5.6 Reaction Energy Profile

1. AIt falls, because energy is given out as the A–B bond stretches and begins to break.
   A student who thinks that breaking bonds releases energy picks this. Pulling bonded atoms apart requires energy, so the energy rises as the A–B bond starts to break.
2. BIt rises, because energy is absorbed as the new B–C bond starts to form.
   A student who thinks that forming a bond requires energy picks this. Bond formation releases energy; the rise toward the transition state comes from the A–B bond starting to break.
3. CIt falls, because the reaction releases energy overall, so the release starts at once.
   A student who thinks that a reaction releasing energy overall has no barrier picks this. Here the A–B bond must begin to break, so the energy first rises to the transition state; energy is released only after the peak is passed.
4. DIt rises, because energy is absorbed as the A–B bond stretches and begins to break. Correct
   Breaking a bond requires energy. On the way to the transition state the A–B bond is stretching and partly breaking, and the B–C bond is only beginning to form, so the energy of the system rises to the peak even though the reaction releases energy overall.

CED 5.6.A.1 · Read this in Fix

1. AThe time that has passed since the reactants were first mixed together
   A student who thinks the horizontal axis is time picks this. The reaction coordinate follows the rearrangement of atoms in one event; a profile says nothing about how long the reaction takes.
2. BThe fraction of the reactant molecules in the sample that have become products
   A student who reads the profile as a description of the whole sample picks this. The profile describes one elementary reaction event, not how much of a sample has reacted.
3. CThe progress of the atoms' rearrangement from the reactants into the products Correct
   The reaction coordinate is the axis along which the complex set of motions that rearranges the reactants into products is plotted: it follows how far one rearrangement has gone.
4. DThe distance that the reactant particles travel before they collide
   A student who reads 'coordinate' as a position in space picks this. The reaction coordinate is not a distance traveled; it stands for the progress of the rearrangement of the atoms.

CED 5.6.A.2 · Read this in Fix

1. AArrow 1 Correct
   The activation energy for the forward reaction is the energy difference between the reactants and the transition state, so it is the arrow from the reactant level up to the peak.
2. BArrow 2
   A student who thinks that the activation energy is the energy difference between reactants and products picks this arrow. That arrow shows the overall energy change; Eₐ is measured from the reactants up to the transition state.
3. CArrow 3
   A student who thinks that the activation energy is the energy of the transition state read from the axis picks this arrow. Eₐ is a difference: the energy of the transition state minus the energy of the reactants, not the whole height of the peak.
4. DArrow 4
   A student who thinks that the activation energy is the energy the reactants already have picks this arrow. That arrow shows the energy of the reactants; Eₐ is the further rise needed to reach the transition state.

CED 5.6.A.3 · Read this in Fix

1. AIt will be slightly lower, because the only effect of cooling is that the particles collide less often.
   A student who thinks that temperature changes the rate only through collision frequency picks this. Collisions are a little less frequent in the cooler mixture, but the main effect is that a smaller fraction of collisions are energetic enough to reach the transition state, so the rate is much lower, not slightly lower.
2. BIt will be lower, because the activation energy of the reaction is greater at the lower temperature.
   A student who thinks that Eₐ changes with temperature picks this. The rate is lower, but Eₐ is the same; fewer collisions have enough energy to reach the transition state at the lower temperature.
3. CIt will be lower, because a smaller fraction of collisions are energetic enough to reach the transition state. Correct
   Just after mixing, the solution is still well below 35 °C. At the lower temperature a smaller proportion of collisions are energetic enough to reach the transition state, so the measured initial rate is lower.
4. DIt will be about the same, because the rate constant for a reaction is the same at every temperature.
   A student who thinks that a rate constant does not depend on temperature picks this. k is constant only at a given temperature; at the lower temperature k is smaller, so the initial rate is lower.

CED 5.6.A.4 · Read this in Fix

1. A+250 kJ/mol
   A student who thinks that a reaction that releases energy has a positive energy change picks this. ΔH = 50 − 300 = −250 kJ/mol: the products are lower in energy, so the sign is negative.
2. B−250 kJ/mol Correct
   ΔH is the energy of the products minus the energy of the reactants: 50 kJ/mol − 300 kJ/mol = −250 kJ/mol. It is negative because the products are lower in energy than the reactants.
3. C+100 kJ/mol
   A student who takes the rise from the reactants to the peak as the energy change picks this, confusing the activation energy (400 − 300 = 100 kJ/mol) with ΔH, which compares the products with the reactants.
4. D−350 kJ/mol
   A student who measures the energy change from the transition state down to the products picks this: 50 − 400 = −350 kJ/mol. That drop is the activation energy of the reverse reaction; ΔH starts at the reactant level.

Working ΔH = energy of products − energy of reactants = 50 kJ/mol − 300 kJ/mol = −250 kJ/mol. Distractors: +250 kJ/mol (sign taken as positive because energy is released); +100 kJ/mol (Eₐ = 400 − 300 taken as the energy change); −350 kJ/mol (measured from the peak: 50 − 400).

