SME
Organic Chemistry I · Problem type
How the rate changes with [RX] and [Nu]
Use this for “In an SN2 reaction [RX] is halved and [Nu] is made 4 times larger. How does the rate change?”
This is an example. Type your own values to start a new problem.
fᵣₐₜₑ
=
fRX
×
fNu
Formulas
fᵣₐₜₑ
=
fRX
×
fNu
p
6
=
2
×
3
1
Assumptions
•
SN2 is one step with both in it, so rate = k[RX][Nu] and p = 1; the slow step of SN1 is RX alone, so rate = k[RX] and p = 0.
•
Each factor is the new concentration ÷ the old one: 2 means doubled, 0.5 halved.
•
The temperature and the solvent stay the same, so k does not change.
Step-by-step
WE KNOW
p = 1, fRX = 2, fNu = 3
FIND
factor on the rate (fᵣₐₜₑ)
Step 1 · Find factor on the rate (fᵣₐₜₑ)
Use fᵣₐₜₑ = fRX × fNuᵖ
Multiply the [RX] factor by the [Nu] factor raised to the order in the nucleophile.
fᵣₐₜₑ
=
2
×
3
1
fᵣₐₜₑ = 2 × 3
fᵣₐₜₑ = 6 → SN2: both meet in its one step, so the rate follows each concentration.
Check
Put the numbers back in. Both sides should match.
6
=
2
×
3
1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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