2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
At 25°C, a reaction has ΔG° = +120 kJ/molrxn, and when its reactants are mixed in the laboratory almost no product forms. Which statement about obtaining a useful amount of product at 25°C, without coupling the reaction to any other reaction, is correct?
Answer and reasoning
AEnergy must be supplied from an external source for as long as product is formingCorrect A positive ΔG° means the reaction is thermodynamically unfavorable, not impossible. An external energy source, such as electrical energy or light, can drive it, but only while that energy is being supplied.
BNo method can produce a useful amount of the product, because the reaction's ΔG° is positive A student who thinks an unfavorable reaction can never be made to occur picks this. Electrolysis, the charging of a battery and photosynthesis are all unfavorable processes driven by an external energy source.
CAdding a suitable catalyst will give a useful amount, because it lowers the activation energy A student who thinks lowering the activation energy can make an unfavorable reaction form product picks this. A catalyst speeds up the forward and reverse reactions equally and leaves ΔG° and K unchanged.
DHeating the mixture briefly will start the reaction, which then continues by itself at 25°C A student who thinks energy is needed only to start an unfavorable reaction picks this. That is true of a favored reaction with a high activation energy; an unfavored reaction stops as soon as the energy input stops.
The table gives ΔG° for two reactions at 298 K. In a coupled system, the two reactions share the intermediate Q, and the conversion of W to V drives the formation of XY according to the overall equation X + Y + 3 W → XY + 3 V. Based on the table, what is ΔG° for the overall reaction at 298 K?
Answer and reasoning
A+22 kJ/molrxn A student who thinks ΔG° does not change when a reaction's coefficients are multiplied picks this: +40 + (−18) = +22 kJ/molrxn. Three W react, so the second reaction contributes 3(−18) = −54 kJ/molrxn.
B−14 kJ/molrxnCorrect The overall equation is X + Y → XY + 3 Q plus three times W + Q → V, with the intermediate Q canceling, so ΔG° = +40 + 3(−18) = −14 kJ/molrxn. The overall coupled reaction is thermodynamically favored.
C−94 kJ/molrxn A student who thinks coupling makes the formation of XY itself favorable, changing its ΔG° to −40 kJ/molrxn, picks this: −40 + 3(−18) = −94 kJ/molrxn. The reaction that forms XY keeps ΔG° = +40 kJ/molrxn; only the sum is negative.
D+40 kJ/molrxn A student who thinks the favorable reaction only supplies energy and is not part of the overall reaction picks this. The overall equation includes the conversion of 3 W to 3 V, so its ΔG° is +40 + 3(−18) = −14 kJ/molrxn.
Working The overall equation is (X + Y → XY + 3 Q) + 3 × (W + Q → V); the 3 Q formed in the first reaction are used up in the second and cancel, giving X + Y + 3 W → XY + 3 V. ΔG° = (+40 kJ/molrxn) + 3(−18 kJ/molrxn) = +40 − 54 = −14 kJ/molrxn.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.7.A.1 Thermodynamically unfavorable process Fix
Thermodynamically unfavorable process
A process with ΔG° > 0: under standard conditions reactants are favored at equilibrium (K < 1). Energy supplied from outside the system, or coupling to a favorable process, can drive it further toward products.
External source of energy
Energy supplied to a reacting system from outside it, such as electrical energy from a power supply or light from the Sun or a lamp. A thermodynamically unfavorable process driven this way continues only while the energy is being supplied.
Electrolytic cell
An electrochemical cell in which electrical energy from an external power supply drives a thermodynamically unfavorable redox reaction, such as the decomposition of water into H₂(g) and O₂(g).
Charging a battery
Using electrical energy from an external source to drive a battery's cell reaction in the reverse, thermodynamically unfavorable direction, regenerating the reactants that the favorable discharge reaction uses.
Photosynthesis as a light-driven process
The overall conversion of carbon dioxide and water to glucose and oxygen, 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂, has a large positive ΔG°; in plants it occurs only while light supplies the energy.
Students often think A reaction with a positive ΔG° can never be made to form a useful amount of product by any method. In fact Yes. A positive ΔG° means the process is thermodynamically unfavorable on its own, not that it is impossible. Energy supplied continuously from an external source, or coupling to a favorable reaction, can make it form product.
Students often think Lowering the activation energy, for example with a catalyst or with an electric current acting like one, is enough to make a thermodynamically unfavorable reaction form product. In fact No. A catalyst lowers the activation energy and speeds up both the forward and reverse reactions, but it does not change ΔG° or K, so it cannot make an unfavorable reaction give a useful amount of product.