CED 5.6.A.3 · Read this in Fix

1. AReaction 1, because it releases energy overall, whereas reaction 2 absorbs energy.
   A student who thinks that the reaction that releases more energy is the faster one picks this. The claim is right but the reason is not: the overall energy change does not set the rate; reaction 1 is faster because its Eₐ (100 kJ/mol) is smaller than that of reaction 2 (150 kJ/mol).
2. BReaction 1, because its rise from reactants to transition state is the smaller one. Correct
   Eₐ is measured from the reactants to the transition state: 250 − 150 = 100 kJ/mol for reaction 1 and 200 − 50 = 150 kJ/mol for reaction 2. With the same temperature, collision frequency and orientation factor, the smaller Eₐ lets a larger fraction of collisions react, so reaction 1 is faster.
3. CReaction 2, because its transition state is at a lower energy on the axis.
   A student who reads the activation energy as the height of the transition state on the axis picks this. Reaction 2's peak is lower (200 vs 250 kJ/mol), but its reactants are much lower too, so its Eₐ is 150 kJ/mol against 100 kJ/mol for reaction 1.
4. DReaction 2, because its reactants and products are closer together in energy.
   A student who thinks that Eₐ is the difference between reactants and products picks this (50 kJ/mol for reaction 2 against 100 kJ/mol for reaction 1). Eₐ is measured from the reactants to the transition state: 150 kJ/mol for reaction 2 and 100 kJ/mol for reaction 1.

Working Eₐ(1) = 250 − 150 = 100 kJ/mol; Eₐ(2) = 200 − 50 = 150 kJ/mol. With equal collision frequencies and orientation factors at the same temperature, the smaller Eₐ gives the larger fraction of collisions able to reach the transition state, so reaction 1 is faster. Distractor readings: ΔH(1) = −100 kJ/mol vs ΔH(2) = +50 kJ/mol (release of energy taken as the cause); peak heights 250 vs 200 kJ/mol read from the axis; |reactants − products| 100 vs 50 kJ/mol taken as Eₐ.

CED 5.6.A.4 · Read this in Fix

1. AThe rate of Y rises by a greater factor than the rate of X does. Correct
   The larger Eₐ is, the more steeply the fraction of collisions able to reach the transition state rises with temperature. Reaction Y has the larger Eₐ, so its rate rises by the greater factor (and Y becomes the faster reaction at 310 K).
2. BThe rate of X goes up by a greater factor than the rate of Y does.
   A student who thinks that a reaction with a smaller barrier speeds up more on heating picks this. The fraction of collisions beyond a large Eₐ grows by a larger factor on heating, so it is Y, with the larger Eₐ, whose rate rises more.
3. CEach rate doubles, because each mixture is 10 K warmer than it was.
   A student who applies the rule of thumb that rates double every 10 degrees to every reaction picks this. The factor depends on Eₐ, so the two rates do not rise by the same factor.
4. DEach rate rises by the same small factor, as collisions become more frequent.
   A student who thinks that heating speeds a reaction only by making collisions more frequent picks this. The main effect is the larger fraction of collisions able to reach the transition state, which depends on Eₐ, so the factors differ.

Working In the Arrhenius relationship k = A e^−Eₐ/RT, the factor by which k (and so the rate, at fixed concentrations) increases for a given temperature rise grows with Eₐ: the fraction of collisions able to reach the transition state grows by a larger factor when Eₐ is larger. So the rate of Y rises by a greater factor than that of X, and Y is faster at 310 K. No numerical Arrhenius calculation is needed.

CED 5.6.A.4 · Read this in Fix

1. A340 kJ/mol
   A student who thinks that a positive energy change means energy is released places the products 110 kJ/mol below the reactants and gets 230 + 110 = 340 kJ/mol. ΔH = +110 kJ/mol means the products are 110 kJ/mol above the reactants.
2. B230 kJ/mol
   A student who thinks that the forward and reverse reactions have the same activation energy picks this. They share the transition state but start from different energies, so Eₐ(reverse) = 230 − 110 = 120 kJ/mol.
3. C110 kJ/mol
   A student who thinks that the activation energy is the energy difference between reactants and products applies it to the reverse reaction and picks 110 kJ/mol. Eₐ(reverse) is measured from the products up to the transition state: 120 kJ/mol.
4. D120 kJ/mol Correct
   The transition state is 230 kJ/mol above the reactants and the products are 110 kJ/mol above the reactants, so the transition state is 230 − 110 = 120 kJ/mol above the products: Eₐ(reverse) = Eₐ(forward) − ΔH.

Working The products lie 110 kJ/mol above the reactants, and the transition state lies 230 kJ/mol above the reactants, so the transition state lies 230 − 110 = 120 kJ/mol above the products: Eₐ(reverse) = Eₐ(forward) − ΔH = 120 kJ/mol. Distractors: 340 kJ/mol (products placed 110 kJ/mol below the reactants, i.e. ΔH = +110 read as energy released); 230 kJ/mol (same Eₐ both ways); 110 kJ/mol (Eₐ taken as the reactant–product difference).

CED 5.6.A.3 · Read this in Fix

1. ABonds that are breaking are partly broken, and bonds that are forming are partly formed. Correct
   At the transition state the old bonds have weakened but not fully broken and the new bonds have started but not fully formed: this arrangement has the highest energy along the reaction coordinate.
2. BEvery bond in the reactants has broken, and the atoms are briefly separate from each other.
   A student who thinks that all reactant bonds break before new ones form picks this. At the transition state bonds are partly broken while new bonds are partly formed; separating every atom would take far more energy.
3. CThe atoms form a stable compound that could be isolated and then stored like the products.
   A student who thinks that the transition state is a stable substance picks this. It is the arrangement at the energy maximum, which the atoms pass through in an instant, so it cannot be isolated.
4. DThe colliding particles are moving faster than at any other point in the reaction.
   A student who reads the profile's energy as the kinetic energy of the particles picks this. The profile plots potential energy, which is greatest at the transition state; kinetic energy has been used to climb the barrier.

CED 5.6.A.3 · Read this in Fix