SME
Chemical Reaction Engineering
Multiple reactions
Use this for “A → B → C with k₁ = 0.3 per min and k₂ = 0.1 per min, fed at 1.5 mol/L of A. When should the batch stop to get the most B, and how much B is that?”
Diagram
A → B → C with k₁ = 0.5 h⁻¹ and k₂ = 0.2 h⁻¹, from CA0 = 2 mol/L.
CA + CB + CC = CA0 at every t (the dotted line).
B
peaks
at
tₘₐₓ
=
ln(k₁ ÷ k₂)
k₁ − k₂
=
3.05
h,
CB
=
1.09
mol/L.
This is an example. Type your own values to start a new problem.
tₘₐₓ (h)
=
ln(
k₂ (h⁻¹)
÷
k₁ (h⁻¹)
)
÷
(
k₂ (h⁻¹)
−
k₁ (h⁻¹)
)
CB,max (mol/L)
=
CA0 (mol/L)
×
k₁ (h⁻¹)
k₂ (h⁻¹)
h⁻¹
÷
(
h⁻¹
−
h⁻¹
)
Y
Yield of B at the peak
calculated · %
Formulas
tₘₐₓ
=
ln(
k₂
÷
k₁
)
k₂
−
k₁
3.0543
=
ln(0.2 ÷ 0.5)
0.2 − 0.5
CB,max
=
CA0
×
(
k₁
k₂
)
k₂
÷ (
k₂
−
k₁
)
1.0858
=
2
×
(
0.5
0.2
)
0.2 ÷ (0.2 − 0.5)
Y
=
100
×
CB,max
CA0
54.2884
=
100
×
1.0858
2
Assumptions
•
Both steps are first order, in a batch reactor or a plug-flow reactor.
•
The feed holds no B; t is the batch time or the PFR residence time, in the time unit of k₁ and k₂.
•
B peaks when it is made as fast as it is used; with k₁ = k₂, tₘₐₓ = 1 ÷ k and CB,max = CA0 ÷ e.
Step-by-step
WE KNOW
k₁ = 0.5 h⁻¹, k₂ = 0.2 h⁻¹, CA0 = 2 mol/L
FIND
time of the most B (tₘₐₓ), most B (CB,max), yield of B at the peak (Y)
Step 1 · Find time of the most B (tₘₐₓ)
Use tₘₐₓ = ln(k₂ ÷ k₁) ÷ (k₂ − k₁)
B
stops
rising
when
k₁CA
=
k₂CB;
with
both
exponentials
in,
that
time
is
ln(
k₂
÷
k₁
)
k₂
−
k₁
.
0.5 ≠ 0.2: different rate constants
tₘₐₓ
=
ln(0.2 ÷ 0.5)
0.2 − 0.5
tₘₐₓ
=
ln(0.4)
−0.3
tₘₐₓ
=
−
0.91629
−0.3
tₘₐₓ = 3.0543 h
Step 2 · Find most B (CB,max)
Use CB,max = CA0 × (k₁ ÷ k₂)k₂ ÷ (k₂ − k₁)
Raise
k₁
k₂
to
the
power
k₂
k₂
−
k₁
for
the
share
of
the
feed
that
is
B
at
its
peak,
then
scale
by
CA0.
CB,max
=
2
×
(
0.5
0.2
)
0.2 ÷ (0.2 − 0.5)
CB,max
=
2
×
2.5
−0.66667
CB,max = 2 × 0.54288
CB,max = 1.085767 mol/L
Step 3 · Find yield of B at the peak (Y)
Use Y = 100 × CB,max ÷ CA0
The share of the fed A that is B at the peak, as a percent.
Y
=
100
×
1.085767
2
Y
=
108.5767
2
Y
=
54.2884%
Check
Put the numbers back in. Both sides should match.
3.0543
=
ln(0.2 ÷ 0.5)
0.2 − 0.5
✓
1.085767
=
2
×
(
0.5
0.2
)
0.2 ÷ (0.2 − 0.5)
✓
54.2884
=
100
×
1.085767
2
✓
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Instantaneous selectivity of parallel reactions
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