9.7.A.2 Coupled reactions Fix
Coupled reactions
A thermodynamically unfavorable reaction that makes a desired product is linked to a favorable reaction so that the overall reaction, the sum of the two, has ΔG° < 0 and forms the desired product.
Common intermediate
A species formed in one of the coupled reactions and used up in the other, so it cancels from the overall equation. Sharing one or more intermediates is what links the two reactions; two reactions merely occurring in the same container are not coupled.
Adding ΔG° values of coupled reactions
When reactions are added to give an overall reaction, their ΔG° values are added; a reaction used c times contributes c × ΔG°, and a reaction written in reverse contributes −ΔG°.
Conversion of ATP to ADP
In biological systems, the thermodynamically favorable conversion of ATP to ADP is coupled to unfavorable reactions, sharing intermediates with them, so that the overall reactions have ΔG° < 0.
Students often think Coupling changes the unfavorable reaction itself, so its ΔG° becomes negative, or its ΔG° becomes the negative ΔG° of the coupled system. In fact No. The unfavorable reaction keeps its own positive ΔG°. It is the overall reaction, the sum of the unfavorable and favorable reactions, that has ΔG° < 0.
Students often think In a coupled system the favorable reaction only supplies energy and is not part of the overall reaction, so the overall ΔG° equals the ΔG° of the reaction that forms the desired product. In fact No. The overall reaction is the sum of both reactions, so it includes the reactants and products of the favorable reaction as well, and its ΔG° is the sum of the two ΔG° values.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
An electrolytic cell containing aqueous Na₂SO₄ is used to carry out the reaction 2 H₂O(l) → 2 H₂(g) + O₂(g), for which ΔG° = +474 kJ/molrxn at 25°C. The graph shows the volume of H₂(g) collected at one electrode; the power supply was disconnected at the time marked by the dashed line. Which statement best explains the shape of the graph after that time?
Answer and reasoning
AThe decomposition of water had reached equilibrium by the time marked on the graph A student who thinks a reaction that stops has reached equilibrium picks this. With ΔG° = +474 kJ/molrxn, K is extremely small, so a mixture containing H₂ and O₂ is far from equilibrium; the gas stopped forming because the energy input stopped.
BThe Na₂SO₄ dissolved in the cell had all been used up by the time marked A student who thinks the dissolved electrolyte is the substance consumed in electrolysis picks this. The H₂ and O₂ come from water, as the equation shows; Na₂SO₄ lets the solution conduct and is not used up.
CThe current had been acting as a catalyst that lowered the activation energy A student who thinks lowering the activation energy can drive an unfavorable reaction picks this. Electrical energy is an energy input that drives the reaction; a catalyst could not make a reaction with ΔG° = +474 kJ/molrxn form product.
DThe unfavorable decomposition of water occurs only while energy is being suppliedCorrect The decomposition of water has a large positive ΔG°. Electrical energy from the power supply drives it, so H₂ forms only while current flows; once the supply is disconnected the volume stays constant.
A student places a sprig of an aquatic plant in water containing dissolved CO₂ under a lamp and collects the O₂(g) produced, 6 CO₂(aq) + 6 H₂O(l) → C₆H₁₂O₆(aq) + 6 O₂(g), for 30 minutes. In a second trial with an identical plant, the lamp is on for the first 10 minutes and the flask is then wrapped in foil that blocks all light for the remaining 20 minutes. How will the volume of O₂ collected in the second trial compare with the first, and why?
Answer and reasoning
AThe same volume is collected, because the light was needed only to start the formation of glucose A student who thinks an external energy source is needed only to start an unfavorable process picks this. Photosynthesis stops when the light is removed, so less O₂ is collected.
BLess O₂ is collected, because light acts as a catalyst that speeds up a thermodynamically favored reaction A student who thinks photosynthesis is a favored reaction that light merely speeds up picks this. The prediction of less O₂ is right, but the reason is wrong: the reaction is unfavorable, and light is the energy input that drives it.
CLess O₂ is collected, because forming glucose from CO₂ is unfavorable and needs the energy of lightCorrect The overall conversion of CO₂ and water to glucose and O₂ has a large positive ΔG°. Light supplies the energy that drives it, so O₂ is produced only during the first 10 minutes of the second trial.
DThe same volume is collected, because breaking the bonds in CO₂ and H₂O releases the energy to make glucose A student who thinks energy is released when bonds break picks this, and so sees the reactants as the energy source. Breaking the bonds in CO₂ and H₂O requires energy; in the dark nothing supplies it, so O₂ forms only during the first 10 minutes.
