For the reaction, $2 \mathrm{~A}+\mathrm{B} \longrightarrow$ products , when the concentrations of A and B both were doubled, the rate of the reaction increased from $0.3 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ to $2.4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. When the concentration of A alone is doubled, the rate increased from $0.3 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ to 0.6 $\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. Which one of the following statements is correct?

(B) Order of the reaction with respect to $B$ is 2

(A) Total order of the reaction is 4

(C) Order of the reaction with respect to B is 1

(D) Order of the reaction with respect to A is 2

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Chemical Kinetics

For the reaction,

2 A + B ⟶

products , when the concentrations of A and B both were doubled, the rate of the reaction increased from

0.3 mol L^{- 1} s^{- 1}

2.4 mol L^{- 1} s^{- 1}

. When the concentration of A alone is doubled, the rate increased from

0.3 mol L^{- 1} s^{- 1}

to 0.6

mol L^{- 1} s^{- 1}

.
Which one of the following statements is correct?

Answer: (B)

Set the rate law as

r = k [A]^{x} [B]^{y}

. Doubling only

A

sent the rate from

0.3

0.6

, a factor

2 = 2^{x}

, so

x = 1

. Doubling both sent it from

0.3

2.4

, a factor

8

; since doubling

A

alone supplies a factor

2

, the

B

doubling must account for the remaining factor

4 = 2^{y}

, so

y = 2

. The reaction is thus first order in

A

, second order in

B

, and third order overall -- so the order with respect to

B

2

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option