SME
Chemical Reaction Engineering · Problem type
Reaction order from initial rates
Use this for “Doubling CA from 0.25 to 0.5 mol/L raises the initial rate from 0.002 to 0.008 mol/(L·s). Find the order n and k.”
Diagram
−rA
=
k
×
(CA)ⁿ
=
0.032
×
CA
2
.
r₂
r₁
=
0.008
0.002
=
4
=
(
0.5
0.25
)
2
.
The curve starts at 0: no reactant, no rate.
This is an example. Type your own values to start a new problem.
r₂ (mol/(L·s))
÷
r₁ (mol/(L·s))
=
(
CA2 (mol/L)
÷
CA1 (mol/L)
)
r₁ (mol/(L·s))
=
k
×
(
CA1 (mol/L)
)
Formulas
r₂
r₁
=
(
CA2
CA1
)
n
0.008
0.002
=
(
0.5
0.25
)
2
r₁
=
k
×
(CA1)
n
0.002
=
0.032
×
(0.25)
2
r₂
=
k
×
(CA2)
n
0.008
=
0.032
×
(0.5)
2
Assumptions
•
The rate law is −rA = kCAⁿ; anything else in it is held the same in both runs.
•
Each rate is an initial rate, read before CA has changed much.
•
k’s unit depends on n: (L/mol)ⁿ⁻¹ per second, so L/(mol·s) when n = 2.
Step-by-step
WE KNOW
CA1 = 0.25 mol/L, r₁ = 0.002 mol/(L·s), CA2 = 0.5 mol/L, r₂ = 0.008 mol/(L·s)
FIND
reaction order (n), rate constant (k)
Step 1 · Find reaction order (n)
Use r₂ ÷ r₁ = (CA2 ÷ CA1)ⁿ
Divide
one
run’s
rate
law
by
the
other:
k
cancels.
Take
ln
of
both
sides
and
divide.
n
=
ln(
r₂
÷
r₁
)
ln(CA2 ÷ CA1)
n
=
ln(0.008 ÷ 0.002)
ln(0.5 ÷ 0.25)
n
=
ln(4)
ln(2)
n
=
1.386
0.6931
n
=
2
Step 2 · Find rate constant (k)
Use r₁ = k × (CA1)ⁿ
Put
run
₁
into
the
rate
law:
divide
its
rate
by
its
concentration
raised
to
the
order.
With
n
=
2,
k
is
in
L/(mol·s).
k
=
r₁
÷
(CA1)
n
k
=
0.002
÷
(0.25)
2
k
=
0.002
0.0625
k
=
0.032
Check
Put the numbers back in. Both sides should match.
0.008
0.002
=
(
0.5
0.25
)
2
✓
0.002
=
0.032
×
(0.25)
2
✓
0.008
=
0.032
×
(0.5)
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.