SME
Chemical Engineering Thermodynamics · Problem type
Equilibrium conversion of A ⇌ B
Use this for “An isomerization A ⇌ B has K = 0.65 at the reactor temperature. Pure A enters at 1.5 mol/L. Find the equilibrium conversion X.”
Diagram
At equilibrium Q = K = 3: [A] = 0.5 M, [B] = 1.5 M.
This is an example. Type your own values to start a new problem.
K
=
1
−
CA (mol/L)
=
CA0 (mol/L)
×
(1
−
X
)
CB (mol/L)
=
CA0 (mol/L)
×
X
Formulas
K
=
X
1 −
X
3
=
0.75
1 − 0.75
CA
=
CA0
×
(1
−
X
)
0.5 = 2 × (1 − 0.75)
CB
=
CA0
×
X
1.5 = 2 × 0.75
Assumptions
•
One mole of A makes one mole of B, so the total moles and the volume do not change.
•
Ideal mixture: K = CB ÷ CA, and with pure A fed, K = X ÷ (1 − X).
•
Conversion X is the fraction of the fed A turned into B once equilibrium is reached.
Step-by-step
WE KNOW
K = 3, CA0 = 2 mol/L
FIND
equilibrium conversion (X), equilibrium concentration of A (CA), equilibrium concentration of B (CB)
Step 1 · Find equilibrium conversion (X)
Use K = X ÷ (1 − X)
Multiply
out
K
(1
−
X
)
=
X
,
gather
the
X
terms:
X
(1
+
K
)
=
K
,
then
divide
by
1
+
K
.
X
=
K
1 +
K
X
=
3
1 + 3
X
=
3
4
X
=
0.75
Step 2 · Find equilibrium concentration of A (CA)
Use CA = CA0 × (1 − X)
The
share
of
A
left,
1
−
X
,
times
the
feed
concentration.
CA = 2 × (1 − 0.75)
CA = 2 × 0.25
CA = 0.5 mol/L
Step 3 · Find equilibrium concentration of B (CB)
Use CB = CA0 × X
Each mole of A converted makes one mole of B.
CB = 2 × 0.75
CB = 1.5 mol/L
Check
Put the numbers back in. Both sides should match.
3
=
0.75
1 − 0.75
✓
0.5 = 2 × (1 − 0.75) ✓
1.5 = 2 × 0.75 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.