Question

The rate expression for independent deactivation for batch solids is ____

(a) ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt

(b) ln \(\frac{C_A}{C_{A∞}}\)  = ln \(\frac{Wk’}{Vk_d}\) – kdt

(c) lnln \(\frac{C_A}{C_{A∞}}\)  = ln \(\frac{Wk’}{Vk_d}\)

(d) lnln \(\frac{C_A}{C_{A∞}}\)  =  \(\frac{Wk’}{Vk_d}\) – kdt

I had been asked this question during an online interview.

This interesting question is from Mechanisms of Catalyst Deactivation in section Suspended Solid Reactors, Catalyst Deactivation, Diffusion and Reaction of Chemical Reaction Engineering

Answer 1

Right option is (a) ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt

To explain I would say: For batch solids – fluid, –\(\frac{-dC_A}{dt} = \frac{W}{V}\)k’CAa and –\(\frac{da}{dt}\) = kda^d. Integrating and combining the equations, ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt.