chemical reaction engineeing

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 $(−ra)=−1/Vfrac{dNa}{dt}$.

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

http://www.the-seventh-dimension.com/images/textlev/LEVENSPIEL%20Chemical%20reaction%20engineering-ch4-ch5.pdf

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.

Foundations

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

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

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

Examples

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.