What is the definition of rate $(−ra)=−\frac{dCa}{dt}$ or $(−ra)=−1/V\frac{dNa}{dt}$ ? I think the general one is the second one and first one is for constant volume reaction system. Is the above written rate equation only valid to batch reactor? If not can we use this in a PFR or MFR? In PFR we know the rate comes out to be $(−ra)=−\frac{dFa}{dV}$ ,can we equate $−1/V\frac{dNa}{dt}=−\frac{dFa}{dV}$? If not then why? Is the rate equation not valid everywhere? Can we equate with CSTR as well ? eg $-\frac{dCa}{dt}=\frac{Fa_o-Fa}{V}$ (considering it a constant volume reaction system CSTR)



The rate $r$ of a reaction on a molar ($n$) basis is

$$ r_A \equiv \frac{dn_A}{dt} $$

In a consumption reaction, $r_A$ is negative.

In general, $n_A = C_A\ V$, where $C_A$ is the molar concentration and $V$ is volume. Expanding from this, we obtain

$$ r_A \equiv \frac{d(C_A\ V)}{dt} = \left(\frac{dC_A}{dt} + \frac{dV}{dt}\right) $$

In a system of ideal gases, we write $n_A = p_A\ V/RT$. Expanding on this, we obtain

$$ r_A \equiv \frac{d(p_A\ V/RT)}{dt} = \frac{1}{R}\ \left(\frac{dp_A}{dt} + \frac{dV}{dt} + \frac{d\ln T}{dt}\right)$$

Armed with these two expansions, we can handle any cases.


Constant volume, isothermal.

$$r_A = \frac{dC_A}{dt} \ \ \mathrm{or} \ \ r_A = \frac{1}{R}\ \frac{dp_A}{dt} $$

Batch reactors with liquids are generally presumed to be systems with constant volume. However, if the density of the liquid changes as the reaction proceeds, the assumption of constant volume is invalid.

The integration over a plug-flow reactors with ideal gases can be solved by keeping the differential volume constant and allowing pressure to change.


You are correct in saying that reaction rate $(−ra)=−\frac{dCa}{dt}$ is for constant volume systems and that the more general form is or $(−ra)=−1/V\frac{dNa}{dt}$.

The general reaction rate equation is the basis for all the types of reactors you mentioned, see the link below for the derivation explanation.


The form of the rate equation as shown should be valid everywhere within the vessel, so long as the limits of integration used to describe the situation are correct. If for example, a vessel has a significant dead volume due to the flows not being properly directed through the vessel (i.e. the "well mixed" assumption is not being satisfied), then the ideal results generated from this equation may deviate significantly from actual data.

  • $\begingroup$ I am certain that posting a reference to a photocopied page from a textbook is a violation of copyright. I have downvoted the answer and will vote to remove the answer unless the page reference is removed. $\endgroup$ – Jeffrey J Weimer Oct 31 '20 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.