VITEEE 2010 Solved Paper

let the initial rate be R and order with respect to A be x and B be y. Thus, rate law can be written as,
rate, R=[A]x[B]y...(i)
After doubling the concentration of A, rate becomes 4R,
4R=[2A]x[B]y ...(ii)
After doubling the concentration of B, rate remains R,
R=[A]x[2B]y ...(iii)
From Eq. (i) and (ii), we get

R

4R

=[

1

2

]x⇒[

1

2

]2=[

1

2

]x
So, x = 2
From Eq. (i) and (iii), we get

R

R

=[

1

2

]y⇒[

1

1

]0=[

1

2

]y
So, y = 0
Hence, the rate law is, rate R=[A]2[B]0
This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are mol−1Ls−1.