2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
The diagram represents a mixture of X₂(g) and Y₂(g) kept in a sealed container at 298 K, when the mixture is prepared and one year later. When a spark is passed through an identical mixture, the gases react rapidly and almost completely to form XY(g). Which statement correctly describes what the diagram shows about the mixture kept at 298 K?
Answer and reasoning
AThe composition is the same at both times, so the mixture has reached equilibrium A student who takes a constant composition as proof of equilibrium picks this. An equilibrium mixture for this reaction would consist almost entirely of XY, as the sparked mixture shows. A mixture of unreacted X₂ and Y₂ is not at equilibrium; it is reacting too slowly to detect.
BOnly X₂ and Y₂ are present at both times, so no measurable reaction has occurredCorrect Both boxes contain four X₂ and four Y₂ molecules and no XY, so no measurable amount of product has formed in a year. Because the same mixture reacts almost completely once a spark starts it, the formation of XY is favored; the unreacted mixture is an example of a favored process that does not occur at a measurable rate.
CNo XY is present at either time, so the formation of XY is not favored at 298 K A student who thinks a favored reaction must be seen to happen picks this. The absence of XY shows only that the reaction is extremely slow at 298 K. A spark supplies activation energy, not favorability, and once started the reaction goes almost to completion.
DThe molecules are unchanged at both times, so X₂ and Y₂ molecules are not colliding A student who thinks every collision leads to reaction picks this. Gas molecules collide constantly; the molecules are unchanged because almost none of the collisions has enough energy to overcome the activation energy.
The energy profile shown is for a reaction that is thermodynamically favored at 298 K but does not occur at a measurable rate at that temperature. What is the value of the quantity, read from the profile, that best accounts for the absence of a measurable reaction at 298 K?
Answer and reasoning
A450 kJ/mol A student who measures the barrier from the lowest level on the profile picks this: 550 − 100 = 450 kJ/mol. That is the activation energy of the reverse reaction; for the forward reaction the barrier is measured from the reactants, 550 − 200 = 350 kJ/mol.
B550 kJ/mol A student who reads the activation energy as the axis value at the top of the profile picks this. Activation energy is a difference: the energy at the peak, 550 kJ/mol, minus the energy of the reactants, 200 kJ/mol, which is 350 kJ/mol.
C100 kJ/mol A student who thinks the energy difference between reactants and products controls the rate picks this: 200 − 100 = 100 kJ/mol. That difference is related to whether the products are favored; the rate depends on the barrier, 550 − 200 = 350 kJ/mol.
D350 kJ/molCorrect A favored reaction that does not occur at a measurable rate is under kinetic control, and a high activation energy is the common reason. The activation energy of the forward reaction is the height of the peak above the reactants: 550 kJ/mol − 200 kJ/mol = 350 kJ/mol.
Working The quantity is the activation energy of the forward reaction, Ea = E(peak) − E(reactants). From the profile, E(peak) = 550 kJ/mol and E(reactants) = 200 kJ/mol, so Ea = 550 kJ/mol − 200 kJ/mol = 350 kJ/mol.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.4.A.1 Thermodynamically favored process Fix
Thermodynamically favored process
A process for which ΔG° < 0. The sign of ΔG° shows that the products are favored; it gives no information about how fast the process occurs.
Favored processes that are not observed
Many thermodynamically favored processes do not occur to any measurable extent, or occur at extremely slow rates. Examples are the reaction of H₂(g) with O₂(g) at room temperature and the conversion of diamond to graphite.
Students often think A thermodynamically favored reaction happens readily, so if the reactants are mixed and nothing happens, the reaction is not favored under those conditions. In fact No. Many thermodynamically favored processes do not occur to any measurable extent because they are extremely slow. Whether a reaction is favored is decided by ΔG°; whether it is observed also depends on its rate.
Students often think A thermodynamically favored reaction starts by itself, so a reaction that needs a spark or heating to begin is not favored until that energy is supplied. In fact No. A spark supplies the activation energy for a small part of the mixture; it does not change ΔG°. The reaction of H₂ with O₂ is thermodynamically favored at room temperature before any spark is applied.
9.4.A.2 Kinetic control Fix
Kinetic control
A process that is thermodynamically favored but does not proceed at a measurable rate is said to be under kinetic control: its rate, not its thermodynamic favorability, decides what is observed.
High activation energy as the cause of kinetic control
A high activation energy is a common reason for kinetic control. When the activation energy is large compared with the kinetic energies of the particles, only a very small fraction of collisions has enough energy to lead to reaction, so the rate is extremely low.
No observable change is not the same as equilibrium
The fact that a process does not proceed at a noticeable rate does not mean that the system is at equilibrium. A system under kinetic control has not reached equilibrium: the favored products have not formed to the extent that equilibrium requires, because the change is too slow to detect.
Concluding that a process is under kinetic control
If a process is known to be thermodynamically favored and yet does not occur at a measurable rate, it is reasonable to conclude that it is under kinetic control. A catalyst, which provides a path with a lower activation energy, can make such a process proceed at a measurable rate without changing ΔG°; raising the temperature also increases the rate.
