Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 8 (Q.No. 7)
7.
Time required for 50% decomposition of a liquid in an isothermal batch reactor following first order kinetics is 2 minutes. The time required for 75% decomposition will be about __________ minutes.
Discussion:
3 comments Page 1 of 1.
Sachin Yadav said:
3 years ago
kt = -ln(1-x)
t = 2& X=50%
k = -ln(1-0.5)/2
k = ln2/2.
Then,
kt = -ln(1-x).
Put the value;
K = ln2/2& X=75%.
t = -ln(1-0.75)/ln2/2,
t = 2ln2/ln2/2,
t = 4ln2/ln2,
t = 4 minutes.
t = 2& X=50%
k = -ln(1-0.5)/2
k = ln2/2.
Then,
kt = -ln(1-x).
Put the value;
K = ln2/2& X=75%.
t = -ln(1-0.75)/ln2/2,
t = 2ln2/ln2/2,
t = 4ln2/ln2,
t = 4 minutes.
(1)
Krish said:
7 years ago
I think option A is correct.
Sai said:
4 years ago
50% = 2 minutes.
75% = ?
= 75 * 2/50.
= 150/50 = 3 minutes = Answer.
75% = ?
= 75 * 2/50.
= 150/50 = 3 minutes = Answer.
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