Reaction Energy Profile反应能量图

Big Idea · Transformations (TRA)AP Science Practice 3 · Representing Data and PhenomenaAP link · Unit 5~45 min

Learning goals

  • Identify the reactants, transition state (activated complex), products, activation energy, and ΔH on the energy profile of a one-step reaction
  • Find the forward and reverse activation energies from an energy profile, and convert between them with Ea(reverse) = Ea(forward) − ΔH
  • Explain qualitatively that a larger Ea means a smaller k and that a higher temperature means a larger k (the Arrhenius relationship); quantitative calculations are enrichment

Before you start

  • Endothermic and exothermic reactions (2.10.2)
  • Collision theory (2.11.1)
  • Collision model (U5.5)

How energy changes during a reaction

Lesson 2.11.1 drew energy diagrams for reactions. AP calls them reaction energy profiles反应能量图. The horizontal axis is the reaction coordinate反应进程, which tracks the progress from reactants to products; the vertical axis is the system’s potential energy. The profile of a one-step (elementary) reaction has a single “hill.”

  • Reactants and products sit at the two ends, and their energy difference is ΔH = E(products) − E(reactants). Products lower than reactants means exothermic; higher means endothermic.
  • The top of the hill is the transition state过渡态, and the arrangement of atoms there is the activated complex活化络合物: old bonds are partly broken and new bonds partly formed. It has the highest energy, lasts an extremely short time, and can’t be isolated.
  • The height from the reactants to the top is the forward activation energy (Ea)活化能, the minimum collision energy from U5.5.

Forward and reverse activation energies

The reverse reaction follows the same path in the opposite direction, so the height from the products to the top is the reverse activation energy. The three quantities are related by:

Ea(reverse)=Ea(forward)−ΔHE_a(\text{reverse}) = E_a(\text{forward}) - \Delta H

For an exothermic reaction (ΔH < 0), the reverse activation energy is larger than the forward one; for an endothermic reaction (ΔH > 0), the forward activation energy is at least ΔH. With a catalyst (U5.11), the reaction takes a pathway with a lower activation energy: the energies of the reactants and products don’t change, so ΔH stays the same and the forward and reverse activation energies drop by the same amount. Move the Ea and ΔH sliders below to see how the three change together, then add the catalyst to compare:

Try it: energy changes during a reaction

    Activation energy, temperature, and the rate constant

    U5.5 showed that the larger Ea is, the smaller the fraction of collisions with enough energy and the slower the reaction; and the higher the temperature, the larger that fraction. The Arrhenius equation阿伦尼乌斯方程 sums up both:

    k=Ae−Ea/RTk = A e^{-E_a/RT}

    A depends on collision frequency and orientation. AP asks only for a qualitative understanding:

    • at the same temperature, of two reactions with similar A, the one with the larger Ea has the smaller k and is slower;
    • for one reaction, k increases as the temperature rises;
    • the larger Ea is, the more sensitive k is to temperature.

    ΔH doesn’t appear in the equation: how much heat a reaction releases has nothing to do with how fast it is.

    Enrichment (AP doesn’t ask you to calculate with it): taking the natural log of both sides gives ln k = −(Ea/R)(1/T) + ln A, so a plot of ln k vs. 1/T is linear with slope −Ea/R. Comparing two temperatures cancels A:

    ln⁡k2k1=EaR(1T1−1T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

    For example, with Ea = 50 kJ/mol, going from 300 K to 310 K: ln(k₂/k₁) = (50000 ÷ 8.314) × (1/300 − 1/310) = 0.647, so k₂/k₁ = 1.91. This fits the rule of thumb that a 10 °C rise multiplies the rate by about 2 to 4.

    Enrichment: the Arrhenius equation (AP doesn’t ask you to calculate with it)

    The gray lines are references for Ea = 25, 75, and 125 kJ/mol. k₃₀₀ is the rate constant at 300 K; the pre-exponential factor A cancels in the ratio.

    Example 1: Reading an energy profile

    On the energy profile of a one-step reaction, the reactants are at 40 kJ/mol, the transition state at 125 kJ/mol, and the products at 15 kJ/mol. Find the forward activation energy, ΔH, and the reverse activation energy.

    Step-by-step solution

    1. Identify the three energies. Reactants 40, transition state 125, products 15 (all in kJ/mol). Activation energy is the climb from the start to the peak; ΔH is end minus start.

    2. Forward activation energy. From the reactants up to the transition state:

      Ea(fwd)=125−40=85 kJ/mol\begin{aligned} E_a(\text{fwd}) &= 125 - 40 \\ &= \boxed{85\ \text{kJ/mol}} \end{aligned}

    3. ΔH. Products minus reactants:

      ΔH=15−40=−25 kJ/mol\begin{aligned} \Delta H &= 15 - 40 \\ &= \boxed{-25\ \text{kJ/mol}} \end{aligned}

      Negative, so the reaction is exothermic.

