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 $100\ \text{g}$ of saline (approximate as water, $c=4.18\ \text{J/(g\cdot ^\circ C)}$) from $25^\circ\text{C}$ to $20^\circ\text{C}$ in a calorimeter. The concept is heat exchange with $q=mc\Delta T$. Which sign and magnitude of $q$ for the saline is most consistent?
$q \approx -2.1\ \text{kJ}$
$q \approx +2.1\ \text{kJ}$
$q \approx -21\ \text{kJ}$
$q \approx +0.21\ \text{kJ}$
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. Heat exchange uses q = m c ΔT, with negative q for cooling. In this saline cooling, ΔT = -5 °C. The correct answer B follows because q = 100 × 4.18 × (-5) ≈ -2.1 kJ. Distractor A fails by ignoring the negative sign for heat loss. To approach similar questions, include ΔT sign in calculations. Remember that in calorimetry, signs indicate direction of heat flow.
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?
$\Delta U = +210\ \text{J}$
$\Delta U = +90\ \text{J}$
$\Delta U = -210\ \text{J}$
$\Delta U = -90\ \text{J}$
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.
A sealed, rigid bioreactor (no volume change) contains a buffered solution where an enzyme converts a single substrate into two smaller products. The reactor is thermally insulated (no heat exchange) and no matter enters or leaves. The key thermodynamic concept is entropy change in an isolated system. Which conclusion is most consistent with the second law of thermodynamics?
Total entropy decreases if the reaction is exothermic, regardless of products formed.
Total entropy of the reactor must decrease because the enzyme increases reaction rate.
Total entropy of the reactor can increase because forming more particles increases microstates.
Total entropy must remain constant because temperature is constant.
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. The second law of thermodynamics states that the entropy of an isolated system increases or remains constant for spontaneous processes. In this scenario, the sealed, insulated bioreactor is an isolated system where an enzyme converts one substrate into two products, potentially altering entropy. The correct answer B follows because producing more particles increases the number of microstates, allowing total entropy to increase consistent with the second law. Distractor A fails by assuming enzymes decrease entropy, misapplying the role of catalysts which affect rate but not thermodynamic direction. To approach similar questions, identify if the system is isolated and assess changes in particle number or disorder. Remember that in biological reactions, entropy increases can drive processes in closed systems.
A membrane ATPase hydrolyzes ATP to pump ions against a concentration gradient. The system is the ATPase plus transported ions; the surroundings are the cytosol. The key concept is coupling and the first law of thermodynamics. Which statement is most consistent with energy conservation?
Energy released by ATP hydrolysis is converted into work required to move ions uphill plus heat.
If the pump is 100% efficient, it must violate the first law by producing extra work.
ATP hydrolysis provides matter, not energy, so it cannot power transport.
Ion pumping can occur without ATP hydrolysis if entropy increases in the cytosol.
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. The first law of thermodynamics ensures energy conservation in coupled processes like ATP hydrolysis and transport. In this ion pumping, ATP energy drives uphill transport. The correct answer B follows because ATP hydrolysis energy becomes work for ion movement plus heat, conserving total energy. Distractor A fails by ignoring energy needs, assuming entropy alone suffices. To approach similar questions, trace energy flows in coupled systems. Remember that in cells, exergonic reactions power endergonic ones without violating conservation.
A student states that because a cell maintains internal order (e.g., ion gradients and macromolecular structure), it must violate the second law of thermodynamics. The concept is entropy in open biological systems. Which statement is most consistent with thermodynamics?
Entropy applies only to gases, not to biochemical systems.
Cells are isolated systems, so their entropy must always increase.
Cells violate the second law because living systems can decrease entropy without compensation.
Cells can decrease internal entropy by increasing the entropy of the surroundings through heat and waste export.
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. Entropy in open systems allows local decreases if surroundings increase more. In cells, order is maintained by exporting disorder via heat and waste. The correct answer B follows because cells increase surroundings entropy to decrease internal. Distractor A fails by claiming violation of the second law. To approach similar questions, distinguish open from isolated systems. Remember that life obeys thermodynamics through openness.
