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Order of Reaction 2 Part

1) The order of a reaction was determined by comparing how the rate of the reaction changed when the concentration of the reactant was doubled. When the concentration was doubled, the rate also doubled, indicating the reaction was first order. 2) The order of a reaction was examined with respect to reactants A and B. Doubling the concentration of A quadrupled the rate, showing the reaction was second order with respect to A. Doubling the concentration of B doubled the rate, showing the reaction was first order with respect to B. 3) The overall order of the reaction was determined to be third order, which is the sum of the individual orders with respect to each reactant (2 + 1 = 3

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66 views2 pages

Order of Reaction 2 Part

1) The order of a reaction was determined by comparing how the rate of the reaction changed when the concentration of the reactant was doubled. When the concentration was doubled, the rate also doubled, indicating the reaction was first order. 2) The order of a reaction was examined with respect to reactants A and B. Doubling the concentration of A quadrupled the rate, showing the reaction was second order with respect to A. Doubling the concentration of B doubled the rate, showing the reaction was first order with respect to B. 3) The overall order of the reaction was determined to be third order, which is the sum of the individual orders with respect to each reactant (2 + 1 = 3

Uploaded by

Ihtisham Ul Haq
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Case II:

When concentration of reactants and rate of the


reaction is given and asking for order of the
reaction.
(Change in Conc)^m = Rate----(1)
Here
m=order of reaction
Now let
1) Conc in mole/dm3 ---------Rate
O.170------------------0.050
0.340------------------0.10
0.680------------------0.20
So what will be the order of the reaction..??????
Solution:
When conc is 0.170 then rate is 0.050. When conc
becomes double that is 0.340 then rate is also
double that is 0.10
Thus
(Change in Conc)^m = rate
(2)^m = 2
m=1
Its a 1st order reaction.

2) A+B--->P
R=k[A]^m [B]^n
Now
Concentration -----------Rate
A : B ----------------R
x : y -----------------z
2x : y -----------------4z
2x : 2y -----------------8z
Then what will be the order of reaction..??????
So
If we double the concentration of A from x to 2x
and keep the concentration of B then rate will
becomes 4 times
Its clear that reaction is of 2nd order with respect
to A.
Proof:
(Change in Conc)^m = Rate
(2)^m = 4
(2)^m = 2^2
m=2
Now
If we double the concentration of B from y to 2y by
keep the concentration of A constant then rate
becomes double from 4z to 8z. Its clear also that
reaction is of 1st order with respect to B.
Proof:
(Change in Conc)^n = Rate
(2)^n = 2
n=1
Now
Overall order of the reaction is
m+n=2+1=3

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