process.

Q8. A reaction has $\Delta H=+18$ kJ/mol and $\Delta S=+60$ J/(mol·K). A student calculates $\Delta G=18-(300)(60)= -17,982$ kJ/mol. What is the primary error?

A.The student used Celsius instead of kelvin

B.The student forgot that enthalpy must be negative

C.The entropy unit was not converted to kJ/(mol·K) ✅

D.The equation should contain

+T\Delta S

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The entropy must be expressed in the same energy units as enthalpy. Since $60$ J equals $0.060$ kJ, the correct calculation is $18-(300)(0.060)=0$ kJ/mol. The large negative value results from a unit inconsistency.

Q9. A graph of $\Delta G$ versus temperature for a process is a straight line that slopes upward as temperature increases. What combination of signs is most consistent with this graph?

\Delta H>0,\Delta S>0

\Delta H<0,\Delta S>0

\Delta H>0,\Delta S<0

\Delta H<0,\Delta S<0

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Because $\Delta G=\Delta H-T\Delta S$ , the slope of a $\Delta G$ -versus- $T$ graph is $-\Delta S$ . An upward slope therefore means $\Delta S<0$ . The intercept is related to $\Delta H$ , which can be positive or negative; among the choices, D is consistent with an upward slope and negative intercept.

Q10. Two researchers evaluate the same reaction. Researcher A reports $\Delta H=-25$ kJ/mol and $\Delta S=-50$ J/(mol·K), while Researcher B reports $\Delta G=-10$ kJ/mol at 300 K. Which conclusion is best supported?

A.The measurements are inconsistent because

\Delta G

must equal

\Delta H

B.The values are consistent because the entropy term changes

\Delta G

by +15 kJ/mol ✅

C.The values are inconsistent because entropy must always be positive

D.The values are consistent only if temperature is zero

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Using the reported enthalpy and entropy, $\Delta G=-25-(300)(-0.050)=-25+15=-10$ kJ/mol. Therefore both researchers' results are internally consistent. The entropy term raises $\Delta G$ by 15 kJ/mol because the entropy change is negative.

Q11. A process has $\Delta H=+40$ kJ/mol and $\Delta S=+150$ J/(mol·K). It becomes favorable only above a certain temperature. Which temperature is the closest threshold?

A.167 K

B.267 K ✅

C.367 K

D.467 K

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The threshold occurs when $\Delta G=0$ : $40-T(0.150)=0$ . Thus $T=40/0.150\approx267$ K. Therefore option B is correct. Below this temperature $\Delta G$ is positive, while above it the favorable entropy contribution makes $\Delta G$ negative.

Q12. For a hypothetical reaction, $\Delta H=24$ kJ/mol and $\Delta S=80$ J/(mol·K). Another reaction has $\Delta H=-4$ kJ/mol and $\Delta S=-20$ J/(mol·K). At 300 K, which statement correctly compares them?

A.Both have positive

\Delta G

B.The first has

\Delta G=0

, while the second has

\Delta G=+2

kJ/mol ✅

C.The first has

\Delta G=-0.0

kJ/mol, while the second has

\Delta G=-10

kJ/mol

D.The first has

\Delta G=-2

kJ/mol, while the second has

\Delta G=+2

kJ/mol

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For the first reaction, $\Delta G=24-(300)(0.080)=0$ kJ/mol. For the second, $\Delta G=-4-(300)(-0.020)=+2$ kJ/mol. Thus the first is exactly at the thermodynamic threshold, while the second is unfavorable at 300 K.

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