SME
Physical Chemistry I: Thermodynamics & Kinetics
Rate laws and mechanisms
Use this for “In A → B → C, k₁ = 0.10 s⁻¹ and k₂ = 0.050 s⁻¹, starting with 1.00 M of A. Find [A], [B] and [C] after 10 s, and when [B] peaks.”
Diagram
A → B → C with k₁ = 0.1 s⁻¹ and k₂ = 0.05 s⁻¹, from [A]₀ = 1 M.
At t = 10 s: [A] = 0.368, [B] = 0.477, [C] = 0.155 M, which add to 1.
B
peaks
at
tₘₐₓ
=
ln(k₁ ÷ k₂)
k₁ − k₂
=
13.9
s,
[B]
=
0.5
M.
This is an example. Type your own values to start a new problem.
k₁
Rate constant of A → B
example · s⁻¹
k₂
Rate constant of B → C
example · s⁻¹
[A]₀
Starting concentration of A
example · M
t
Time
example · s
[A]
Concentration of A at time t
calculated · M
[B]
Concentration of B at time t
calculated · M
[C]
Concentration of C at time t
worked out (not typed) · M
tₘₐₓ
Time of the most B
calculated · s
Formulas
[A]
=
[A]₀
×
e
−
k₁
×
t
0.368
=
1
×
e
−0.1 × 10
[B]
=
[A]₀
×
k₁
×
e
−
k₁
×
t
−
e
−
k₂
×
t
k₂
−
k₁
0.477
=
1
×
0.1
×
(
e
−0.1 × 10
−
e
−0.05 × 10
)
÷
(0.05
−
0.1)
[C] = [A]₀ − [A] − [B]
0.155 = 1 − 0.368 − 0.477
tₘₐₓ
=
ln(
k₁
÷
k₂
)
k₁
−
k₂
13.9
=
ln(0.1 ÷ 0.05)
0.1 − 0.05
Assumptions
•
Both steps are first order, and only A is there at the start.
•
Nothing else is made or lost, so [A] + [B] + [C] = [A]₀ at every moment.
•
B peaks when it is made as fast as it is used, k₁[A] = k₂[B]. The formulas need k₁ ≠ k₂.
•
k₁, k₂ and t share one time unit, seconds here.
Step-by-step
WE KNOW
k₁ = 0.1 s⁻¹, k₂ = 0.05 s⁻¹, [A]₀ = 1 M, t = 10 s
FIND
concentration of A at time t ([A]), concentration of B at time t ([B]), concentration of C at time t ([C]), time of the most B (tₘₐₓ)
Step 1 · Find concentration of A at time t ([A])
Use [A] = [A]₀ × e−k₁ × t
A
only
decays,
first
order:
scale
[A]₀
by
e
raised
to
−k₁
t
.
[A]
=
1
×
e
−0.1 × 10
[A]
=
e
-1
[A] = 0.368 M
Step 2 · Find concentration of B at time t ([B])
Use [B] = [A]₀ × k₁ × (e−k₁ × t − e−k₂ × t) ÷ (k₂ − k₁)
B
is
made
from
A
at
k₁
and
used
at
k₂
:
the
gap
between
the
two
exponentials,
scaled
by
[A]₀
k₁
k₂
−
k₁
.
[B]
=
1
×
0.1
×
(
e
−0.1 × 10
−
e
−0.05 × 10
)
÷
(0.05
−
0.1)
[B]
=
0.1
×
e
-1
−
e
−0.5
0.05 − 0.1
[B]
=
0.1
×
0.36788 − 0.60653
−0.05
[B]
=
0.1
×
−0.23865
−0.05
[B]
=
−
0.023865
−0.05
[B] = 0.477 M
Step 3 · Find concentration of C at time t ([C])
Use [C] = [A]₀ − [A] − [B]
Nothing is lost: C holds whatever A and B no longer hold.
[C] = 1 − 0.368 − 0.477
[C] = 0.155 M
Step 4 · Find time of the most B (tₘₐₓ)
Use tₘₐₓ = ln(k₁ ÷ k₂) ÷ (k₁ − k₂)
Set
the
slope
of
[B]
to
zero:
k₁
e
−
k₁t
=
k₂
e
−
k₂t
,
then
solve
for
t
.
tₘₐₓ
=
ln(0.1 ÷ 0.05)
0.1 − 0.05
tₘₐₓ
=
ln(2)
0.05
tₘₐₓ
=
0.6931
0.05
tₘₐₓ = 13.9 s
Check
Put the numbers back in. Both sides should match.
0.368
=
1
×
e
−0.1 × 10
✓
0.477
=
1
×
0.1
×
(
e
−0.1 × 10
−
e
−0.05 × 10
)
÷
(0.05
−
0.1)
✓
0.155 = 1 − 0.368 − 0.477 ✓
13.9
=
ln(0.1 ÷ 0.05)
0.1 − 0.05
✓
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Ionic strength and activity (Debye–Hückel)
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Rate constant from the barrier (Eyring)
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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