MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5e Thermodynamics Energy Changes

What this quiz covers

This quiz focuses on 5e Thermodynamics Energy Changes, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $q>0$ because the tissue lost thermal energy to the cold plate.
2. $q<0$ because the tissue released heat to the cold plate. (correct answer)
3. $q=0$ because temperature changed due to decreased entropy, not heat flow.
4. $q$ must be positive because specific heat is positive.

Explanation: This question assesses understanding of heat flow sign conventions from the system's perspective. When the tissue cools from 37°C to 35°C, it loses thermal energy to the cold plate. From the tissue's perspective, heat flows out, making q negative according to standard sign conventions. The heat lost can be calculated as q = mcΔT = 100 g × 3.5 J/(g·°C) × (35-37)°C = -700 J. Choice B correctly identifies the negative sign for heat release. Choice A incorrectly assigns positive q to heat loss, choice C incorrectly denies heat flow, and choice D incorrectly links sign to specific heat value. Remember that q < 0 when a system releases heat (temperature decreases) and q > 0 when it absorbs heat (temperature increases).

Question 2

1. Because \Delta G^\circ' is negative, the reaction is spontaneous only if the temperature is above 100°C.
2. Because \Delta G^\circ' is negative, ATP hydrolysis is thermodynamically favorable under standard biochemical conditions; actual spontaneity depends on $\Delta G$ at the given concentrations. (correct answer)
3. A negative \Delta G^\circ' means the reaction must be exothermic, so $\Delta H$ is necessarily negative.
4. If enzymes are absent, \Delta G^\circ' becomes positive, so hydrolysis is no longer thermodynamically favorable.

Explanation: This question tests understanding of Gibbs free energy and reaction spontaneity in biochemical systems. The Gibbs free energy change determines whether a reaction is thermodynamically favorable, with ΔG < 0 indicating spontaneity under the actual conditions. The standard free energy change ΔG°' = -30.5 kJ/mol indicates ATP hydrolysis is favorable under standard biochemical conditions (1 M concentrations, pH 7, 25°C). The correct answer B properly distinguishes between ΔG°' (standard conditions) and ΔG (actual conditions), noting that while the negative ΔG°' indicates thermodynamic favorability, the actual spontaneity depends on the reaction quotient Q through ΔG = ΔG°' + RT ln Q. Answer C incorrectly assumes negative ΔG°' requires negative ΔH, ignoring that entropy contributions (-TΔS) can make reactions with positive ΔH still have negative ΔG. When evaluating biochemical reactions, always consider both standard and actual conditions to determine true spontaneity.

Question 3

1. Because ATP is produced, the mitochondria must have created new energy beyond that contained in the substrate.
2. The chemical energy stored in ATP must be less than or equal to the chemical energy released from the substrate, with the remainder transferred as heat and/or other work. (correct answer)
3. If the process is efficient, entropy of the mitochondria must decrease to zero so that all substrate energy can be converted to ATP.
4. Energy conservation implies the Gibbs free energy change of ATP synthesis must be negative in the mitochondria.

Explanation: This question evaluates understanding of energy conservation in biological systems, specifically oxidative phosphorylation. The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or transformed between different forms. In this mitochondrial system, 1.0 μmol glucose equivalents provide 1200 kJ/mol × 1.0 μmol = 1.2 kJ total energy, while 30 μmol ATP stores 30 kJ/mol × 30 μmol = 900 kJ. Since only 900 kJ is stored in ATP from 1200 kJ available, the remaining 300 kJ must be accounted for as heat released or other work, making choice B correct. Choice A violates the first law by suggesting energy creation, while C incorrectly assumes perfect efficiency is possible by eliminating entropy, and D misapplies Gibbs free energy concepts to an endergonic process. When analyzing biological energy conversions, always verify that total energy input equals the sum of all energy outputs including heat.

Question 4

1. $\Delta U = +210\ \text{J}$
2. $\Delta U = -90\ \text{J}$ (correct answer)
3. $\Delta U = +90\ \text{J}$
4. $\Delta U = -210\ \text{J}$

Explanation: This question evaluates understanding of thermodynamic energy changes in biological contexts. The first law, ΔU = q + w, tracks internal energy with signs for heat and work. In this expansion, q = +60 J and w = -150 J. The correct answer B follows because ΔU = 60 - 150 = -90 J. Distractor A fails by adding without signs. To approach similar questions, apply conventions consistently. Remember that in systems with pistons, work affects energy changes.

