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?
Ans
Yes , I need Ans !
To answer this question, we can use the concept of stoichiometry and the equation of the irreversible elementary reaction provided.
Let's denote the molar feed rate of A as Fa and the molar feed rate of B as Fb. According to the equimolar ratio, we have Fa = Fb. Given that the conversion of A is 80%, we can write the reaction rate equation as:
r = k * Ca^2
where r is the reaction rate, k is the rate constant, and Ca is the molar concentration of A.
Since the reactor is isothermal and ideal, we can assume constant volume and pressure, which means that molar concentrations are directly proportional to molar flow rates.
Given that the conversion of A is 80%, the molar flow rate of A exiting the reactor (Fa') is 20% of the initial molar flow rate of A (Fa):
Fa' = 0.2 * Fa
If the molar feed rate of A is cut in half, the new molar flow rate of A (Fa_new) is:
Fa_new = 0.5 * Fa
However, the molar feed rate of C is left unchanged. Therefore, the new molar flow rate of C (Fc_new) is the same as the initial molar flow rate of C (Fc):
Fc_new = Fc
Now, we need to determine the conversion of A with the new molar flow rates.
Using the reaction rate equation, we can write:
r' = k * (Ca_new)^2
Since the reactor is isothermal, the rate constant remains the same:
r' = k * Ca_new^2
Since the stoichiometry of the reaction is 2A -> B, we can write:
Cc_new = 0.5 * Ca_new
Given that the molar flow rate of B (Fb) is the same as the molar flow rate of A (Fa), we can write:
Fb_new = Fa_new
Now, let's substitute the expressions for Fa_new, Ca_new, and Fb_new into the reaction rate equation:
k * (Ca_new)^2 = Fb_new / V
where V is the volume of the reactor.
Since Ca = Fa / V, we can write:
k * (Fa_new / V)^2 = Fb_new / V
Simplifying the equation, we get:
k * (Fa_new)^2 = Fb_new
Substituting the expressions for Fa_new and Fb_new, we get:
k * (0.5 * Fa)^2 = 0.5 * Fa
Simplifying the equation further, we get:
0.5 * Fa = Fa^2 * k / 4
Rearranging the equation, we get:
Fa^2 * k / 4 - 0.5 * Fa = 0
Now, we can solve this quadratic equation to find the molar flow rate of A (Fa) and then calculate the conversion of A.
However, the problem does not provide the value of the rate constant (k), so we cannot determine the exact value of the conversion of A with the given information. Therefore, the point of this problem may be to practice using the concept of stoichiometry and solving mathematical equations in chemical engineering problems.
To determine the conversion of A when the molar feed rate is reduced, we need to apply the concept of reaction stoichiometry and the molar balance in the reactor.
Let's start by understanding the given information. We have an irreversible elementary reaction:
2A ---+ B
This reaction occurs in a gas-phase isothermal tubular reactor, where reactant A and a diluent C are fed in equimolar ratio. The conversion of A is stated to be 80% under these conditions.
Now, let's consider the scenario when the molar feed rate of A is halved, while the feed rate of C remains unchanged. We want to find the new conversion of A under these conditions.
To solve this, we can set up a molar balance equation based on the stoichiometry of the reaction. The reaction stoichiometry tells us that for every 2 moles of A, we get 1 mole of B.
Let's assume the molar feed rate of A under the initial conditions is Fa0 (moles/time). Since A and C are fed in equimolar ratio, the molar feed rate of C is also Fa0.
The initial conversion of A, X, is given as 80%, which means 80% of A is converted to B. Therefore, the molar flow rate of A leaving the reactor, Fa, is given by:
Fa = Fa0 * (1 - X)
Now, when the molar feed rate of A is halved, the new molar feed rate becomes Fa0/2. However, the feed rate of C remains unchanged at Fa0.
We want to calculate the new conversion of A, X_new, under these conditions. Similar to before, the molar flow rate of A leaving the reactor, Fa_new, is given by:
Fa_new = (Fa0/2) * (1 - X_new)
Since the feed rate of C remains unchanged, the molar flow rate of C leaving the reactor, Fc, remains Fa0.
Now, we can use the molar balance equation to relate the molar flow rates of A, B, and C:
2 * Fa = Fb
Fa + Fc = Fa0
Using the above equations, we can substitute for Fa, Fa_new, and Fc:
2 * Fa0 * (1 - X) = Fb
(Fa0/2) * (1 - X_new) + Fa0 = Fa0
Simplifying the equation:
2 * (1 - X) = Fb / Fa0
(1 - X_new) + 1 = 2
The first equation gives us the relationship between X (initial conversion) and Fb (flow rate of B). However, since we are not given the value of Fb, we cannot solve for X directly.
The second equation tells us that the sum of the conversion under the new conditions, X_new, and the conversion under the initial conditions, X, is equal to 1. Assuming ideal behavior and a constant reactor temperature, this equation holds.
Therefore, the point of this problem is to highlight the relationship between the conversion of a reactant and the molar feed rates in a reactor. It demonstrates how changes in the molar feed rate of a reactant can affect the conversion of that reactant, while keeping other conditions constant.