SME
Materials Science & Material Properties · Problem type
Diffusion coefficient at a temperature
Use this for “Carbon in iron has D₀ = 2.3 × 10⁻⁵ m²/s and Qd = 148 kJ/mol. Find D at 900 °C.”
Chart
D(1000/T) = 2.3 × 10⁻⁵·e−148/8.314·1000/T
A log y-axis: an exponential draws as a straight line.
This is an example. Type your own values to start a new problem.
D (m²/s)
=
D₀ (m²/s)
×
e
−
kJ/mol
×
1000
÷
(8.314
×
K
)
Formulas
D
=
D₀
×
e
−Qd × 1,000 ÷ (8.314 ×
T
)
5.9011 ×
10
-12
=
2.3 ×
10
-5
×
e
−148 × 1,000 ÷ (8.314 × 1,173)
Assumptions
•
D₀ and Qd belong to one diffusing species in one host, read from a table for that pair.
•
R = 8.314 J/(mol·K), so Qd in kJ/mol is multiplied by 1000.
•
T is in kelvins (°C + 273): 900 °C is 1173 K.
•
Plotted on a log axis against 1 ÷ T, D falls on a straight line of slope −Qd ÷ R.
Step-by-step
WE KNOW
D₀ = 2.3 × 10⁻⁵ m²/s, Qd = 148 kJ/mol, T = 1,173 K
FIND
diffusion coefficient (D)
Step 1 · Find diffusion coefficient (D)
Use D = D₀ × e−Qd × 1,000 ÷ (8.314 × T)
Only
the
share
e
−Qd ÷ RT
of
atom
jumps
clears
the
energy
barrier.
Scale
D₀
by
that
share.
D
=
2.3 ×
10
-5
×
e
−148 × 1,000 ÷ (8.314 × 1,173)
D
=
(2.3
×
10
-5
)
×
e
−148,000 ÷ 9,752.322
D
=
(2.3
×
10
-5
)
×
e
−15.1759
D
=
(2.3
×
10
-5
)
×
(2.5656
×
10
-7
)
D
=
5.9011 ×
10
-12
m
2
/s
Check
Put the numbers back in. Both sides should match.
5.9011 ×
10
-12
=
2.3 ×
10
-5
×
e
−148 × 1,000 ÷ (8.314 × 1,173)
✓
Topic check
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