Question 5

1. Because heat is released, $\Delta S$ must be positive and large.
2. An exothermic $\Delta H<0$ can favor spontaneity even if $\Delta S$ is small or negative. (correct answer)
3. If particle number is unchanged, then $\Delta G$ must be zero.
4. Heat release implies $\Delta G$ is always positive because energy leaves the system.

Explanation: This question evaluates understanding of thermodynamic energy changes in biological contexts. Spontaneity via ΔG balances entropy and enthalpy, even with minimal ΔS. In this pathway, heat release suggests negative ΔH despite unchanged particles. The correct answer B follows because exothermic ΔH can drive spontaneity with small or negative ΔS. Distractor C fails by assuming ΔG = 0 for unchanged particles. To approach similar questions, evaluate both terms in ΔG. Remember that in metabolism, enthalpy often dominates.

Question 6

1. The system absorbed $16\ \text{kJ}$ of heat, so $q$ is positive.
2. The surroundings gained $16\ \text{kJ}$ of heat from the system. (correct answer)
3. The internal energy of the system must increase by $16\ \text{kJ}$.
4. Because a bomb calorimeter is rigid, the reaction cannot release heat.

Explanation: This question evaluates understanding of thermodynamic energy changes in biological contexts. The first law uses sign conventions where heat released by the system is q < 0. In this bomb calorimetry, q_v = -16 kJ for glucose oxidation. The correct answer B follows because surroundings gain 16 kJ from the system. Distractor A fails by misassigning positive q for absorption. To approach similar questions, apply consistent signs for q and w. Remember that in calorimetry, negative q indicates exothermic reactions.

Question 7

1. Binding becomes more favorable at higher $T$ because $\Delta S<0$.
2. Binding becomes less favorable at higher $T$ because $-T\Delta S$ becomes more positive. (correct answer)
3. Binding favorability is independent of $T$ because $\Delta H$ is constant.
4. Binding becomes less favorable at higher $T$ because $\Delta H<0$.

Explanation: This question evaluates understanding of thermodynamic energy changes in biological contexts. The Gibbs free energy, ΔG = ΔH - TΔS, governs binding favorability, with temperature affecting the -TΔS term. In this ligand-protein binding, negative ΔH and ΔS indicate exothermic ordering. The correct answer B follows because with ΔS < 0, higher T makes -TΔS more positive, increasing ΔG and reducing favorability. Distractor D fails by attributing the effect solely to ΔH < 0, ignoring entropy's role. To approach similar questions, analyze signs of ΔH and ΔS to predict temperature effects on ΔG. Emphasize that in biomolecular interactions, entropy-enthalpy compensation is common.

Question 8

1. The body gained about $58\ \text{kJ}$ of thermal energy. (correct answer)
2. The body gained about $5.8\ \text{kJ}$ of thermal energy.
3. The body lost thermal energy because temperature increased (energy left as heat).
4. No thermal energy change occurred because specific heat is constant for water.

Explanation: This question evaluates application of heat capacity to calculate thermal energy changes in biological systems. Heat capacity relates temperature change to thermal energy change through q = mcΔT, where q is heat absorbed, m is mass, c is specific heat, and ΔT is temperature change. For the 70 kg body with temperature increase of 0.20°C: q = (70 kg)(4.18 kJ/kg·K)(0.20 K) = 58.52 kJ ≈ 58 kJ. The positive temperature change indicates the body gained thermal energy. Choice B contains a calculation error (missing a factor of 10), while Choice C incorrectly assumes temperature increase means energy loss. Choice D misunderstands that constant specific heat doesn't prevent energy changes. To solve heat capacity problems, carefully track units and remember that positive ΔT means heat gain for the system.

Question 9

1. The substrate's chemical energy decrease is $150\ \text{J}$, matching heat released plus energy stored in ATP. (correct answer)
2. Because ATP is produced, no heat should be released if the process is energy-conserving.
3. The mitochondria must have gained $150\ \text{J}$ of internal energy because they produced ATP and released heat.
4. The enthalpy change must equal $30\ \text{J}$ because only ATP formation counts as useful energy.