A biology text states: "ATP stores energy in a high-energy phosphate bond, and this energy is released when the bond is broken, driving reactions coupled to the conversion of ATP to ADP." Which statement best evaluates how well this description represents the particulate-level origin of the energy released?
Answer and reasoning
AIt is accurate, because energy is stored in the P–O bond and is released when that bond is broken A student who thinks energy is stored in bonds and released when they break picks this. Bond breaking is endothermic; energy is released only as bonds and attractions form.
BIt is accurate, because the P–O bond in ATP is unusually strong and so releases more energy as it breaks A student who reads 'high-energy bond' as 'strong bond' picks this. A stronger bond needs more energy to break, not less; the term refers to the overall reaction's large negative ΔG°.
CIt is misleading, because breaking a bond absorbs energy; the release comes from new bonds and interactions formingCorrect Breaking any bond, including the P–O bond in ATP, requires energy. The conversion of ATP to ADP is favorable overall because the bonds and interactions formed in the products are stronger overall than those broken, so the description misplaces the source of the energy.
DIt is misleading, because breaking a bond absorbs energy, so converting ATP to ADP is unfavorable overall A student who applies 'bond breaking absorbs energy' to the whole reaction picks this. The conversion of ATP to ADP is favorable overall, which is why it can be coupled to unfavorable reactions; the products' new bonds and interactions must be counted.
A student knows ΔG° at 298 K for the thermodynamically unfavorable reaction X + Y → XY + Q. The student wants to decide whether coupling it to the reaction W + Q → V, with which it shares the intermediate Q, gives an overall reaction, X + Y + W → XY + V, that is thermodynamically favored under standard conditions at 298 K. Which additional quantity does the student need?
Answer and reasoning
AΔG° for the reaction W + Q → VCorrect The overall reaction is the sum of the two reactions, so its ΔG° is the sum of their ΔG° values. With ΔG° for X + Y → XY + Q known, ΔG° for W + Q → V at 298 K is all that is needed to find the sign of the overall ΔG°.
BΔH° of the reaction W + Q → V A student who thinks the sign of ΔH° decides favorability picks this. Favorability under standard conditions depends on ΔG° = ΔH° − TΔS°; ΔH° alone, without ΔS°, cannot give ΔG°.
CActivation energy for W + Q → V A student who thinks the activation energy decides whether a reaction is favored picks this. Activation energy affects the rate, not the sign of ΔG° for the overall reaction.
DInitial concentration of W added A student who thinks ΔG° and K depend on the amount of reactant used picks this. ΔG° refers to standard conditions and does not change with the amount of W used.
Working The Q formed in the first reaction is used up in the second and cancels, so the coupled reactions add to X + Y + W → XY + V, and ΔG°(overall) = ΔG°(X + Y → XY + Q) + ΔG°(W + Q → V). The first is known, so the only quantity needed is ΔG° for W + Q → V at 298 K; the overall reaction is favored under standard conditions if the sum is negative.
The diagram shows standard free energies for a coupled system in which the conversion of WP to W and P drives the formation of XY through the intermediate XP. For the reaction WP → W + P, ΔG° = −38 kJ/molrxn. Based on the diagram, what is ΔG° for the reaction X + Y → XY?
Answer and reasoning
A−20 kJ/molrxn A student who thinks coupling gives the formation of XY itself the coupled system's negative ΔG° picks this. −20 kJ/molrxn is ΔG° for the overall reaction, which includes WP → W + P; X + Y → XY on its own is unfavorable.
B−58 kJ/molrxn A student who removes WP → W + P from the overall reaction without changing the sign of its ΔG° picks this: −20 + (−38) = −58 kJ/molrxn. Removing a reaction means adding its reverse, ΔG° = +38 kJ/molrxn.
C+38 kJ/molrxn A student who thinks the favorable reaction hands over exactly the free energy the unfavorable reaction needs picks this. The overall ΔG° is −20 kJ/molrxn, not zero, so ΔG° for forming XY is +18 kJ/molrxn, smaller in magnitude than the −38 kJ/molrxn of WP → W + P.
D+18 kJ/molrxnCorrect The diagram gives ΔG° = −20 kJ/molrxn for the overall reaction X + Y + WP → XY + W + P. Subtracting WP → W + P (adding its reverse, +38 kJ/molrxn) leaves X + Y → XY with ΔG° = −20 + 38 = +18 kJ/molrxn, an unfavorable reaction.
Working From the diagram, the overall coupled reaction X + Y + WP → XY + W + P has ΔG° = −20 − 0 = −20 kJ/molrxn. X + Y → XY = (overall) + (reverse of WP → W + P): ΔG° = −20 kJ/molrxn + (+38 kJ/molrxn) = +18 kJ/molrxn.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