Students often think A mixture whose composition does not change with time is at equilibrium, with the forward and reverse reactions occurring at equal rates. In fact Not necessarily. A system at equilibrium has a constant composition, but so does a system in which a favored reaction is too slow to detect. A mixture of H₂ and O₂ at room temperature has a constant composition and is far from equilibrium.
Students often think At equilibrium the reaction has stopped, so any system in which nothing is reacting is at equilibrium. In fact No. At equilibrium the forward and reverse reactions continue at equal rates, so there is no net change. A mixture in which no particles are reacting at a measurable rate is not thereby at equilibrium.
3 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 3
A mixture of H₂(g) and O₂(g) can be kept in a sealed container at 25°C for years with no detectable formation of water. For the reaction 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔG° = −474 kJ/molrxn at 25°C. Which statement best explains why no water is detected?
Answer and reasoning
AVery few H₂–O₂ collisions have enough energy to react, so water forms too slowly to detectCorrect The reaction is thermodynamically favored (ΔG° < 0) but has a high activation energy. At 25°C only an extremely small fraction of the collisions between H₂ and O₂ molecules has enough energy to lead to reaction, so the rate is too low for any water to be detected: the reaction is under kinetic control.
BH₂O molecules form and decompose at equal rates, so the amounts of H₂ and O₂ remain constant A student who takes a constant composition as proof of equilibrium picks this. The mixture is far from equilibrium: the products are strongly favored and almost none have formed. The composition is constant because the forward reaction is extremely slow.
CH₂ and O₂ molecules in a gas are too far apart to collide, so the reaction has no way to start A student who thinks every collision leads to reaction, and so reads 'no reaction' as 'no collisions', picks this. The molecules collide extremely often; almost none of the collisions has enough energy to overcome the activation energy.
DA spark is needed to supply energy, so the reaction is favored only after it has been sparked A student who thinks a favored reaction must start without help, and so becomes favored only when energy is supplied, picks this. ΔG° is negative at 25°C for the unsparked mixture; a spark supplies activation energy to start the reaction and does not change ΔG°.
The decomposition 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g) is thermodynamically favored at 25°C, yet a solution of H₂O₂ stored in a dark bottle shows no visible change for weeks. When a small amount of MnO₂(s) is added to the solution, bubbles of O₂(g) form rapidly. Which explanation of the effect of the MnO₂(s) is correct?
Answer and reasoning
AIt makes ΔG° for the decomposition reaction more negative A student who thinks a catalyst makes a reaction more favored picks this. The reactants and products are the same with or without MnO₂, so ΔG° is unchanged; the decomposition was already favored, and only its rate changes.
BIt supplies the energy that the H₂O₂ molecules need to react A student who thinks a catalyst works as heating does picks this. MnO₂ gives no energy to the H₂O₂ molecules; it lowers the energy barrier, so that molecules with the energies they already have at 25°C can react.
CIt provides a reaction path that has a lower activation energyCorrect The decomposition is favored but under kinetic control: its activation energy is high, so it is very slow at 25°C. MnO₂ acts as a catalyst, providing a path with a lower activation energy, so a much larger fraction of the particles can react at the same temperature and the favored reaction proceeds at a measurable rate.
DIt disturbs the equilibrium that the solution had already reached A student who takes the unchanging solution to be at equilibrium picks this. The stored solution was not at equilibrium: almost all of the H₂O₂ remained although the products are favored. It was reacting too slowly to notice, and the catalyst increases the rate.
At 25°C and 1 atm, the conversion C(diamond) → C(graphite) has ΔG° = −2.9 kJ/molrxn, yet diamonds show no measurable conversion to graphite. A student explains this observation by proposing that diamond and graphite are at equilibrium. Which evaluation of the student's explanation is correct?
Answer and reasoning
ACorrect: no net change is observed, so the forward and reverse conversions must have equal rates A student who takes the absence of observable change as proof of equilibrium picks this. No change is observed because the favored conversion is immeasurably slow, not because two opposing processes are occurring at equal rates.
BIncorrect: ΔG° is so close to zero that the conversion has too little driving force to be fast A student who thinks the size of ΔG° sets the rate picks this. The rate of a process depends on its activation energy, not on the magnitude of ΔG°; reactions with ΔG° values close to zero can be fast, and reactions with very negative ΔG° values can be immeasurably slow.
CCorrect: the carbon atoms have stopped reacting, which is what being at equilibrium means A student who thinks equilibrium means that reaction has stopped picks this. At equilibrium the forward and reverse processes continue at equal rates. Diamond is not converting at a detectable rate because of a high activation energy, although the conversion is favored.
DIncorrect: the conversion is favored, but its activation energy is so high that no change is detectedCorrect ΔG° < 0, so the conversion to graphite is thermodynamically favored, and a system at equilibrium would not consist of unconverted diamond. The absence of change is explained by a very high activation energy for rearranging the bonded carbon atoms: the process is under kinetic control. A process that does not proceed at a noticeable rate is not thereby at equilibrium.
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
