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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 n

[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!

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 n

_{D}= V⋅C_{D}, 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!