The irreversible reaction 2A --+ B takes place in the gas phase in a constant
temperature plug flow reactor. Reactant A and diluent gas are fed in equimolar
ratio, and the conversion of A is 85 percent. If the molar feed rate of A is doubled,
what is the conversion of A assuming the feed rate of diluent is
unchanged?
To solve this problem, we need to use the concept of stoichiometry and the definition of conversion.
Given:
- Reactants in equimolar ratio: 2A and B
- Conversion of A is 85%
- Molar feed rate of A is doubled
- Feed rate of diluent gas is unchanged
First, let's calculate the initial moles of A and B:
Since A and B are in equimolar ratio, the initial moles of B will be equal to half the initial moles of A.
Let x be the initial moles of A.
Initial moles of A = x
Initial moles of B = (1/2)x
Now, let's calculate the moles of A and B after the reaction:
Given that the conversion of A is 85%, the final moles of A will be (1 - 0.85) times the initial moles of A.
Final moles of A = (1 - 0.85) * x = 0.15 * x
Since the reaction is irreversible, the final moles of B will be equal to the initial moles of B plus the moles of A consumed in the reaction.
Final moles of B = (1/2)x + 2*(0.15 * x) = (1/2)x + 0.3x = 0.8x
Now, let's consider the effect of doubling the molar feed rate of A:
If the molar feed rate of A is doubled, the new moles of A supplied will be 2x.
The total moles of A at the reactor inlet will be:
Initial moles of A + New moles of A = x + 2x = 3x
Since the feed rate of diluent gas is unchanged, the moles of B at the reactor inlet remain the same:
Initial moles of B = (1/2)x
Now, let's calculate the conversion of A under the new conditions:
The final moles of A will still be 0.15 times the initial moles of A, which is 0.15 * (3x) = 0.45x
The moles of B after the reaction will be the initial moles of B plus the moles of A consumed, which is:
Final moles of B = (1/2)x + 2*(0.45x) = (1/2)x + 0.9x = 1.4x
Finally, the conversion of A under the new conditions is:
Conversion of A = (Initial moles of A - Final moles of A) / Initial moles of A
= (3x - 0.45x) / (3x)
= 2.55 / 3
≈ 0.85 (rounded to two decimal places)
Therefore, the conversion of A assuming the feed rate of diluent gas is unchanged is approximately 85%.
To determine the conversion of A when the molar feed rate of A is doubled while the feed rate of the diluent gas is unchanged, we need to use the concept of mole balance in a constant temperature plug flow reactor.
Mole balance equation for A:
Rate of disappearance = Rate of formation
In this case, the stoichiometry of the irreversible reaction is 2A -> B. Therefore, the rate of disappearance of A is given by -rA = 2 * rB, where rA is the rate of disappearance of A, and rB is the rate of formation of B.
Given that the conversion of A is 85 percent, we can write the mole balance equation as:
Fa0 * (1 - X) = 2 * rB * V
Where:
Fa0 is the molar feed rate of A
X is the conversion of A (0.85 in this case)
V is the volume of the reactor
Now, let's consider the case where the molar feed rate of A is doubled while the feed rate of the diluent gas is unchanged. Let Fa1 be the new molar feed rate of A.
Fa1 = 2 * Fa0
Since the feed rate of the diluent gas is unchanged, we can still use the same volume V for the reactor.
Using the new feed rate Fa1 and the same volume V, the new mole balance equation becomes:
Fa1 * (1 - X') = 2 * rB * V
Substituting Fa1 = 2 * Fa0, we get:
2 * Fa0 * (1 - X') = 2 * rB * V
Simplifying the equation, we find:
Fa0 * (1 - X') = rB * V
Comparing this equation with the previous mole balance equation, we can see that both equations are identical. Therefore, the conversion of A, denoted as X', remains unchanged when the molar feed rate of A is doubled while the feed rate of the diluent gas is unchanged.
In conclusion, the conversion of A will still be 85 percent even if the molar feed rate of A is doubled while the feed rate of the diluent gas is unchanged.
P4-12, The irreversible elementary reaction 2A ---+ B takes place in the gas
phase in an isothermal tubular (plug-Jlow) reactor. Reactant A and a diluent
C are fed in equimolar ratio, and conversion of A is 80%. If the molar feed
rate of A is cut in half, what is the conversion of A assuming that the feed rate
of C is left unchanged? Assume ideal behavior and that the reactor tempera-
ture remains unchanged. What was the point of this problem? (From Califor-
nia Professional Engineers Exam.)