Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 4 (Q.No. 10)
10.
Pure ethanol vapor is fed to a reactor packed with alumina catalyst, at the rate of 100 kmole / hr. The reactor products comprise: ethylene :95 kmole / hr, water vapour: 97.5 k mole / hr and diethyl ether :2.5 kmole/hr. The reactions occuring can be represented by:
C2H5OH
C2H4 + H2O
2C2H5OH C2H5 - O - C2H5 + H2O
The percent conversion of ethanol in the reactor is
C2H5OH
C2H4 + H2O2C2H5OH
C2H5 - O - C2H5 + H2OThe percent conversion of ethanol in the reactor is
Discussion:
3 comments Page 1 of 1.
Zeeshan ulhaq said:
6 years ago
In my opinion 5 mol is required for 2nd reaction and 95 mol for 1st reaction from the balance chemical reaction.
Yogesh Kuriya said:
6 years ago
As per the data the 95 kmole/hr ethylene is produced,
From reaction 1, 1 kmole of ethanol produces 1 kmole of ethylene and 1 kmole of H2O.
So that for 95 kmole ethylene require 95kmole ethanol.
And same form reaction 2, 2.5kmole ether require 5 kmole ethanol.
Total ethanol require is 95+5kmole=100kmole.
And the initial amount is 100 kmole so that the conversion is 100%.
From reaction 1, 1 kmole of ethanol produces 1 kmole of ethylene and 1 kmole of H2O.
So that for 95 kmole ethylene require 95kmole ethanol.
And same form reaction 2, 2.5kmole ether require 5 kmole ethanol.
Total ethanol require is 95+5kmole=100kmole.
And the initial amount is 100 kmole so that the conversion is 100%.
Saleem said:
6 years ago
Please describe the answer.
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