Why Can't I Get the Correct Constants in My Chemical Reaction Calculations?

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SUMMARY

The discussion focuses on the calculation of constants in chemical reaction equations, specifically the determination of the residence time constant (\tau_1) and the heat capacity at constant volume (Cv). The user initially calculated \tau_1 as 0.5148 instead of the expected 2.07, leading to confusion regarding the correct substitution of values in the equation. The correct formulation for the concentration of reactant A is clarified as (e(-E/RT))cA. Additionally, the user derived Cv as -28.10 Btu/lb-°R, raising questions about its validity in the context of an exothermic reaction.

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saltine
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Homework Statement


This is an example from a book. I am having trouble in substituting the numbers to get the given result.

Homework Equations


3144674953_011c70185c_o.jpg

The bar notation is to indicate the operating point. The subscript A is for the reactant A (which reacts and becomes product B in a A->B reaction). The subscript i stands for the input. So [tex]c_{A_i}[/tex] means the concentration of A entering the reactor.

fc is the flow rate of the coolant
Tci is the temperature of the coolant going in
Tc is the temperature of the coolant going out

f is the flow rate of the reactant going in
Ti is the temperature of the reactant going in
cAi is the concentration of the reactant going in
T is the temperature of the reactant going out
cA is the concentration of the reactant going out


The Attempt at a Solution


My problem is that when I plug in the values, I don't get 2.07 for [tex]\tau_1[/tex]. I first assumed that [tex]\bar{f} = f= 1.3364[/tex] and [tex]\bar{T}_{\bar{c}_A} = 678.9[/tex]. But I got [tex]\tau_1 = 0.5148[/tex].

I assumed that my [tex]\bar{f}[/tex] was correct, so I set [tex]\tau_1=2.07[/tex] and rearranged the equation to solve for [tex]\bar{T}_{\bar{c}_A}[/tex].

[tex]\bar{T}_{\bar{c}_A} = -\frac{E}{R}(ln(\frac{1-\frac{\tau_1 \bar{f}}{V}}{2\tau_1 k_o}))^{-1}[/tex] . . . (2)

When I do that I got [tex]\bar{T}_{\bar{c}_A} = 631[/tex], which doesn't match any of the given temperature. So I thought [tex]\bar{f}[/tex] might be 0.8771. And I got [tex]\bar{T}_{\bar{c}_A} = 633[/tex], which still doesn't match any number.

I have tried some other equations but I couldn't get any result to match. Since I don't have a background in studying chemical reactions, I was suspecting that there is something fundamental that I am missing here.

Do you know how the author got 2.07 for [tex]\tau_1[/tex]?

- Thank you.
 
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Hi Saltine! :smile:

Look at the dimensions …

it isn't e(-E/RTcA)

it's (e(-E/RT))cA :wink:
 
That is right!

A second problem:

3145242921_52ddb6c5fa_o.jpg


Now I am trying to verify T2. Here, Cv is the heat capacity at constant volume of the reaction mixture (Unit of Cv: Btu/lb-°R). However, the book did not give its value. This is the same for other examples where it seems that every other value is listed, except the Cv. Is it directly derivable from the other values?

U is the overall heat transfer coefficient, assumed to be constant, Btu/ft²-°R-min
A is heat transfer area, ft²
∆Hr is the heat of reaction, Btu/lbmole of A reacted
rA is the rate of reaction


I first computed [tex]\bar{r}_A = 0.03938[/tex]. Since Cv is just alone in the numerator, I first evaluated T2 without using Cv and got 0.2833. Therefore,

Cv = -7.96 / 0.2833 = -28.10 Btu/lb-°R

There is nowhere else in this particular example where Cv is used, so I couldn't check Cv in another way. Does it make sense that Cv is negative? This is an exothermic reaction. The problem didn't state this, but I suppose it is since coolant is used.
 

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