QJEEM20Chee91: JEE Main 2020 Chemistry - Class 12 JEEM20Chee91 Solution: The results given in the below table were obtained during…

Rate = k[A]^a[B]^b

6 × 10^–3 = k(0.1)^a(0.1)^b ...(1)

2.4 × 10^–3 = k(0.1)^a(0.2)^b ...(2)

1.2 × 10^–3 = k(0.2)^a (0.1)^b ...(3)

(3) ÷ (1) $\Rightarrow$ x = 1

(2) ÷ (3) $\Rightarrow$ x = 2

So, order with respect to A = 1

Order with respect to B = 2

6 × 10^–3 = k(0.1)¹ (0.1)² from (1)

$\Rightarrow$ k = 6

From experiment IV,

7.2 × 10^–2 = 6(x)¹(0.2)² $\Rightarrow$ x = 0.3

From experiment V,

2.88 × 10^–2 = 6(0.3)¹(y)² $\Rightarrow$ y = 0.4