SME
Chemical Reaction Engineering · Problem type
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.”
Chart
y-intercept (0, 29.08)
This is an example. Type your own values to start a new problem.
ln k
=
ln A
+
(
m (K)
)
÷
T (K)
E (kJ/mol)
=
−(
m (K)
)
×
R
A
=
e
Formulas
ln
k
=
ln
A
+
m
T
−0.9888
=
29.08
+
−9,622
320
E
=
−
m
×
8.314
79,997 = −(−9,622) × 8.314
A
=
e
ln
A
4.2588 ×
10
12
=
e
29.08
Formulas are worked in J/mol. The step-by-step shows the conversions.
Assumptions
•
ln k = ln A − (E ÷ R)(1 ÷ T): a straight line in 1 ÷ T, so the slope is −E ÷ R.
•
R = 8.314 J/(mol·K), so E = −mR comes out in J/mol.
•
A carries k’s unit, the one k had when ln k was plotted; T is in kelvins.
Step-by-step
WE KNOW
m = −9,622 K, ln A = 29.08, T = 320 K
FIND
ln k at T (ln k), activation energy (E), pre-exponential factor (A)
Step 1 · Find ln k at T (ln k)
Use ln k = ln A + m ÷ T
Read
the
line
at
1
T
:
start
at
the
intercept
and
add
the
slope
times
1
T
.
ln
k
=
29.08
+
−9,622
320
ln k = 29.08 − 30.0688
ln k = −0.9888
Step 2 · Find activation energy (E)
Use E = −m × 8.314
The
slope
is
−
E
R
,
so
E
is
minus
the
slope
times
R.
E
=
−(−9,622)
×
8.314
E
=
79,997.308
J/mol
Step 3 · Find pre-exponential factor (A)
Use A = eˡⁿᴬ
The
intercept
is
ln
A
,
so
raise
e
to
it.
A
=
e
29.08
A
=
4.2588 ×
10
12
Step 4 · Convert the answers
Change each answer back to the unit you chose.
E = 79,997.308 J/mol = 80 kJ/mol (1 kJ/mol = 1000 J/mol)
Check
Put the numbers back in. Both sides should match.
−0.9888
=
29.08
+
−9,622
320
✓
79,997 = −(−9,622) × 8.314 ✓
4.2588 ×
10
12
=
e
29.08
✓
Topic check
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