SME
Chemical Reaction Engineering
Rate laws
Use this for “E = 80 kJ/mol and k = 0.05 per min at 300 K. Find k at 320 K”, or for E from k at two temperatures.
Diagram
ln
k
against
1
T
is
straight:
ln
k
=
ln
A
−
Eₐ/(RT).
Its
slope
through
the
two
readings
is
ln(k₂ ÷ k₁)
1
T₂
−
1
T₁
=
−9,620
K.
Eₐ = −R × slope = 8.314 J/(mol·K) × 9,620 K = 80 kJ/mol.
This is an example. Type your own values to start a new problem.
ln(
k₂ (min⁻¹)
÷
k₁ (min⁻¹)
)
=
E (kJ/mol)
÷
R
×
(1
÷
T₁ (K)
−
1
÷
T₂ (K)
)
r
Ratio k₂ ÷ k₁
calculated
Formulas
r
=
k₂
k₁
7.424
=
0.3712
0.05
ln(
r
)
=
E
8.314
×
(
1
T₁
−
1
T₂
)
ln(7.424)
=
80,000
8.314
×
(
1
300
−
1
320
)
Formulas are worked in J/mol. The step-by-step shows the conversions.
Assumptions
•
E and the pre-exponential factor A stay the same between the two temperatures.
•
Temperatures are in kelvins (°C + 273.15); R = 8.314 J/(mol·K), so E is worked in J/mol.
•
k₁ and k₂ share one unit; only their ratio sets E, whatever the order.
•
The “k doubles every 10 K” rule is rough: it holds only near one E and T.
Step-by-step
WE KNOW
E = 80 kJ/mol, k₁ = 0.05 min⁻¹, T₁ = 300 K, T₂ = 320 K
FIND
ratio k₂ ÷ k₁ (r), rate constant at T₂ (k₂)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to J/mol).
E = 80 kJ/mol = 80,000 J/mol (1 kJ/mol = 1000 J/mol)
Step 2 · Find ratio k₂ ÷ k₁ (r)
Use ln(r) = E ÷ 8.314 × (1 ÷ T₁ − 1 ÷ T₂)
Divide
k
=
A
e
−
E
÷ RT
at
T₂
by
k
at
T₁
:
A
cancels,
leaving
e
raised
to
(
E
R
)(
1
T₁
−
1
T₂
).
r
=
e^(
E
÷
8.314
×
(
1
T₁
−
1
T₂
))
r
=
e^(80,000
÷
8.314
×
(
1
300
−
1
320
))
r
=
e
80,000 ÷ 8.314 × (0.00333333 − 0.003125)
r
=
e
80,000 ÷ 8.314 × 0.00020833
r
=
e
9,622.32 × 0.00020833
r
=
e
2.00462
r
=
7.424
Step 3 · Find rate constant at T₂ (k₂)
Use r = k₂ ÷ k₁
Scale
k₁
by
how
many
times
faster
the
reaction
runs.
k₂
=
r
×
k₁
k₂
=
7.424
×
0.05
k₂
=
0.3712
min⁻¹
Check
Put the numbers back in. Both sides should match.
7.424
=
0.3712
0.05
✓
ln(7.424)
=
80,000
8.314
×
(
1
300
−
1
320
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
Arrhenius plot: E from the slope
Use this for “A plot of ln k against 1 ÷ T has slope −9622 K and intercept 29.08. Find E, A and ln k at 320 K.”