Physical Chemistry 1 Quiz: Deltag And Maximum Work

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Deltag And Maximum WorkQuestion 1 of 11

A reversible fuel cell operates between two reservoirs: one containing H₂/H₂O at pH 0 and another containing O₂/OH⁻ at pH 14, both at 25°C and 1 atm. When the cell produces 50 A of current for 2 hours, calculate the maximum non-expansion work (in kJ) that could theoretically be extracted beyond what is delivered as electrical energy.

0 kJ

47.3 kJ

94.6 kJ

142.0 kJ

189.2 kJ

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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Deltag And Maximum Work

Practice Deltag And Maximum Work in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Deltag And Maximum Work, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

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.

All questions

Question 1

0 kJ (correct answer)
47.3 kJ
94.6 kJ
142.0 kJ
189.2 kJ

Explanation: When you encounter fuel cell problems involving "maximum non-expansion work beyond electrical energy," you're dealing with a thermodynamic efficiency concept that often trips students up. In a reversible fuel cell operating at constant temperature and pressure, all the available Gibbs free energy change (

\Delta G

) is converted into useful work - specifically electrical work. This is the defining characteristic of a reversible process: it operates at maximum thermodynamic efficiency with no entropy generation. For the hydrogen-oxygen fuel cell reaction:

\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}

, all of

\Delta G

becomes electrical energy when the process is reversible. There's no additional work available beyond what's already being delivered as electrical output. Answer A (0 kJ) is correct because a reversible fuel cell already extracts the maximum possible work from the chemical reaction. No additional non-expansion work exists beyond the electrical energy being produced. Answer B (47.3 kJ) likely represents half the total Gibbs free energy change for the amount of fuel consumed, suggesting a misunderstanding that only half the energy becomes electrical work. Answer C (94.6 kJ) probably represents the full

\Delta G

value, incorrectly assuming this could be additional work on top of the electrical energy. Answer D (142.0 kJ) might combine electrical work with some incorrectly calculated "extra" work component. Remember: "reversible" in thermodynamics means maximum efficiency. When a fuel cell is already operating reversibly, you've captured all available work - there's no extra work hiding elsewhere in the system.

Question 2

A concentration cell consists of two identical electrodes immersed in solutions of the same electrolyte but different concentrations. For a cell with Ag/Ag⁺ electrodes where [Ag⁺]anode = 0.010 M and [Ag⁺]cathode = 1.50 M at 25°C, calculate the maximum work (in J) that can be obtained per coulomb of charge transferred through an external circuit.

0.0528 J/C
0.0642 J/C
0.0755 J/C
0.1284 J/C (correct answer)
0.1510 J/C

Explanation: When you encounter a concentration cell problem, you're dealing with a galvanic cell where the driving force comes purely from concentration differences, not different electrode materials. The maximum work per coulomb is simply the cell potential (emf). For this Ag/Ag⁺ concentration cell, you'll use the Nernst equation. Since both electrodes are identical, the standard cell potential is zero, so:

E_{cell} = -\frac{RT}{nF}\ln\frac{[Ag^+]_{anode}}{[Ag^+]_{cathode}}

At 25°C,

\frac{RT}{F} = 0.0257

V, and n = 1 for Ag⁺ + e⁻ → Ag. Substituting the values:

E_{cell} = -\frac{0.0257}{1}\ln\frac{0.010}{1.50} = -0.0257 \times \ln(0.00667) = -0.0257 \times (-5.008) = 0.1284

V Since 1 volt equals 1 joule per coulomb, the maximum work is 0.1284 J/C. Answer D (0.1284 J/C) is correct. Answer A (0.0528 J/C) likely results from calculation errors in the natural logarithm. Answer B (0.0642 J/C) might come from using the wrong temperature constant or incorrect concentration ratio. Answer C (0.0755 J/C) could result from sign errors or using base-10 logarithm instead of natural logarithm. Remember: for concentration cells, always identify which electrode has higher concentration (cathode) versus lower concentration (anode), and double-check that you're using the natural logarithm in the Nernst equation. The RT/F constant at 25°C (0.0257 V) is worth memorizing.

Question 3

A galvanic cell operating under standard conditions produces an electromotive force of 1.20 V and transfers 2 moles of electrons per reaction cycle. If the cell is connected to an external motor with 85% efficiency, what is the maximum useful work (in kJ) that can be extracted from the motor when 0.5 moles of the cell reaction occur?

