KEAM 2012 Physics and Chemistry Paper

Let the order of reaction with respect to A is n.
therefore Rate law is given as,
rate, ‌‌r=k[A0]n.....(i)
On putting the given values in Eq. (i), we get
5.0×10−5=k[0.01]n.....(i)
2.0×10−4=k[0.02]n.....(iii)
Dividing Eqs. (iii) by (ii) gives,
[‌

2.0×10−4

5.0×10−5

]=[‌

0.02

0.01

]n
[‌

20

5

]=(2)n
or (4)=(2)n
(2)2=(2)n
∴‌‌n=2
On putting the value of n in Eq. (i), we get
rate, r=k[A0]2
∴‌‌5×10−5=k(0.01)2