Question

Engineers Julie and Walter operate a reactor where two unbalanced, gas phase reactions occur: CH3COOH+H2→C2H5OH+H2O and C2H5OH→(C2H5)2O+H2O. In the first reaction, acetic acid (CH3COOH) and hydrogen (H2) react to produce ethanol (C2H5OH) and water (H2O); and ethanol decomposes to diethyl ether ((C2H5)2O) and water in the second reaction. A stream of 67.8 mol/s hydrogen and the balance acetic acid at 147°C is fed to the reactor, which operates at steady state. The stream exiting the reactor contains acetic acid, hydrogen, ethanol, water, and diethyl ether at 417°C. The component flow rate of acetic acid exiting the reactor is 3.81 mol/s. The fraction conversion of acetic acid is 71.6%, and the selectivity, which is the ratio of the amount of ethanol to diethyl ether in the exiting stream, is 3.14. Assume all components are in the gas phase and no phase changes occur in the reactor.
Find the component molar flow rate of acetic acid(AA) entering the reactor.

Answer 1

Answer:

Explanation:

From the given information:

The two reactions occurring in the reactor are:

$\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}$

$\mathtt{2C_2H_5OH \to (C_2H_5)_2O+H_2O}$

Let molar flowrate of $\mathtt{CH_3COOH}$ entering reactor = $\mathtt{x_n}$

From the first reaction:

$\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}$

Since there is a 71.6% conversion of $\mathtt{CH_3COOH}$;

Then, amount of $\mathtt{CH_3COOH}$ remaining unreacted and exiting the reactor is:

= x - 0.716x

= 0.284x

Recall that; the component flow rate exiting the reactor = 3.81 mols

∴

0.284x = 3.81

x = 3.81/0.284

x = 13.42 moles

Thus, the molar flowrate of $\mathtt{CH_3COOH}$ entering the reactor is 13.42 moles