SME
Inorganic Chemistry
Organometallic chemistry and catalysis
Use this for “A hydrogenation of 0.800 mol of alkene uses 0.050 mol% of catalyst and reaches 95% conversion in 2.5 h. Find the TON and the TOF.”
Chart
This is an example. Type your own values to start a new problem.
nS
Substrate
example · mol
xcat
Catalyst loading (mol%)
example · %
ncat
Catalyst
calculated · mol
X
Conversion
example · %
nP
Product
calculated · mol
TON
Turnover number
calculated
t
Reaction time
example · h
TOF
Turnover frequency
calculated · h⁻¹
Formulas
ncat
=
nS
×
xcat
100
5.00 ×
10
-4
=
0.5
×
0.1
100
nP
=
nS
×
X
100
0.46
=
0.5
×
92
100
TON
=
nP
ncat
920
=
0.46
5.00 ×
10
-4
TOF
=
TON
t
230
=
920
4
Assumptions
•
Every catalyst molecule is active for the whole run and none of it decomposes.
•
Each mole of substrate converted gives one mole of product, with no side products.
•
The TOF is averaged over the whole run; the rate at the start is usually higher.
Step-by-step
WE KNOW
nS = 0.5 mol, xcat = 0.1%, X = 92%, t = 4 h
FIND
catalyst (ncat), product (nP), turnover number (TON), turnover frequency (TOF)
Step 1 · Find catalyst (ncat)
Use ncat = nS × xcat ÷ 100
Take the loading, as a percent of the substrate’s moles.
ncat
=
0.5
×
0.1
100
ncat
=
0.05
100
ncat
=
5.00 ×
10
-4
mol
Step 2 · Find product (nP)
Use nP = nS × X ÷ 100
The conversion is the percent of the substrate turned into product.
nP
=
0.5
×
92
100
nP
=
46
100
nP = 0.46 mol
Step 3 · Find turnover number (TON)
Use TON = nP ÷ ncat
Moles of product made for each mole of catalyst: how many times each catalyst molecule turned over.
TON
=
0.46
5.00 ×
10
-4
TON = 920
Step 4 · Find turnover frequency (TOF)
Use TOF = TON ÷ t
Turnovers per hour, averaged over the run.
TOF
=
920
4
TOF = 230 h⁻¹
Check
Put the numbers back in. Both sides should match.
5.00 ×
10
-4
=
0.5
×
0.1
100
✓
0.46
=
0.5
×
92
100
✓
920
=
0.46
5.00 ×
10
-4
✓
230
=
920
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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