SME
Material & Energy Balances
Reactive systems
Use this for “100 mol/h of N₂ and 250 mol/h of H₂ feed an ammonia reactor; 60% of the H₂ reacts. What leaves, and how much N₂ is in excess?”
Diagram
ξ = 50 mol/h; each species leaves at ṅ = ṅ₀ + νξ: N₂ 100 − 50 = 50; H₂ 250 − 150 = 100; NH₃ 0 + 100 = 100 mol/h.
Conversion of H₂ X = 0.6
Excess N₂ e = 20%
This is an example. Type your own values to start a new problem.
ξ (mol/h)
=
×
ṅH₂,0 (mol/h)
3
ṅN₂ (mol/h)
=
ṅN₂,0 (mol/h)
−
ξ (mol/h)
ṅH₂ (mol/h)
=
ṅH₂,0 (mol/h)
−
(3
×
ξ (mol/h)
),
ṅNH₃ (mol/h)
=
2
×
ξ (mol/h)
e
Excess N₂
calculated · %
Formulas
ξ
=
X
×
ṅH₂,0
3
50
=
0.6
×
250
3
ṅN₂
=
ṅN₂,0
−
ξ
50 = 100 − 50
ṅH₂
=
ṅH₂,0
−
3
×
ξ
100 = 250 − 3 × 50
ṅNH₃
=
2
×
ξ
100 = 2 × 50
e
=
100
×
ṅN₂,0 − ṅH₂,0 ÷ 3
ṅH₂,0 ÷ 3
20
=
100
×
100 − 250 ÷ 3
250 ÷ 3
Assumptions
•
One reaction at steady state, N₂ + 3H₂ → 2NH₃, with no NH₃ in the feed.
•
The extent ξ is the reaction’s moles per hour; each species leaves at ṅᵢ = ṅᵢ₀ + νiξ.
•
H₂ is the limiting reactant: X is its fractional conversion, and N₂ is fed in excess.
Step-by-step
WE KNOW
ṅN₂,0 = 100 mol/h, ṅH₂,0 = 250 mol/h, X = 0.6
FIND
extent of reaction (ξ), N₂ out (ṅN₂), H₂ out (ṅH₂), NH₃ out (ṅNH₃), excess N₂ (e)
Step 1 · Find extent of reaction (ξ)
Use ξ = X × ṅH₂,0 ÷ 3
The
H₂
that
reacts,
X
×
ṅH₂,0,
over
its
coefficient
3.
ξ
=
0.6
×
250
3
ξ
=
150
3
ξ
=
50
mol/h
Step 2 · Find N₂ out (ṅN₂)
Use ṅN₂ = ṅN₂,0 − ξ
ν = −1: one N₂ is used each time the reaction runs.
ṅN₂ = 100 − 50
ṅN₂ = 50 mol/h
Step 3 · Find H₂ out (ṅH₂)
Use ṅH₂ = ṅH₂,0 − 3 × ξ
ν = −3: three H₂ are used each time.
ṅH₂ = 250 − 3 × 50
ṅH₂ = 250 − 150
ṅH₂ = 100 mol/h
Step 4 · Find NH₃ out (ṅNH₃)
Use ṅNH₃ = 2 × ξ
ν = +2: two NH₃ are made each time; none was fed.
ṅNH₃ = 2 × 50
ṅNH₃ = 100 mol/h
Step 5 · Find excess N₂ (e)
Use e = 100 × (ṅN₂,0 − ṅH₂,0 ÷ 3) ÷ (ṅH₂,0 ÷ 3)
The
H₂
fed
needs
ṅH₂,0
3
of
N₂;
the
excess
is
the
N₂
beyond
that,
as
a
percent
of
it.
e
=
100
×
100 − 250 ÷ 3
250 ÷ 3
e
=
100
×
100 − 83.333
83.333
e
=
100
×
16.667
83.333
e
=
1,666.7
83.333
e
=
20%
Check
Put the numbers back in. Both sides should match.
50
=
0.6
×
250
3
✓
50 = 100 − 50 ✓
100 = 250 − 3 × 50 ✓
100 = 2 × 50 ✓
20
=
100
×
100 − 250 ÷ 3
250 ÷ 3
✓
Topic check
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