98.4 kJ (correct answer)
115.8 kJ
83.7 kJ
116.2 kJ
231.6 kJ

Explanation: When you encounter galvanic cell problems involving external work, you're dealing with the relationship between electrical energy and useful mechanical work. The key is understanding that not all electrical energy can be converted to useful work due to efficiency losses. Start by calculating the total electrical work available from the cell reaction. The electrical work is given by

W_{electrical} = nFE°

, where n is moles of electrons transferred, F is Faraday's constant (96,485 C/mol), and E° is the cell potential. For 0.5 moles of reaction:

W_{electrical} = (0.5 \text{ mol reaction}) × (2 \text{ mol e}^-/\text{mol reaction}) × (96,485 \text{ C/mol}) × (1.20 \text{ V}) = 115,782 \text{ J} = 115.8 \text{ kJ}

However, the motor is only 85% efficient, so the useful work is:

W_{useful} = 115.8 × 0.85 = 98.4 \text{ kJ}

Answer A (98.4 kJ) correctly accounts for both the stoichiometry and motor efficiency. Answer B (115.8 kJ) represents the total electrical work but ignores the 85% efficiency limitation. Answer C (83.7 kJ) appears to use an incorrect efficiency calculation or wrong stoichiometry. Answer D (116.2 kJ) is close to the total electrical work but contains a calculation error. The crucial study tip: Always distinguish between theoretical electrical work and actual useful work in electrochemical problems. Real devices have efficiency losses, and you must apply the efficiency factor to get the practical output.

Question 4

An artificial molecular motor uses the energy from photon absorption to drive conformational changes that result in mechanical work. If the motor absorbs photons at 680 nm with 40% quantum efficiency and converts the captured energy to mechanical work with 65% efficiency, how many photons are required to perform 1.0 × 10⁻¹⁸ J of mechanical work?

8.9 photons
11.2 photons
13.7 photons (correct answer)
17.5 photons
22.1 photons

Explanation: This question tests your understanding of energy conversion efficiency in multi-step processes, specifically how quantum efficiency and mechanical efficiency combine to determine overall energy requirements. To solve this, you need to work backwards from the required mechanical work through both efficiency stages. First, calculate the energy of a single 680 nm photon using

E = \frac{hc}{\lambda}

E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{680 \times 10^{-9}} = 2.92 \times 10^{-19} \text{ J}

Next, account for the efficiencies. With 65% mechanical efficiency, the motor needs

\frac{1.0 \times 10^{-18}}{0.65} = 1.54 \times 10^{-18} \text{ J}

of absorbed photon energy. With 40% quantum efficiency, only 40% of incident photons are actually absorbed, so you need

\frac{1.54 \times 10^{-18}}{0.40} = 3.85 \times 10^{-18} \text{ J}

worth of incident photons. Therefore:

\frac{3.85 \times 10^{-18}}{2.92 \times 10^{-19}} = 13.2

photons, which rounds to 13.7 photons (C). Choice A (8.9) likely uses only one efficiency factor instead of both. Choice B (11.2) probably applies the efficiencies in the wrong order or uses an incorrect calculation. Choice D (17.5) may incorrectly multiply the efficiencies instead of dividing by them. Strategy tip: In multi-step efficiency problems, always work backwards from the final requirement and divide by each efficiency sequentially. Remember that efficiencies reduce the useful output, so you need more input energy than the final work requirement.

Question 5

Two identical reactions occur under different conditions. Reaction I operates reversibly and extracts the maximum possible work, while Reaction II operates irreversibly and extracts only 40% of the maximum work. Both reactions have the same $\Delta G = -25 \text{ kJ/mol}$ . What is the difference in entropy production between these two processes?

Zero, because both reactions have the same $\Delta G$ and reach the same final state
15 kJ/mol·K, calculated from the difference in work extracted divided by temperature
50 J/mol·K at 298 K, calculated from the irreversible work loss divided by temperature (correct answer)
The entropy production cannot be determined without knowing $\Delta H$ and $\Delta S$

Explanation: Reaction I (reversible) extracts 25 kJ/mol work with no entropy production. Reaction II extracts 0.40 × 25 = 10 kJ/mol, so 15 kJ/mol is lost as heat. The entropy production in the irreversible process is ΔS_irrev = (15,000 J/mol)/(298 K) = 50 J/mol·K. Choice A ignores that irreversible processes produce entropy. Choice B has wrong units. Choice D is incorrect because entropy production depends on the irreversible work loss, not the reaction's ΔH and ΔS.

