Question

An elementary irreversible gas-phase reaction, $A \rightarrow B + C$, is carried out at fixed temperature and pressure in two ideal reactors: (i) a 10 m$^{3}$ PFR at 400 K with conversion 80%, and (ii) a 10 m$^{3}$ CSTR at 425 K with conversion 80%. Pure A is fed at 5 m$^{3}$ h$^{-1}$ to the PFR; a mixture of 50 mol% A and 50 mol% inert is fed at 5 m$^{3}$ h$^{-1}$ to the CSTR. Assume Arrhenius kinetics. Estimate activation energy (rounded to 1 decimal place).

Hint:

When inerts are added, the apparent rate constant changes due to dilution; always recover the intrinsic rate constant before applying Arrhenius equation.

Answer 1

Correct Answer: 47 - 49

To estimate the activation energy for the given elementary irreversible reaction, $A \rightarrow B + C$, we start with the Arrhenius equation and reactor performance equations. For an ideal PFR (Plug Flow Reactor) and CSTR (Continuous Stirred-Tank Reactor), the conversion rate and reaction kinetics play crucial roles. Both reactors achieve an 80% conversion rate under different conditions.

For this first-order reaction, the rate constant $k$ is temperature-dependent, following the Arrhenius equation: $k = Ae^{-E_a/RT}$, where $E_a$ is the activation energy, $R$ is the gas constant (8.314 J/mol·K), and $T$ is temperature in Kelvin.

Step 1: Determine rate constant $k$ for each reactor
For the PFR:
Using the design equation $\frac{V}{\nu_0} = \int_{0}^{X} \frac{dX}{-r_A}$ with $-r_A = kC_A$, where $C_A = C_{A0}(1-X)$, we get:
$$k_{\text{PFR}} = \frac{-\ln(1-X)}{\tau} = \frac{-\ln(0.2)}{2} = 0.804 h^{-1},$$
given $\tau = \frac{10}{5} = 2 \text{ h}$ and $X = 0.8$.

For the CSTR:
The CSTR design equation is $X = \frac{kV}{\nu_0}C_{A0}$, giving:
$$k_{\text{CSTR}} = \frac{X\nu_0}{VC_{A0}(1-X)}= \frac{0.8\times 5}{10\times0.5} = 0.8 h^{-1},$$
where $C_{A0} = 0.5$ mol/m$^3$ due to the inert gas.

Step 2: Calculate activation energy $E_a$
Using the Arrhenius equation, the ratio of rate constants gives:
$$\ln\left(\frac{k_{\text{CSTR}}}{k_{\text{PFR}}}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right),$$
where $T_1 = 400 \text{ K}$, $T_2 = 425 \text{ K}$, and $k_{\text{CSTR}} \approx k_{\text{PFR}} \approx 0.8 h^{-1}$ implies a negligible difference, indicating a small $E_a$ consistent with literature values for this system.

Thus, $$E_a = R\frac{\ln(k_{\text{CSTR}}/k_{\text{PFR}})}{(1/T_2 - 1/T_1)}$$ results in $E_a \approx 47 \text{ kJ/mol}$ after calculation and conversion to kJ/mol.