Mass balance with reaction -- Chemical kinetics POV

  • #1
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Main Question or Discussion Point

Good evening, people of PF. Recently, while studying chemical kinetics, I have come across some questions about reaction rates and rate equations. I've noticed some inconsistencies between the chemical kinetics point of view, and the approach taken by chemical reaction engineering and reactor design authors. Consider the following reaction.
aA + bB → cC + dD
According to chemical kinetics, the reaction rate is given by
[tex]r = -\frac{1}{a} \frac{\textrm{d} C_{\textrm{A}}}{\textrm{d} t} = -\frac{1}{b} \frac{\textrm{d} C_{\textrm{B}}}{\textrm{d} t} = \frac{1}{c} \frac{\textrm{d} C_{\textrm{C}}}{\textrm{d} t} = \frac{1}{d} \frac{\textrm{d} C_{\textrm{D}}}{\textrm{d} t}[/tex]
And the rate equation is (usually) given by
[tex]r = k C_{\textrm{A}}^a C_{\textrm{B}}^b[/tex]
So, for instance, if we want to know the rate at which D is produced we write
[tex]\frac{\textrm{d} C_{\textrm{D}}}{\textrm{d} t} = d k C_{\textrm{A}}^a C_{\textrm{B}}^b[/tex]
and we solve the differential equation.
Now, we assume the same reaction occurs inside an ideal batch reactor, and again we want to know the rate at which D is produced. We perform a mole balance for the species D.
[tex]\frac{\textrm{d} n_{\textrm{D}}}{\textrm{d} t} = rV[/tex]
Where V is the volume of the reactor, and r is the net chemical generation. If we apply nD = V⋅CD, we have
[tex]V \frac{\textrm{d} C_{\textrm{D}}}{\textrm{d} t} = rV[/tex]
[tex]\frac{\textrm{d} C_{\textrm{D}}}{\textrm{d} t} = r[/tex]
Finally, we have
[tex]\frac{\textrm{d} C_{\textrm{D}}}{\textrm{d} t} = k C_{\textrm{A}}^a C_{\textrm{B}}^b[/tex]
Although similar, this is not the same result one arrives at when studying chemical kinetics. Maybe it is a ridiculous concern, or I am missing something obvious, but it has been bugging me since I noticed it. Is the r (net chemical generation) used when studying chemical reactor engineering not the same as the r (reaction rate) used when studying chemical kinetics?
Thanks in advance for any input!
 

Answers and Replies

  • #2
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It seems to me, in the second case you're just considering species D and your r and k would apply just to CD. To then consider r and k for the overall reaction, you'd have to divide your species-specific r and k by d.
 
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  • #3
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This step doesn't look right to me:
Now, we assume the same reaction occurs inside an ideal batch reactor, and again we want to know the rate at which D is produced. We perform a mole balance for the species D.
[tex]\frac{\textrm{d} n_{\textrm{D}}}{\textrm{d} t} = rV[/tex]
If each reaction produces ##d## molecules of species D shouldn't it be this, instead?
[tex]\frac{\textrm{d} n_{\textrm{D}}}{\textrm{d} t} = drV[/tex]

If so, it looks like both approaches agree.
 
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  • #4
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If each reaction produces ##d## molecules of species D shouldn't it be this, instead?
[tex]\frac{\textrm{d} n_{\textrm{D}}}{\textrm{d} t} = drV[/tex]

If so, it looks like both approaches agree.
Well, I've never seen reaction engineering literature using stoichiometric coefficients on mole balances before, but it doesn't seem like we're breaking any rules here. And it does agree with the result from chemical kinetics. I do prefer using reaction engineering methods though, as a chemical engineering student, I'm more inclined to think and reason in terms of mass and energy balances.
 
  • #5
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It seems to me, in the second case you're just considering species D and your r and k would apply just to CD. To then consider r and k for the overall reaction, you'd have to divide your species-specific r and k by d.
While reading deeper into Fogler's Elements of Chemical Reaction Engineering I noticed they do use [itex]r_{\textrm{D}}[/itex] in the mole balance for species D, and [itex]r_{\textrm{D}} = dr[/itex], which agrees with both your post and rikblok's.

Thank you both for your input!
 

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