A CSTR of volume 1 L (constant density liquid phase) is operated at steady state and constant temperature. The following irreversible reaction occurs:
A → C
The feed always contains reactant A at the concentration C_A0 = 1 mol/L. The rate depends only on A, and this dependence is expressible by an nth order rate form. Exit concentrations of un-reacted reactant A were measured in two runs. In each run, steady state was achieved at different volumetric flow rates q listed below.
| Run | q (L/min) | C_A (mol/L) |
|---|---|---|
| Run 1 | q_1 = 0.707 | C_A = 0.5 |
| Run 2 | q_2 = 0.112 | C_A = 0.2 |
(d) Find the order of reaction rounded off to 1 decimal place.
V = \frac{f_A F_{A0}}{-r_A}
Run 1:
1 = \frac{(0.5)(1)(0.707)}{k_A(0.5)^n}
Run 2:
1 = \frac{(0.5)(1)(0.112)}{k_A(0.2)^n}
Equating (1) and (2) and solving for n gives n = 1.498.
Answer (1)
To find the order of reaction, often represented as n , for the irreversible reaction A → C in a Continuous Stirred Tank Reactor (CSTR), we can use the information provided from two experimental runs.
Given Data:
Run 1: Volumetric flow rate, q 1 = 0.707 L/min and exit concentration C A 1 = 0.5 mol/L .
Run 2: Volumetric flow rate, q 2 = 0.112 L/min and exit concentration C A 2 = 0.2 mol/L .
Feed concentration C A 0 = 1 mol/L .
Reactor volume V = 1 L .
The reactor design equation for a CSTR is given by:
V = − r A F A 0 C A 0
where − r A is the rate of reaction. The rate is expressed using the rate law:
− r A = k A C A n
Solving using the given runs:
Run 1:
1 = k A ( 0.5 ) n 0.707 × 1
Run 2:
1 = k A ( 0.2 ) n 0.112 × 1
Equating the two equations:
From Run 1:
k A ( 0.5 ) n = 0.707
From Run 2:
k A ( 0.2 ) n = 0.112
Dividing Run 2 by Run 1 to eliminate k A :
0.707 0.112 = ( 0.5 0.2 ) n
0.1585 = ( 0.4 ) n
Taking the logarithm of both sides:
lo g ( 0.1585 ) = n ⋅ lo g ( 0.4 )
n = lo g ( 0.4 ) lo g ( 0.1585 ) ≈ 1.498
Thus, the order of the reaction, n , is approximately 1.5 when rounded off to one decimal place.