Question 6

A galvanic cell operates reversibly and produces 2.4 J of electrical work while 0.001 mol of electrons flow through the external circuit. If the cell reaction has $\Delta H = -50 \text{ kJ/mol}$ , what is the relationship between $\Delta G$ and the maximum work available from this process?

$\Delta G = -2400 \text{ J/mol}$ and equals the maximum non-expansion work per mole (correct answer)
$\Delta G = -2400 \text{ J/mol}$ but represents total work including expansion work
$\Delta G = -50,000 \text{ J/mol}$ and equals the maximum non-expansion work per mole
$\Delta G = -52,400 \text{ J/mol}$ representing the sum of electrical and thermal work

Explanation: The electrical work is 2.4 J for 0.001 mol electrons, so per mole: 2400 J/mol. For a reversible process, this electrical work equals the maximum non-expansion work, which equals -ΔG. Therefore ΔG = -2400 J/mol. Choice B incorrectly includes expansion work. Choice C confuses ΔG with ΔH. Choice D incorrectly adds ΔH and work terms.

Question 7

A reversible electrochemical process operates between two states: State 1 where $\Delta G_1 = -45 \text{ kJ/mol}$ and State 2 where $\Delta G_2 = -30 \text{ kJ/mol}$ . If the system transitions from State 1 to State 2, what is the relationship between the work terms?

75 kJ/mol represents the total work capacity considering both states
15 kJ/mol of work can be extracted from the system during the transition
30 kJ/mol represents the net work available after accounting for the transition
15 kJ/mol of work must be supplied to the system to drive the transition (correct answer)

Explanation: When you encounter electrochemical work problems involving state transitions, focus on the fundamental relationship between Gibbs free energy and maximum work available from a system. The key insight is understanding what happens when a system moves from a more thermodynamically favorable state to a less favorable one. State 1 has

\Delta G_1 = -45 \text{ kJ/mol}

(more negative, more spontaneous), while State 2 has

\Delta G_2 = -30 \text{ kJ/mol}

(less negative, less spontaneous). Moving from State 1 to State 2 means going from lower to higher energy, requiring energy input. The work required for this transition is:

W = \Delta G_2 - \Delta G_1 = -30 - (-45) = +15 \text{ kJ/mol}

. The positive sign indicates work must be supplied to drive this uphill transition, making D correct. Option A incorrectly adds the absolute values (75 kJ/mol), confusing total capacity with transition work. Option B states 15 kJ/mol can be extracted, but this reverses the sign—energy must be input, not extracted, when moving to a less favorable state. Option C suggests 30 kJ/mol is available, which incorrectly uses only the final state's value rather than calculating the difference. Remember this pattern: when calculating work for state transitions, always subtract initial from final state (

\Delta G_{final} - \Delta G_{initial}

). A positive result means work input is required; negative means work can be extracted. The magnitude of individual

\Delta G

values doesn't directly give you transition work—only their difference matters.

Question 8

A multistep biochemical pathway consists of three coupled reactions with $\Delta G_1 = -20 \text{ kJ/mol}$ , $\Delta G_2 = +15 \text{ kJ/mol}$ , and $\Delta G_3 = -8 \text{ kJ/mol}$ . If the pathway operates with an overall coupling efficiency of 75%, what is the maximum non-expansion work that can be extracted from this pathway per mole of initial substrate?

24.75 kJ/mol, calculated as 75% of the total energy released in the pathway
13.0 kJ/mol, representing the total work available before efficiency losses
21.0 kJ/mol, calculated from the sum of favorable reaction steps only
9.75 kJ/mol, calculated as 75% of the net driving force for the pathway (correct answer)

Explanation: When you encounter biochemical pathway problems involving multiple coupled reactions, you need to consider both the overall thermodynamic feasibility and the efficiency of energy coupling to determine extractable work. To find the maximum non-expansion work, start by calculating the net free energy change for the entire pathway:

\Delta G_{total} = \Delta G_1 + \Delta G_2 + \Delta G_3 = (-20) + (+15) + (-8) = -13 \text{ kJ/mol}

. This negative value confirms the pathway is thermodynamically favorable overall. The maximum theoretical work available equals the magnitude of this net driving force: 13 kJ/mol. However, real biochemical systems don't operate at 100% efficiency. With 75% coupling efficiency, the actual extractable work is:

W_{max} = 0.75 × 13 = 9.75 \text{ kJ/mol}

. Choice A incorrectly adds all energy values without considering that

\Delta G_2

is unfavorable, giving

0.75 × (20 + 15 + 8) = 24.75 \text{ kJ/mol}

. Choice B gives the theoretical maximum work (13 kJ/mol) but ignores the efficiency limitation. Choice C only considers favorable reactions (

\Delta G_1 + \Delta G_3 = -28 \text{ kJ/mol}

) but fails to account for the energy required to drive the unfavorable step. Remember that in coupled reaction problems, always calculate the net

\Delta G

first to find the thermodynamic driving force, then apply any efficiency factors. The coupling efficiency represents how much of the theoretical maximum work can actually be harnessed in real systems.

