order of reaction!

b. (i)
Let Rate = k [I2]^a [CH3COCH3]^b [H^+]^c

3.5 x 10^-5 = k (2.5 x 10^-4)^a (2.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (1)
3.5 x 10^-5 = k (1.5 x 10^-4)^a (2.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (2)
1.4 x 10^-4 = k (2.5 x 10^-4)^a (4.0 x 10^-1)^b (1.0 x 10^-2)^c ...... (3)
7.0 x 10^-5 = k (2.5 x 10^-4)^a (4.0 x 10^-1)^b (5.0 x 10^-3)^c ...... (4)

(1)/(2):
1 = (5/3)^a
a = 0

(3)/(4):
2 = 2^c
c = 1

(1)/(4):
(1/2) = (1/2)^b
b = 1

Hence, Rate = k [CH3COCH3][H^+]


b. (ii)
Put the first set of data into the rate equation :
3.5 x 10^-5 = k (2.0 x 10^-1) (5.0 x 10^-3)
Rate constant, k = 3.5 x 10^-2 (mol dm^-3)^-1 s^-1


(c)
When pH = 4 :
[H^+] = 10^-pH = 1 x 10^-4 M

Rate = (3.5 x 10^-2) x [CH3COCH3] x (1 x 10^-4)
Hence, Rate = (3.5 x 10^-6) x [CH3­COCH3]

It becomes a pseudo-first-order reaction in the form :
Rate = k' [CH3COCH3]

Half life of the pseudo-first-order reaction, t1/2
= ln2/k'
= ln2/(3.5 x 10^-6)
= 1.98x 10^5 s