    4. Reverse activation energy. The reverse reaction starts from the products and climbs to the same transition state:

      Ea(rev)=125−15=110 kJ/mol\begin{aligned} E_a(\text{rev}) &= 125 - 15 \\ &= \boxed{110\ \text{kJ/mol}} \end{aligned}

    5. Check. The three are related by Ea(rev) = Ea(fwd) − ΔH:

      85−(−25)=11085 - (-25) = 110

      This matches step 4. For an exothermic reaction, the reverse activation energy is always the larger one.

    Example 2: Comparing two reactions

    Compare two one-step reactions at 300 K: reaction 1 has Ea = 50 kJ/mol and ΔH = −200 kJ/mol; reaction 2 has Ea = 80 kJ/mol and ΔH = −20 kJ/mol (assume their A values are similar). Which reaction is faster? For the same temperature rise, whose k increases by the larger factor?

    Step-by-step solution

    1. Identify: what sets the rate. With similar A values, the rate is set by the activation energy, not by ΔH (how much heat is released). ΔH is only the height difference between the two ends; it doesn’t change how high the peak is.

    2. Compare Ea: which is faster? Reaction 1’s Ea of 50 kJ/mol is smaller than reaction 2’s 80 kJ/mol, so a larger fraction of collisions has enough energy and k is larger: reaction 1 is faster. Its large heat release is not the reason.

    3. Compare sensitivity to temperature. The larger Ea is, the farther out it sits in the tail of the distribution, where the area is small to begin with and heating increases it by a larger factor: reaction 2’s k increases by the larger factor.

    4. Enrichment check (not required on the AP exam). Use the Arrhenius equation to estimate the factor from 300 K to 310 K:

      k310k300=eEaR(1300−1310)\frac{k_{310}}{k_{300}} = e^{\frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{310}\right)}
      • Reaction 1 (50 kJ/mol): about 1.9 times
      • Reaction 2 (80 kJ/mol): about 2.8 times

      This agrees with step 3.

    5. Conclusion. Reaction 1 is faster; reaction 2 is more sensitive to temperature.

    Example 3: An energy profile from bond enthalpies

    Use average bond enthalpies (this site’s table, from Brown et al., Chemistry: The Central Science: H–H 436, Br–Br 193, H–Br 366 kJ/mol) to estimate ΔH for HX2(g)+BrX2(g)→2 HBr(g)\ce{H2(g) + Br2(g) -> 2HBr(g)}, and describe what its energy profile should look like.

    Step-by-step solution

    1. Identify: how ΔH relates to bond enthalpies. Breaking bonds absorbs energy and forming bonds releases it:

      ΔH=∑(broken)−∑(formed)\Delta H = \sum(\text{broken}) - \sum(\text{formed})

    2. Energy absorbed to break bonds. 1 mol of H–H and 1 mol of Br–Br (in kJ):

      436+193=629436 + 193 = 629

    3. Energy released forming bonds. 2 mol of H–Br:

      2×366=7322 \times 366 = 732

    4. ΔH.

      ΔH=629−732=−103 kJ/mol\begin{aligned} \Delta H &= 629 - 732 \\ &= \boxed{-103\ \text{kJ/mol}} \end{aligned}

      Negative, so exothermic: on the profile, the products are about 103 kJ/mol below the reactants.

    5. Connect to the profile: how high is the peak? Bond enthalpies set only the difference between the two ends, not the height of the peak. At the transition state, old bonds are only partly broken while new ones are partly formed, so Ea is usually far less than the 629 kJ needed to break the old bonds completely; Ea has to be measured. Draw reactants on the left, one peak, and products on the right about 103 kJ/mol lower.

    Practice: from each description, is the reaction endothermic or exothermic?
    1. Reactants at 20 kJ/mol, products at 60 kJ/mol
    2. Ea(forward) = 100 kJ/mol, Ea(reverse) = 60 kJ/mol
    3. Ea(forward) = 50 kJ/mol, Ea(reverse) = 120 kJ/mol
    4. ΔH = −92 kJ/mol
    5. On the profile, the products are lower than the reactants
    6. The transition state is 80 kJ/mol above the reactants and 30 kJ/mol above the products
    7. The transition state is 40 kJ/mol above the reactants and 140 kJ/mol above the products
    8. The forward activation energy is smaller than the reverse activation energy
    9. The beaker feels colder as the reaction proceeds
    10. Combustion of methane

    Common mistakes

    Mistake 1: “The more exothermic a reaction, the faster it is.” The rate depends on the activation energy, not ΔH. A very exothermic reaction can be extremely slow if Ea is large; wood doesn’t burn in air at room temperature, for example.