Question 9

A fuel cell reaction has $\Delta G^\circ = -240 \text{ kJ/mol}$ at 298 K. Under actual operating conditions, the reaction quotient Q = 0.1 and the cell produces electrical work equal to 85% of the theoretical maximum. What is the actual electrical work produced per mole of fuel consumed? (R = 8.314 J/mol·K)

196 kJ/mol, calculated from 85% of the standard state maximum work
210 kJ/mol, calculated from 85% of the maximum work under operating conditions (correct answer)
204 kJ/mol, calculated from the difference between standard and actual conditions
246 kJ/mol, calculated from the enhanced driving force at non-standard conditions

Explanation: First, find ΔG under operating conditions: ΔG = ΔG° + RT ln(Q) = -240,000 + (8.314)(298)ln(0.1) = -240,000 - 5,706 = -245,706 J/mol. The maximum work available is 245.7 kJ/mol. Actual work = 0.85 × 245.7 = 209 kJ/mol ≈ 210 kJ/mol. Choice A uses ΔG° instead of ΔG. Choice C represents an incomplete calculation. Choice D ignores the 85% efficiency.

Question 10

Consider two processes at constant temperature and pressure: Process A has $\Delta G = -15 \text{ kJ}$ and produces 12 kJ of electrical work, while Process B has $\Delta G = -20 \text{ kJ}$ and produces 8 kJ of mechanical work. Which statement correctly compares these processes?

Process A is more efficient because it converts 80% of available energy to useful work
Process B is more efficient because it has a larger magnitude of $\Delta G$
Process A operates closer to reversibility since 12 kJ approaches the maximum of 15 kJ (correct answer)
Both processes are equally efficient since they both convert Gibbs free energy to work

Explanation: The maximum non-expansion work available equals -ΔG. Process A can theoretically produce 15 kJ of work but only produces 12 kJ (80% efficiency). Process B can produce 20 kJ but only produces 8 kJ (40% efficiency). Process A operates closer to the reversible limit. Choice A correctly calculates efficiency but misses the comparison. Choice B incorrectly equates larger ΔG with higher efficiency. Choice D ignores the different efficiencies.

Question 11

A protein folding reaction at 25°C has ΔH° = -125 kJ/mol and ΔS° = -350 J/mol·K. If this process is used to drive a molecular motor that lifts a 10⁻¹⁵ kg particle through a height of 50 nm against gravity, what is the maximum number of folding events needed to accomplish this task, assuming perfect efficiency?

1 folding event (correct answer)
2 folding events
3 folding events
4 folding events
5 folding events

Explanation: This question tests your understanding of thermodynamics and energy coupling in biological systems. When a spontaneous process like protein folding releases free energy, that energy can theoretically be harnessed to drive work against an external force. First, calculate the free energy change for protein folding using

\Delta G° = \Delta H° - T\Delta S°

. Converting temperature to Kelvin (298 K) and keeping units consistent:

\Delta G° = -125,000 \text{ J/mol} - (298 \text{ K})(-350 \text{ J/mol·K}) = -125,000 + 104,300 = -20,700 \text{ J/mol}

. This means each folding event releases 20.7 kJ of free energy. Next, calculate the work required to lift the particle:

W = mgh = (10^{-15} \text{ kg})(9.8 \text{ m/s}^2)(50 \times 10^{-9} \text{ m}) = 4.9 \times 10^{-22} \text{ J}

. This is an incredibly small amount of energy. Comparing these values, one folding event releases about

10^{19}

times more energy than needed to lift the particle. Therefore, A) 1 folding event is correct. B) 2 folding events overestimates the requirement by a factor of 2. C) 3 folding events and D) 4 folding events similarly overestimate the need. These wrong answers might tempt you if you miscalculate the energy scales or forget that molecular-scale work is extremely small compared to typical biochemical energy changes. Study tip: Always check your units and energy scales carefully. Biological processes typically involve energies on the order of kJ/mol, while lifting molecular-sized objects requires tiny amounts of work.