Question

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answer 1

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is$1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$