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On an energy profile, the activation energy is a transition state's height above the start of its step and sets the rate; the overall energy change is products less reactants and sets which side is lower; a catalyst lowers the hill but not the ends.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will read forward and reverse activation energies and the overall energy change from a profile, decide exothermic or endothermic, find each step's barrier in a two-step reaction, and say what a catalyst changes.
You can find the hills and valleys of an energy profile. Their heights carry two separate pieces of information.
The activation energy, Ea, is the height of a transition state above the species its step starts from. The overall energy change, ΔH, is the products' energy less the reactants'. Exothermic: products lower, energy released. Endothermic: products higher, energy absorbed.
Take a profile with reactants at 100 kJ/mol, transition state at 180, products at 60:
Forward and reverse barriers always differ by exactly the overall change. A very exothermic reaction can still be slow if its hill is high; a slightly endothermic one can be fast if its hill is low.
Another way: steps
From any profile:
A catalyst offers a different route with a lower transition state, so more collisions succeed and the reaction is faster. It does not touch the reactants or the products, so ΔH is unchanged — and because it lowers the forward and reverse barriers by the same amount, it speeds both directions equally and cannot move an equilibrium. Acid catalysis of alkene hydration, enzyme catalysis in cells: in every case the ends stay put and the hill comes down.
The highest point on a profile is a stable intermediate. It is a transition state.
A big energy release means a fast reaction. Rate depends on Ea, not ΔH.
Measure every barrier from the reactants. Measure it from where that step starts.
A catalyst makes a reaction more exothermic. ΔH is unchanged.
Reactants 50, transition state 140, products 90 kJ/mol.
Read the three heights.
Ea = 140 − 50 = 90; ΔH = 90 − 50 = +40: endothermic.
Two differences.
The reverse barrier is 140 − 90 = 50, smaller by exactly the 40 of ΔH.
They always differ by ΔH.
Forward activation energy?
50 kJ/mol.
Overall change, and its sign?
−90 kJ/mol: strongly exothermic, with a modest barrier.
On a one-step reaction's energy profile, the reactants sit at $220$ kJ/mol, the transition state at $270$ kJ/mol and the products at $240$ kJ/mol. What is the activation energy, in kJ/mol?
Answer:
On a one-step reaction's energy profile, the reactants sit at $230$ kJ/mol, the transition state at $390$ kJ/mol and the products at $250$ kJ/mol. What is the overall energy change, in kJ/mol? (A decrease is negative.)
Answer:
On a one-step reaction's energy profile, the reactants sit at $120$ kJ/mol, the transition state at $180$ kJ/mol and the products at $75$ kJ/mol. Is the reaction exothermic or endothermic?
On a one-step reaction's energy profile, the reactants sit at $260$ kJ/mol, the transition state at $350$ kJ/mol and the products at $185$ kJ/mol. What is the activation energy of the reverse reaction, in kJ/mol?
Answer:
An acid catalyst speeds up the hydration of propene to propan-2-ol. What does it change on the energy profile?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A two-step reaction, in kJ/mol: reactants 0, first transition state $90$, intermediate $40$, second transition state $60$, products −20. Fill in each step's activation energy.
| activation energy, kJ/mol | |
|---|---|
| step 1 | |
| step 2 |
You can read rate and energy change separately off a profile. Tell someone why a very exothermic reaction can still be slow. Next: which step of a mechanism controls how fast the whole thing goes.
8. Your turn: reactants 120, transition state 170, products 30 kJ/mol, step 3