SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Rate constant from the barrier (Eyring)
Use this for “A reaction has ΔH‡ = 65.1 kJ/mol and ΔS‡ = −50 J/(mol·K). Find ΔG‡ and the rate constant at 298.15 K.”
Diagram
ΔH‡: Reactants → Enthalpy alone, +65.1 kJ/mol
−TΔS‡: Enthalpy alone → Transition state, +14.9 kJ/mol
ΔG‡ = ΔH‡ + −TΔS‡ = 65.1 + 14.9
ΔG‡ = +80 kJ/mol
This is an example. Type your own values to start a new problem.
ΔH‡
Enthalpy of activation
example · kJ/mol
ΔS‡
Entropy of activation
example · J/(mol·K)
T
Temperature
example · K
ΔG‡
Free energy of activation
calculated · kJ/mol
kBT/h
Crossing frequency
worked out (not typed) · s⁻¹
k
Rate constant
calculated · s⁻¹
Formulas
ΔG‡
=
ΔH‡
−
T
×
ΔS‡
÷
1,000
80
=
65.1
−
298.15
×
−50
1,000
kBT/h
=
1.381 ×
10
-23
×
T
6.626 ×
10
-34
6.21 ×
10
12
=
1.381 ×
10
-23
×
298.15
6.626 ×
10
-34
k
=
(kBT/h)
×
e
−ΔG‡ × 1,000 ÷ (8.314 ×
T
)
0.0597
=
6.21 ×
10
12
×
e
−80 × 1,000 ÷ (8.314 × 298.15)
Assumptions
•
Transition-state theory: reactants are in balance with the activated complex, which crosses the top at the rate kBT ÷ h.
•
ΔG‡ = ΔH‡ − TΔS‡; ΔS‡ is in J/(mol·K), so TΔS‡ is divided by 1000 for kJ/mol. A negative ΔS‡ (a tighter complex) raises the barrier.
•
kB = 1.381 × 10⁻²³ J/K, h = 6.626 × 10⁻³⁴ J·s, R = 8.314 J/(mol·K); the transmission coefficient is 1.
Step-by-step
WE KNOW
ΔH‡ = 65.1 kJ/mol, ΔS‡ = −50 J/(mol·K), T = 298.15 K
FIND
free energy of activation (ΔG‡), crossing frequency (kBT/h), rate constant (k)
Step 1 · Find free energy of activation (ΔG‡)
Use ΔG‡ = ΔH‡ − T × ΔS‡ ÷ 1,000
Take TΔS‡, in kJ/mol, from ΔH‡.
ΔG‡
=
65.1
−
298.15
×
−50
1,000
ΔG‡
=
65.1
−
−14,907.5
1,000
ΔG‡ = 65.1 − (−14.9075)
ΔG‡ = 80.008 kJ/mol
Step 2 · Find crossing frequency (kBT/h)
Use kBT/h = 1.381 × 10⁻²³ × T ÷ (6.626 × 10⁻³⁴)
How often the activated complex crosses the top each second at this temperature.
kBT/h
=
1.381 ×
10
-23
×
298.15
6.626 ×
10
-34
kBT/h
=
4.1175 ×
10
-21
6.626 ×
10
-34
kBT/h
=
6.2141 ×
10
12
s
-1
Step 3 · Find rate constant (k)
Use k = (kBT/h) × e−ΔG‡ × 1,000 ÷ (8.314 × T)
Only
the
share
e
−ΔG‡/RT
of
molecules
is
at
the
top:
multiply
the
crossing
frequency
by
it.
k
=
6.2141 ×
10
12
×
e
−80.008 × 1,000 ÷ (8.314 × 298.15)
k
=
(6.2141
×
10
12
)
×
e
−80.008 × 1,000 ÷ 2,478.8191
k
=
(6.2141
×
10
12
)
×
e
−80,008 ÷ 2,478.8191
k
=
(6.2141
×
10
12
)
×
e
−32.2767
k
=
(6.2141
×
10
12
)
×
(9.603
×
10
-15
)
k
=
0.0597
s⁻¹
Check
Put the numbers back in. Both sides should match.
80.008
=
65.1
−
298.15
×
−50
1,000
✓
6.2141 ×
10
12
=
1.381 ×
10
-23
×
298.15
6.626 ×
10
-34
✓
0.0597
=
6.2141 ×
10
12
×
e
−80.008 × 1,000 ÷ (8.314 × 298.15)
✓
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