Problem 4: Equilibrium conversions -B is carried out in a The reversible reactio
ID: 701109 • Letter: P
Question
Problem 4: Equilibrium conversions -B is carried out in a The reversible reaction 2A flow reactor where pure A is fed at a concentration of 4 mol/ L. The equilibrium conversion is found to be 60%. Assume that the reactor is isothermal and isobaric. a. What is the equilibrium constant, Kc, if the reaction is a gas phase reaction? [0.33 b. What is Kc if it is a liquid phase reaction? [0.47] The gas phase reaction A3C is carried out in a flow reactor with no pressure drop. Pure A enters at a temperature of 400K and a pressure of 10 atm. At this temperature, Kc=0.25 M2. c. What is the equilibrium conversion if the reaction is carried out in a constant volume batch reactor? Assume isothermal operation. [0.4] What is the equilibrium conversion if the reaction is carried out in a constant pressure batch reactor? Assume isothermal operation. [0.6] d.Explanation / Answer
Problem 4.
a.
rate of reaction of A = -rA = kfCA2 - kbCB
at equilibrium -rA = 0
kfCA2 - kbCB = 0
Kc = kf/kb = CB/CA2
volume expansion factor = (final - initial )/initial = (1-2)/2 = -0.5
Also,
CA = CAo{1-X}/{1+X} = 4{1-0.6}/{1-0.5 x 0.6} = 2.2857
CB = CAo{0.5X}/{1+X} = 4{0.5 x 0.6}/{1-0.5 x 0.6} = 1.7143
Kc = CB/CA2 = 1.7143 / 2.28572 = 0.328 ~ 0.33
b.
we have reaction:
2A <------> B
Kc = [B]/[A]2
[ ] = erpresents molar concentration
2A <------> B
4.0 mol/L 0 initially
-4.0 mol/L x 0.6 +4.0 mol/L x 0.6 / 2 change
4.0 mol/L - 4.0 mol/L x 0.6 0+ 4.0 mol/L x 0.6 / 2 equilibrium
= 1.6 mol/L 1.2 mol/L
Kc = [1.2]/[1.6]2 = 0.46875 ~ 0.47
c.
for constant volume
[A] = P/RT
[A] = 10 atm / 0.08206 atm.L/mol.L x 400K = 0.305 mol/L
We have reaction:
A <-----> 3C
0.305 0 initially
-x +3x change
0.305 - x +3x equilibrium
Kc = [C]3 / [A] = 0.25 = [3x]3/[0.305 - x]
x is < 1 assume 0.305 -x ~ 0.305
0.07625 = 27x3
x = 0.141
lets verify if error is <5%
Kc = [3 x 0.141]3/[0.305-0.141] = 0.465 seems our assumption to take x << 0.305 was wrong!
we have to find the roots of the equation
27x3 - 0.305 x 0.25 + 0.25x = 0
doing in calculator we get roots as follows:
we have got only one real root = 0.1197 ~ 0.12
Conversion = (initial concentration - final concentration) / initial concentration = x / 0.305 = 0.12 / 0.305 = 0.39344
~ = 0.4
d.
for constant pressure batch reactor
at equilibrium -rA = 0
kfCA - kbCc3 = 0
Kc = kf/kb = Cc3/CA
volume expansion factor = (final - initial )/initial = (3-1)/1 = 2
Also,
CA = CAo{1-X}/{1+X} = 0.305{1-X}/{1+2 x X} = 0.305{1-X}/{1+2X}
Cc = CAo{3X}/{1+X} = 0.305{3 x X}/{1+2 x X} = 0.305{3X}/{1+2X}
0.25 = CC3/CA = {0.305{3X}/{1+2X}}3 / {0.305{1-X}/{1+2X}} =
which can be simplified as:
3.52x3-0.75x-0.25 = 0
we have got a single real root = 0.58 ~ = 0.6
conversion = 0.6
kf and kb are forward and backward reaction constant