A researcher observes that a reaction in a closed flask proceeds spontaneously and releases heat to the surroundings at constant pressure. The key concept is second law of thermodynamics for the universe ($\Delta S_{univ} = \Delta S_{sys}+\Delta S_{surr}$). Which statement is most consistent?
Spontaneity requires $\Delta S_{univ}>0$, which can occur even if $\Delta S_{sys}<0$ when $\Delta S_{surr}$ is sufficiently positive.
If the reaction is exothermic, then $\Delta S_{univ}=0$.
Because the system releases heat, $\Delta S_{sys}$ must be positive.
Spontaneity requires $\Delta S_{sys}>0$ regardless of surroundings.
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. The second law requires ΔS_univ > 0 for spontaneity, including system and surroundings. In this exothermic reaction, heat to surroundings increases ΔS_surr. The correct answer B follows because ΔS_univ > 0 even if ΔS_sys < 0. Distractor C fails by requiring positive ΔS_sys always. To approach similar questions, calculate universe entropy. Remember that exothermic processes increase surroundings entropy.
A reaction has $\Delta S = -200\ \text{J/(mol\cdot K)}$ and $\Delta H = -20\ \text{kJ/mol}$. The concept is Gibbs free energy and temperature effects. At sufficiently high temperature, which outcome is most consistent for $\Delta G$?
$\Delta G$ becomes more negative because the reaction is exothermic.
$\Delta G$ can become positive because the $-T\Delta S$ term becomes large and positive.
$\Delta G$ becomes zero because entropy terms cancel enthalpy for any $T$.
$\Delta G$ must remain negative at all temperatures because $\Delta H<0$.
Explanation
This question evaluates understanding of thermodynamic energy changes, specifically the effects of temperature on Gibbs free energy in chemical reactions. The Gibbs free energy change is given by the equation ΔG = ΔH - TΔS, where ΔG determines the spontaneity of a reaction: negative ΔG indicates spontaneity, positive indicates non-spontaneity, and zero indicates equilibrium. In this scenario, the reaction has a negative ΔH (-20 kJ/mol) and a negative ΔS (-200 J/mol·K), meaning it is exothermic but decreases in entropy, and the impact of the entropy term grows with temperature. The correct answer, B, follows because at high temperatures, the -TΔS term becomes large and positive (since ΔS is negative, -TΔS = +T|ΔS|), which can outweigh the negative ΔH, making ΔG positive. A common distractor like C fails by assuming that a negative ΔH ensures ΔG remains negative at all temperatures, which misapplies the equation by ignoring the temperature-dependent entropy contribution that can dominate at high T. To approach similar questions, always plug in the signs of ΔH and ΔS into the Gibbs equation and consider how increasing T amplifies the -TΔS term, especially when ΔS is negative. This transferable check ensures consistent evaluation of temperature effects on reaction spontaneity across various thermodynamic scenarios.
A researcher reports that a particular ATP-driven process in a cell has an overall $\Delta G < 0$ due to coupling, even though one substep has $\Delta G > 0$. The concept is Gibbs free energy additivity. Which statement is most consistent?
If any substep has $\Delta G>0$, the overall pathway must be nonspontaneous.
Coupling works because enzymes can change the sign of $\Delta G$ for any substep.
A positive-$\Delta G$ substep becomes spontaneous only if temperature decreases.
The overall process can be spontaneous if the sum of $\Delta G$ values for coupled steps is negative.
Explanation
This question evaluates understanding of thermodynamic energy changes in biological contexts. Gibbs free energy is additive in coupled pathways, allowing overall spontaneity. In this ATP-driven process, one substep has positive ΔG but overall negative. The correct answer B follows because summed ΔG < 0 enables the pathway. Distractor D fails by requiring all substeps negative. To approach similar questions, sum ΔG for coupled steps. Remember that in cells, exergonic reactions drive endergonic ones.
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?
If particle number is unchanged, then $\Delta G$ must be zero.
An exothermic $\Delta H<0$ can favor spontaneity even if $\Delta S$ is small or negative.
Because heat is released, $\Delta S$ must be positive and large.
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.
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 surroundings gained $16\ \text{kJ}$ of heat from the system.
Because a bomb calorimeter is rigid, the reaction cannot release heat.
The internal energy of the system must increase by $16\ \text{kJ}$.
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.