Explanation: This question evaluates understanding of energy conservation in biological systems using the first law of thermodynamics. The first law states that energy cannot be created or destroyed, only transferred or transformed, meaning the total energy change must balance. In this mitochondrial system, the substrate loses chemical energy that is converted into two forms: heat released to surroundings (120 J) and chemical energy stored in ATP (30 J). Therefore, the substrate must have lost 120 J + 30 J = 150 J of chemical energy to account for both outputs. Choice B incorrectly assumes that energy-conserving processes produce no heat, ignoring that energy conservation refers to total energy, not prevention of heat release. To verify energy accounting in biological systems, always sum all energy outputs (heat, work, chemical storage) to find the required input energy change.

Question 10

1. The enzyme lowers the activation energy but does not change the reaction's state function $\Delta G$. (correct answer)
2. The enzyme makes $\Delta G$ more negative by stabilizing products relative to reactants.
3. The enzyme increases entropy of the universe by converting heat directly into work, changing $\Delta G$.
4. The enzyme changes $\Delta H$ but not $\Delta S$, so $\Delta G$ must change at constant $T$.

Explanation: This question tests understanding of enzyme effects on reaction thermodynamics versus kinetics. Enzymes are catalysts that lower activation energy, increasing reaction rates without changing the thermodynamic state functions (ΔG, ΔH, ΔS) between reactants and products. The Gibbs free energy change depends only on the initial and final states, not the pathway, so enzyme presence doesn't affect ΔG. Choice B incorrectly suggests enzymes change product stability and thus ΔG. Choice C proposes an impossible mechanism of converting heat to work. Choice D wrongly claims enzymes change ΔH without affecting ΔS, which would still alter ΔG. When analyzing enzyme effects, remember they affect kinetics (rates) but not thermodynamics (equilibrium position and energy differences between states).

Question 11

1. Entropy of the universe must decrease because a highly ordered molecule (glucose) is consumed.
2. Entropy of the universe must increase for the reaction to proceed spontaneously. (correct answer)
3. Entropy of the universe is unchanged because the container is insulated.
4. Entropy change depends only on enzyme concentration, not on reactants or products.

Explanation: This question tests application of the second law to spontaneous biological reactions. The second law states that the entropy of the universe must increase for any spontaneous process. Although the container is insulated (preventing heat exchange with surroundings), the universe includes both the system and its surroundings. Glucose oxidation converts one large, ordered molecule plus O2 into many smaller CO2 and H2O molecules, increasing system entropy. The reaction's spontaneity requires that total entropy (universe) increases. Choice A incorrectly focuses on glucose order without considering products. Choice C wrongly assumes insulation prevents entropy changes in the universe. Choice D incorrectly relates entropy to enzyme concentration rather than reactant/product states. For spontaneous reactions, universe entropy must increase regardless of container properties.

Question 12

1. At least $40\ \text{J}$ must be transferred as heat to the surroundings and/or stored as increased internal energy of the system. (correct answer)
2. Exactly $40\ \text{J}$ must be converted into new chemical energy because work cannot produce heat.
3. The mechanical work must equal the chemical energy released, so the $60\ \text{J}$ value implies measurement error.
4. The missing $40\ \text{J}$ indicates a violation of the first law because biological systems are not closed.

Explanation: This question evaluates understanding of energy conservation during biological energy conversions. The first law of thermodynamics requires that all energy be accounted for: energy in must equal energy out plus any stored energy. The muscle cell releases 100 J from ATP hydrolysis and performs 60 J of mechanical work, leaving 40 J unaccounted for. This remaining energy must either be released as heat to surroundings or stored as increased internal energy (such as in ion gradients or molecular conformations). Choice A correctly identifies these possibilities. Choice B incorrectly claims work cannot produce heat, choice C incorrectly requires exact equality between chemical energy and work, and choice D incorrectly suggests biological systems violate conservation laws. In biological systems, energy conversion efficiency is never 100%, with excess energy typically dissipated as heat.

