Question

If the rate of a reaction is given as (-rA) = \(\frac{k_1+C_A^2}{k_2×C_A} \frac{mol}{m^3 min} \), then the units of k1 and k2 respectively are ____

(a) \(\frac{mol^2}{m^6} \) and min

(b) min and \(\frac{mol^2}{m^6} \)

(c) mol and min

(d) m^3 and mol

This question was addressed to me during an interview.

Enquiry is from Conversion and Reactor Sizing topic in portion Ideal Reactors for a Single Reaction, Conversion and Reactor Sizing of Chemical Reaction Engineering

Answer 1

The correct answer is (a) \(\frac{mol^2}{m^6} \) and min

To explain: Two quantities of the same dimensions can only be added. As the unit of concentration is \(\frac{mol^2}{m^3} \) , k1 has the units \(\frac{mol^2}{m^6} \).

Equating the units of all known quantities on both sides, \(\frac{mol}{m^3 min} = \frac{mol^2}{m^6} × \frac{1}{k_2×(\frac{mol}{m^3})}. \)

Hence, k2 has the unit of min.