    Mistake 2: “The transition state is a substance that can be isolated.” The transition state is the highest point on the profile and lasts an extremely short time, so it can’t be isolated. In a multistep reaction, the “valleys” on the profile are the intermediates (U5.10).

    Mistake 3: “Raising the temperature (or a flame or spark) lowers the activation energy.” Temperature changes neither Ea nor the shape of the profile; heating lets more collisions reach Ea. A flame or spark supplies energy so that some molecules reach Ea; it doesn’t lower Ea either. A catalyst is what lowers Ea (U5.11).

    Quick check

    On the energy profile of a one-step reaction, the reactants are at 60 kJ/mol, the transition state at 150 kJ/mol, and the products at 20 kJ/mol. What is the forward activation energy?

    A one-step reaction has Ea(forward) = 75 kJ/mol and ΔH = −30 kJ/mol. What is the reverse activation energy?

    Which statement about the transition state (activated complex) is correct?

    Two reactions at 300 K have similar A values; reaction 1 has Ea = 40 kJ/mol and reaction 2 has Ea = 60 kJ/mol. Which has the larger rate constant?

    Raising the temperature increases the rate constant of a one-step reaction. Which explanation is best?

    A one-step endothermic reaction has ΔH = +60 kJ/mol. Which could be its forward activation energy?

    A mixture of H₂ and O₂ can be stored for a long time at room temperature, but reacts rapidly when a spark passes through it. What does the spark do?

    When a catalyst is added to a one-step reaction, which set of changes is correct?

    AP-style free response

    AP-style free response

    On the energy profile of the one-step reaction A + B → C + D, the reactants are at 30 kJ/mol, the transition state at 110 kJ/mol, and the products at 70 kJ/mol.

    (a) Find the forward activation energy and ΔH, and state whether the reaction is endothermic or exothermic.

    (b) Find the reverse activation energy.

    (c) At the same temperature, assuming similar A values for the forward and reverse reactions, which direction has the larger rate constant? Justify your answer.

    (d) Describe the particles at the transition state.

    (e) A student says, “Raising the temperature lowers this reaction’s activation energy, so the reaction speeds up.” Evaluate the claim.

    Model answer and scoring

    (a) Ea(forward) = 110 − 30 = 80 kJ/mol; ΔH = 70 − 30 = +40 kJ/mol, endothermic.

    (b) Ea(reverse) = 110 − 70 = 40 kJ/mol (or 80 − 40 = 40 kJ/mol).

    (c) The reverse reaction. Its activation energy (40 kJ/mol) is smaller than the forward one (80 kJ/mol), so a larger fraction of collisions has enough energy, and k is larger.

    (d) The old bonds between A and B are partly broken and the new bonds forming C and D are partly formed; the system is at its highest potential energy, and this arrangement is extremely unstable and short-lived.

    (e) The claim is incorrect. Temperature changes neither the activation energy nor the profile; heating shifts the molecules’ energy distribution right, so a larger fraction of collisions has energy at least Ea, which speeds the reaction up.

    Scoring: (a) values, sign, and endothermic; (b) 40 kJ/mol; (c) compares activation energies and links them to the fraction of collisions with enough energy; (d) bonds partly broken and partly formed; (e) states that Ea is unchanged and the fraction with enough energy rises.

    Summary

    • A one-step energy profile has one peak: the top is the transition state (activated complex); the height from the reactants to the top is Ea(forward), from the products to the top is Ea(reverse), and the difference between the ends is ΔH.
    • Ea(reverse) = Ea(forward) − ΔH; exothermic reactions have the larger activation energy in the reverse direction.
    • The Arrhenius relationship (qualitative): the larger Ea, the smaller k; the higher the temperature, the larger k; the larger Ea, the more sensitive k is to temperature. The rate doesn’t depend on ΔH.
    • Temperature doesn’t change the energy profile, only the fraction of collisions with enough energy; a catalyst lowers the forward and reverse activation energies by the same amount and leaves ΔH unchanged.
    • Bond enthalpies can estimate ΔH (the difference between the two ends of the profile), but they don’t give the activation energy.

    AP tip: When reading a profile, first mark the energies of the reactants, transition state, and products, then take differences; remember that the reverse activation energy is measured from the products. When comparing rates, look only at Ea and don’t be misled by ΔH.

    Practice this chapter: Unit 5 · Kinetics · 12 questionsPractice this topic: U5.6 · Reaction Energy Profile · 5 questions

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