Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 1 (Q.No. 6)
6.
What is the order of a chemical reaction, 
, if the rate of formation of 'C', increases by a factor of 2.82 on doubling the concentration of 'A' and increases by a factor of 9 on trebling the concentration of 'B'?
, if the rate of formation of 'C', increases by a factor of 2.82 on doubling the concentration of 'A' and increases by a factor of 9 on trebling the concentration of 'B'?Discussion:
8 comments Page 1 of 1.
Mohamed said:
2 years ago
Simply;
Log of 2.82 = n* log of 2.
Such that,
N = 1.5 of reactant A.
2 is reactant B.
So, 1.5 + 2 = 3.5 or 7/2.
Log of 2.82 = n* log of 2.
Such that,
N = 1.5 of reactant A.
2 is reactant B.
So, 1.5 + 2 = 3.5 or 7/2.
(5)
Kefale Abebe said:
3 years ago
Very helpful, Thanks for explaining it @Gedefaw.
(1)
Gedefaw Dememew said:
7 years ago
RC =kCc*n =k[CC]^y= -k[CA]^n[CB]^m ------> 1
2.82RC= -K[2CA]^n[CB]^m ------> 2
9RC= -k[CA]^n[3CB]^m ------> 3.
Eq1/Eq2 = 1/2.82 = 1/2^n use common natural logarithm. ln2.82=nln2. n=1.5.
Eq1/Eq3 = 1/9 =1/3^m. use common natural logarithm we have mln3=ln9. m= ln9/ln3=2.
Overall order of reaction = n+m = 1.5 + 2 = 3.5 = 7/2.
2.82RC= -K[2CA]^n[CB]^m ------> 2
9RC= -k[CA]^n[3CB]^m ------> 3.
Eq1/Eq2 = 1/2.82 = 1/2^n use common natural logarithm. ln2.82=nln2. n=1.5.
Eq1/Eq3 = 1/9 =1/3^m. use common natural logarithm we have mln3=ln9. m= ln9/ln3=2.
Overall order of reaction = n+m = 1.5 + 2 = 3.5 = 7/2.
(8)
Gedefaw Dememew said:
7 years ago
RC =kCc*n =k[CC]^y= k[CA]^n[CB]^m ----> 1
2.82RC= K[2CA]^n[CB]^m ----> 2
9RC= k[CA]^n[3CB]^m ----> 3.
Eq1/Eq2 = 1/2.82=1/2^n use common natural logarithm. ln2.82=nln2. n=1.5.
Eq1/Eq3= 1/9=1/3^m. use common natural logarithm we have mln3=ln9. m= ln9/ln3=2.
Overall order of reaction= n+m = 1.5+2 = 3.5 = 7/2.
2.82RC= K[2CA]^n[CB]^m ----> 2
9RC= k[CA]^n[3CB]^m ----> 3.
Eq1/Eq2 = 1/2.82=1/2^n use common natural logarithm. ln2.82=nln2. n=1.5.
Eq1/Eq3= 1/9=1/3^m. use common natural logarithm we have mln3=ln9. m= ln9/ln3=2.
Overall order of reaction= n+m = 1.5+2 = 3.5 = 7/2.
(4)
Rahul said:
7 years ago
What is order of A or B? Please explain about it.
RAJIV KUMAR said:
8 years ago
You explained in a simple way. Thanks a lot @Amit Soni.
(2)
Amit soni said:
1 decade ago
Rc = C^n.
In Rc = n.In.c.
1st.2nd
In 2.82 = n* In2.In9 = n In*3.
n = 1.49 ::::: n = 2 (Total = 3.5 = n over all = 7/2).
In Rc = n.In.c.
1st.2nd
In 2.82 = n* In2.In9 = n In*3.
n = 1.49 ::::: n = 2 (Total = 3.5 = n over all = 7/2).
(5)
B.Naga Sandhya said:
1 decade ago
Rc = k(Ca^n)(Cb^2) from the given reaction.
2.82Rc = k((2Ca)^n)(Cb^2)------A.
9Rc = k(Ca^n)((3Cb)^2).
Solving Rc = k(Ca^n)(Cb^2)--------B.
A/B==> 2.82=2^n.
Applying log on both sides
log(2.82) = n log(2)
n=1.5 order of reactant A.
2 is order of reactant B.
Over all order 2+1.5 = 3.5 = 7/2.
2.82Rc = k((2Ca)^n)(Cb^2)------A.
9Rc = k(Ca^n)((3Cb)^2).
Solving Rc = k(Ca^n)(Cb^2)--------B.
A/B==> 2.82=2^n.
Applying log on both sides
log(2.82) = n log(2)
n=1.5 order of reactant A.
2 is order of reactant B.
Over all order 2+1.5 = 3.5 = 7/2.
(5)
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