SME
Organic Chemistry I · Problem type
Reading a two-step energy diagram
Use this for “Step 1 climbs 85 kJ/mol to a cation at 50 kJ/mol; step 2 climbs 15 more and ends at −30. Which step is slow, and what is ΔH?”
Diagram
T₁ = 0 + 90 = 90 kJ/mol
T₂ = 60 + 10 = 70 kJ/mol
The highest transition state is T₁, so step 1 is rate-determining.
ΔH = products − reactants = −20 − 0 = −20 kJ/mol
This is an example. Type your own values to start a new problem.
Eₐ₁
Barrier of step 1
example · kJ/mol
I
Intermediate energy
example · kJ/mol
Eₐ₂
Barrier of step 2, from I
example · kJ/mol
P
Product energy
example · kJ/mol
T₁
First transition state
calculated · kJ/mol
T₂
Second transition state
calculated · kJ/mol
T(max)
Highest transition state
worked out (not typed) · kJ/mol
ΔH
Enthalpy change
calculated · kJ/mol
Formulas
T₁
=
Eₐ₁
90 = 90
T₂
=
I
+
Eₐ₂
70 = 60 + 10
T(max)
=
greatest
of
T₁
,
T₂
90 = greatest of 90, 70
ΔH
=
P
−20 = −20
Assumptions
•
Energies are in kJ/mol with the reactants at 0; each step climbs its Eₐ from the level before it.
•
The intermediate sits in the dip between the two tops; a transition state is a top and is never isolated.
•
The step with the higher transition state is rate-determining; in SN1 it is usually the first, forming the cation.
Step-by-step
WE KNOW
Eₐ₁ = 90 kJ/mol, I = 60 kJ/mol, Eₐ₂ = 10 kJ/mol, P = −20 kJ/mol
FIND
first transition state (T₁), second transition state (T₂), highest transition state (T(max)), enthalpy change (ΔH)
Step 1 · Find first transition state (T₁)
Use T₁ = Eₐ₁
The reactants are at 0, so the first top is Eₐ₁ up.
T₁
=
90
kJ/mol
Step 2 · Find second transition state (T₂)
Use T₂ = I + Eₐ₂
Climb Eₐ₂ from the intermediate.
T₂
=
60
+
10
T₂
=
70
kJ/mol
Step 3 · Find highest transition state (T(max))
Use T(max) = greatest of T₁, T₂
The higher transition state marks the rate-determining step.
T(max) = greatest of 90, 70
T(max) = 90 kJ/mol → Step 1 has the higher top, so it is rate-determining: forming the intermediate is slow.
Step 4 · Find enthalpy change (ΔH)
Use ΔH = P
Products − reactants, with the reactants at 0.
ΔH = −20 kJ/mol → ΔH < 0: the products sit below the reactants, so heat is given off.
Check
Put the numbers back in. Both sides should match.
90 = 90 ✓
70 = 60 + 10 ✓
90 = greatest of 90, 70 ✓
−20 = −20 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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