Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 2 (Q.No. 29)
29.
For a gaseous phase reaction, rate of reaction is equal to K. CA . CB. If the volume of the reactor is suddenly reduced to l/4th of its initial volume, then the rate of reaction compared to the original rate will be __________ times.
Discussion:
2 comments Page 1 of 1.
Pavi said:
9 years ago
Concentration = (moles)/(volume),
Here moles remain same but volume becomes 1/4 so new concentration becomes 4 times the original.
K * 4CA * 4CB = 16 (K CA CB) = 16 (original rate).
Here moles remain same but volume becomes 1/4 so new concentration becomes 4 times the original.
K * 4CA * 4CB = 16 (K CA CB) = 16 (original rate).
(4)
Akhil said:
9 years ago
Can anyone explain this?
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