Question 13

1. Total entropy of the chamber must decrease because enzymes create order during catalysis.
2. Total entropy of the chamber is expected to increase because the number of accessible microstates increases upon product formation and mixing. (correct answer)
3. Entropy change cannot be inferred unless the reaction enthalpy is known.
4. Entropy must remain exactly constant because the temperature is constant.

Explanation: This question tests understanding of entropy changes in closed systems according to the second law of thermodynamics. The second law states that the total entropy of an isolated system can never decrease and typically increases for spontaneous processes. In this sealed chamber, the enzyme converts one substrate molecule into two smaller product molecules, increasing the number of particles and thus the number of possible microstates (arrangements) of the system. This increase in microstates corresponds to an increase in entropy, making choice B correct. Choice A incorrectly claims entropy decreases due to enzymatic ordering, choice C incorrectly requires enthalpy information for entropy determination, and choice D incorrectly assumes constant temperature means constant entropy. Remember that entropy changes depend on the number of accessible microstates, not temperature alone, and breaking molecules into smaller pieces generally increases entropy.

Question 14

1. At least one step in the coupled process must have $\Delta G < 0$, and the sum of stepwise $\Delta G$ values is negative. (correct answer)
2. Every individual step in the coupled process must have $\Delta G < 0$.
3. The coupled process cannot involve ATP synthesis because ATP synthesis always has $\Delta G < 0$.
4. The negative $\Delta G$ implies the process absorbs heat, so it must be endothermic overall.

Explanation: This question evaluates understanding of coupled reactions and overall Gibbs free energy. When reactions are coupled, the overall ΔG equals the sum of individual ΔG values. For the coupled process to have ΔG = -10 kJ/mol (spontaneous), at least one component reaction must have negative ΔG, and the sum must be negative. This allows coupling of unfavorable reactions (positive ΔG) with favorable ones (negative ΔG) as long as the overall process is thermodynamically favorable. Choice A correctly states these requirements. Choice B incorrectly requires all steps to be favorable, choice C incorrectly claims ATP synthesis always has negative ΔG, and choice D incorrectly equates negative ΔG with heat absorption. Biological systems commonly couple unfavorable processes to favorable ones to drive essential reactions.

Question 15

1. The reaction cannot proceed in either direction because the activation energy is infinite.
2. The system is at equilibrium with no net driving force, though forward and reverse reactions can still occur. (correct answer)
3. The reaction is maximally exothermic, so heat release prevents net change.
4. The reaction is spontaneous in the forward direction because $\Delta H$ must be negative.

Explanation: This question tests understanding of the relationship between Gibbs free energy and chemical equilibrium. When ΔG = 0 for a reaction under specific conditions, the system is at equilibrium with no net driving force in either direction. At equilibrium, forward and reverse reactions occur at equal rates, producing no net change in concentrations, though individual molecular transformations continue. Choice B correctly identifies this equilibrium condition. Choice A incorrectly confuses thermodynamic barriers with kinetic barriers, choice C incorrectly links zero ΔG to maximum heat release, and choice D incorrectly claims spontaneity with zero driving force. Zero ΔG indicates equilibrium, not the absence of molecular reactions or infinite activation energy.

Question 16

1. $\Delta U$ is $+15\ \text{J}$ because heat enters and some energy leaves as work. (correct answer)
2. $\Delta U$ is $-15\ \text{J}$ because work reduces internal energy more than heat can increase it.
3. $\Delta U$ is $+35\ \text{J}$ because heat and work both add energy to the system.
4. $\Delta U$ is $0\ \text{J}$ because biological systems maintain homeostasis.

Explanation: This question evaluates application of the first law of thermodynamics with proper sign conventions. The first law states ΔU = Q - W, where Q is heat absorbed by the system and W is work done by the system. The organ absorbs 25 J of heat (Q = +25 J) and does 10 J of work on surroundings (W = +10 J). Therefore, ΔU = 25 J - 10 J = +15 J, indicating the internal energy increases by 15 J. Choice A correctly calculates this increase. Choice B incorrectly suggests a decrease, choice C incorrectly adds work instead of subtracting it, and choice D incorrectly invokes homeostasis to claim zero change. When applying the first law, carefully track sign conventions: absorbed heat is positive, released heat is negative, work done by the system is positive.

