Thermodynamics and Energy Changes (5E)
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MCAT Chemical and Physical Foundations of Biological Systems › Thermodynamics and Energy Changes (5E)
A researcher cools a $100\ \text{g}$ tissue sample (assume $c = 3.5\ \text{J},\text{g}^{-1},^{\circ}\text{C}^{-1}$) from $37^{\circ}\text{C}$ to $35^{\circ}\text{C}$ by placing it in contact with a cold plate. Using heat exchange reasoning, which statement is most consistent with the sign of $q$ for the tissue sample?
$q<0$ because the tissue released heat to the cold plate.
$q$ must be positive because specific heat is positive.
$q=0$ because temperature changed due to decreased entropy, not heat flow.
$q>0$ because the tissue lost thermal energy to the cold plate.
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).
A researcher measures the standard Gibbs free energy change for ATP hydrolysis in a buffered aqueous solution at $25^\circ\text{C}$: $\Delta G^\circ' = -30.5\ \text{kJ/mol}$. The solution contains initially 1 mM ATP, 1 mM ADP, and 1 mM $\mathrm{P_i}$, and pH is held constant. This question tests reaction spontaneity via Gibbs free energy. How does the concept of Gibbs free energy apply to this biochemical reaction under these conditions?
Because $\Delta G^\circ'$ is negative, the reaction is spontaneous only if the temperature is above 100°C.
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.
A negative $\Delta G^\circ'$ means the reaction must be exothermic, so $\Delta H$ is necessarily negative.
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.
In isolated mitochondria, researchers measure oxidative phosphorylation under steady-state conditions. The mitochondria consume $1.0,\text{\mu mol}$ glucose equivalents and produce $30,\text{\mu mol}$ ATP. Assume ATP synthesis stores $30,\text{kJ/mol}$ of free energy per mole ATP under these conditions, and the total chemical energy available from the substrate is $1200,\text{kJ/mol}$ glucose equivalent. Using the first law of thermodynamics (energy conservation), which statement is most consistent with this system’s energy accounting?
Because ATP is produced, the mitochondria must have created new energy beyond that contained in the substrate.
Energy conservation implies the Gibbs free energy change of ATP synthesis must be negative in the mitochondria.
If the process is efficient, entropy of the mitochondria must decrease to zero so that all substrate energy can be converted to ATP.
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.
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.
A researcher runs a reaction in a closed container with a movable piston at constant external pressure. The system expands, doing $150\ \text{J}$ of work on the surroundings (take work done by the system as $w<0$), and absorbs $60\ \text{J}$ of heat from the surroundings ($q>0$). The concept is first law of thermodynamics, $\Delta U=q+w$. Which internal energy change is most consistent?
$\Delta U = +210\ \text{J}$
$\Delta U = -210\ \text{J}$
$\Delta U = +90\ \text{J}$
$\Delta U = -90\ \text{J}$
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.
In a metabolic pathway step, one reactant molecule is converted into one product molecule with no net change in the number of dissolved particles, but the reaction releases heat to the surroundings. The key concept is entropy vs enthalpy in determining spontaneity ($\Delta G=\Delta H-T\Delta S$). Which statement is most consistent?
Heat release implies $\Delta G$ is always positive because energy leaves the system.
Because heat is released, $\Delta S$ must be positive and large.
An exothermic $\Delta H<0$ can favor spontaneity even if $\Delta S$ is small or negative.
If particle number is unchanged, then $\Delta G$ must be zero.
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.
A researcher measures the heat released when $1.0\ \text{g}$ of glucose is completely oxidized in a bomb calorimeter and obtains $q_v = -16\ \text{kJ}$. The thermodynamic concept is sign conventions and the first law (heat released by the system is negative). Which statement is most consistent?
The system absorbed $16\ \text{kJ}$ of heat, so $q$ is positive.
The internal energy of the system must increase by $16\ \text{kJ}$.
The surroundings gained $16\ \text{kJ}$ of heat from the system.
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.
For a ligand binding reaction $\text{P} + \text{L} \rightleftharpoons \text{PL}$ at $310\ \text{K}$, a researcher estimates $\Delta H = -40\ \text{kJ/mol}$ and $\Delta S = -120\ \text{J/(mol\cdot K)}$. Using Gibbs free energy, $\Delta G = \Delta H - T\Delta S$, which conclusion about temperature dependence is most consistent?
Binding becomes less favorable at higher $T$ because $\Delta H<0$.
Binding becomes less favorable at higher $T$ because $-T\Delta S$ becomes more positive.
Binding favorability is independent of $T$ because $\Delta H$ is constant.
Binding becomes more favorable at higher $T$ because $\Delta S<0$.
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.
During a brief sprint, a subject’s core temperature rises by $0.20^\circ\text{C}$. Approximating the body as $70\ \text{kg}$ of water with specific heat $c = 4.18\ \text{kJ},\text{kg}^{-1},\text{K}^{-1}$ (assume $1\ \text{K} = 1^\circ\text{C}$), using heat exchange and heat capacity, which statement best describes the thermal energy change of the body?
The body lost thermal energy because temperature increased (energy left as heat).
The body gained about $5.8\ \text{kJ}$ of thermal energy.
The body gained about $58\ \text{kJ}$ of thermal energy.
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.
In an isolated preparation, mitochondria oxidize a substrate while synthesizing ATP. Over a short interval, calorimetry detects $120\ \text{J}$ of heat released to the surroundings, and assays indicate $30\ \text{J}$ of chemical energy is stored in newly formed ATP. Using the first law of thermodynamics (energy conservation), which statement is most consistent with the energy accounting for the mitochondria during this interval?
Because ATP is produced, no heat should be released if the process is energy-conserving.
The enthalpy change must equal $30\ \text{J}$ because only ATP formation counts as useful energy.
The substrate’s chemical energy decrease is $150\ \text{J}$, matching heat released plus energy stored in ATP.
The mitochondria must have gained $150\ \text{J}$ of internal energy because they produced ATP and released heat.
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.
An enzyme-catalyzed reaction is studied at constant temperature. Adding enzyme increases the observed reaction rate but does not change measured values of $\Delta G$ between reactants and products. Applying thermodynamics to enzyme catalysis, which statement is most consistent with this observation?
The enzyme lowers the activation energy but does not change the reaction’s state function $\Delta G$.
The enzyme increases entropy of the universe by converting heat directly into work, changing $\Delta G$.
The enzyme makes $\Delta G$ more negative by stabilizing products relative to reactants.
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).