Second order reaction, conversion in a CSTR

[MichelV]

Homework Statement

We have a second order reaction: A + B → C + D with -r_A = k[A][ B]
[A]₀ = [ B]₀ = 300 mol / m³; τ = 11 minutes.
k = 4.0 × 10^-4 m³ / (mol × minutes)

What is the conversion in a CSTR?

Homework Equations

I think:

τ = ([A]₀ - [A]₁) / -r_A,1

τ = (X_A × [A]₀) / -r_A,1

But since I don't know the conversion, I don't know the concentrations at the CSTR output. And because I don't have a volume or flow rate I don't know how to proceed.

The Attempt at a Solution

1 / [A] = 1 / [A]₀ + kt

1 / [A] = 1 / 300 + 4.0×10^-4×11
1 / [A] = 0.0077333

[A] = [ B] = 129.3 mol

-r_A = 4.0 × 10^-4 × 129.3²
-r_A = 6.688

Rearranging the second formula in the relevant equations for X gives:

X_A = (-r_A,1 × τ) / [A]₀

X_A = (6.688 × 11) / 300

X_A = 0.25

But the answer should be 0.43 and I don't know how to get there.

Thanks in advance.

 

[BvU]

Hello Michel,

You have to sort out your equations a little better. These really won't do; the dimensions don't fit.

Set up a differntial equation for [A] (or ) and solve it. The 0.43 is correct.

[edit]sorry, bad response. The 129.3 is correct. So: what is the conversion ?

 

[BvU]

1 / [A] = 1 / [A]0 + kt

Is the solution of the differential equation I mentioned. (It's good to check how this was deducted!) In your excercise, t = $\tau$ and so you can calculate [A] = 0.007733 mol / m³ after 11 min. And the conversion (it is not 0.43).

 

[BvU]

And because I don't have a volume or flow rate I don't know how to proceed.

You only need their ratio, $\tau$.

 

[MichelV]

Is the solution of the differential equation I mentioned. (It's good to check how this was deducted!) In your excercise, t = $\tau$ and so you can calculate [A] = 0.007733 mol / m³ after 11 min. And the conversion (it is not 0.43).

Thank you for the help. I've looked at the differential equation and it makes much more sense now.

If we know that [A] = 129,3 mol / m³ at the output of the CSTR and we know that X = 1 - ([A] / [A]₀) this means that:

X = 1 - (129.3 / 300) = 0.57

So the conversion for this reaction in a CSTR is 57%.

I believe this should be the correct answer.

 

[Chestermiller]

$\tau$ is supposed to be the mean residence time in the CSTR. If F is the feed rate, then the steady state mass balance on the reactor is $$F(A_{in}-A_{out})=-rV$$ where V is the volume of liquid in the reactor and $$r=kA_{out} B_{out}$$So, $$(A_{in}-A_{out})=-k\tau A_{out} B_{out}$$where $\tau=V/F$. If X represents the number of moles of A (per unit volume) converted, then $$A_{out}=A_{in}-X$$and $$B_{out}=B_{in}-X$$So, combining these equations, we get: $$k\tau(A_{in}-X)(B_{in}-X)=X$$
Solve this quadratic equation for X.

This gives a 43% conversion.

 

[MichelV]

$\tau$ is supposed to be the mean residence time in the CSTR. If F is the feed rate, then the steady state mass balance on the reactor is $$F(A_{in}-A_{out})=-rV$$ where V is the volume of liquid in the reactor and $$r=kA_{out} B_{out}$$So, $$(A_{in}-A_{out})=-k\tau A_{out} B_{out}$$where $\tau=V/F$. If X represents the number of moles of A (per unit volume) converted, then $$A_{out}=A_{in}-X$$and $$B_{out}=B_{in}-X$$So, combining these equations, we get: $$k\tau(A_{in}-X)(B_{in}-X)=X$$
Solve this quadratic equation for X.

This gives a 43% conversion.

Thank you for the clear explanation! I got it now.

 

[BvU]

My apologies for leading you astray. Too quick to respond and too slow thinking. Thanks Chet !