Question 17

1. Increase the temperature, since $-T\Delta S$ becomes more negative.
2. Decrease the temperature, since $-T\Delta S$ becomes less positive when $\Delta S<0$. (correct answer)
3. Increase the pressure, since pressure always makes $\Delta G$ more negative for binding.
4. Decrease the temperature, since exothermic processes are always more favorable at high temperature.

Explanation: This question tests understanding of temperature effects on Gibbs free energy for reactions with negative entropy changes. With ΔG = ΔH - TΔS, when both ΔH < 0 (favorable) and ΔS < 0 (unfavorable), the entropic term -TΔS is positive and opposes spontaneity. Decreasing temperature reduces the magnitude of this unfavorable entropic contribution, making ΔG more negative and binding more favorable. Choice B correctly identifies that lower temperature reduces the positive -TΔS term. Choice A incorrectly suggests increasing temperature, choice C incorrectly invokes pressure effects, and choice D incorrectly states exothermic processes favor high temperature. For entropy-opposed processes, lower temperatures enhance spontaneity by minimizing the unfavorable entropic penalty.

Question 18

1. The gas internal energy decreases because work is done on the surroundings.
2. The gas internal energy increases because work is done on the gas with no heat loss. (correct answer)
3. The gas temperature must remain constant because the compression is slow.
4. The gas internal energy is unchanged because the syringe is sealed.

Explanation: This question tests understanding of work-energy relationships in thermally isolated systems. When a gas is compressed in an insulated container (adiabatic process), work is done on the gas by the surroundings. According to the first law with Q = 0 (no heat exchange), ΔU = -W, where W is work done by the gas. Since the gas is compressed, work is done on it (W < 0), making ΔU positive. This increase in internal energy manifests as increased molecular kinetic energy and thus higher temperature. Choice B correctly identifies this energy increase from compression work. Choice A incorrectly suggests energy decrease, choice C incorrectly assumes constant temperature, and choice D incorrectly claims no energy change. In adiabatic compression, work input increases internal energy and temperature.

Question 19

1. The student is correct: $\Delta G$ directly equals heat released at constant pressure.
2. The student is incorrect: $\Delta G$ reflects maximum non-expansion work, not necessarily heat released. (correct answer)
3. The student is correct only if $\Delta S = 0$ for the reaction.
4. The student is incorrect because negative $\Delta G$ implies the reaction is endothermic.

Explanation: This question evaluates understanding of what Gibbs free energy represents versus other thermodynamic quantities. Gibbs free energy (ΔG) represents the maximum useful work available from a process at constant temperature and pressure, not the heat released. A reaction with ΔG = -3 kJ/mol can perform up to 3 kJ/mol of non-expansion work, but the actual heat released depends on how efficiently this energy is harnessed and the entropy changes involved. Choice B correctly distinguishes between free energy and heat. Choice A incorrectly equates ΔG with heat, choice C incorrectly requires ΔS = 0 for the equivalence, and choice D incorrectly claims negative ΔG implies endothermic behavior. Free energy, enthalpy, and heat are related but distinct thermodynamic quantities that should not be confused.

Question 20

1. Folding cannot occur spontaneously because the protein's entropy decreases.
2. Folding can be spontaneous if the entropy increase of the surroundings (e.g., from released heat) outweighs the protein's entropy decrease. (correct answer)
3. Folding must increase the protein's entropy because spontaneous processes always increase system entropy.
4. Spontaneity depends only on $\Delta H$, so entropy changes are irrelevant.

Explanation: This question examines understanding of entropy changes in protein folding and the second law of thermodynamics. While protein folding decreases the entropy of the polypeptide chain by reducing conformational freedom, the second law requires that total entropy (system plus surroundings) must increase for spontaneous processes. The heat released during folding increases the thermal motion and entropy of surrounding water molecules. If this entropy increase of the surroundings exceeds the entropy decrease of the protein, the overall process can be spontaneous. Choice B correctly identifies this entropy compensation mechanism. Choice A incorrectly claims folding cannot be spontaneous, choice C incorrectly states spontaneous processes always increase system entropy, and choice D incorrectly ignores entropy contributions. Spontaneity requires total entropy increase, not necessarily system entropy increase.