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JEE Main 20199 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2019 · 9 January · Evening Shift · Chemistry · Chemical Kinetics

For the reaction, 2A + B → products, when the concentration of A...

For the reaction, 2A+B→2A + B \to products, when the concentration of AA and BB both were doubled, the rate of the reaction increased from 0.3molL−1s−10.3mol{L}^{- 1}{s}^{- 1} to 2.4molL−1s−1.2.4mol{L}^{- 1}{s}^{- 1} . When the concentration of AA alone is doubled, the rate increased from 0.3molL−1s−10.3mol{L}^{- 1}{s}^{- 1} to 0.6molL−1s−1.0.6mol{L}^{- 1}{s}^{- 1} .

Which one of the following statements is correct?

Options

  1. AOrder of the reaction with respect to BB is 22
  2. BTotal order of the reaction is 44
  3. COrder of the reaction with respect to AA is 22
  4. DOrder of the reaction with respect to BB is 11

Correct answer

A. Order of the reaction with respect to BB is 22

Step-by-step solution

2A+B→C2A + B \to C

Initial r1=k[A]x[B]y=0.3m/sec{r}_{1} = k{\left[A\right]}^{x}{\left[B\right]}^{y} = 0.3m / sec ...(i)

r2=k[2A]x[2B]y=2.4{r}_{2} = k{\left[2A\right]}^{x}{\left[2B\right]}^{y} = 2.4 ...(ii)

r3=k[2A]x[B]y=0.6{r}_{3} = k{\left[2A\right]}^{x}{\left[B\right]}^{y} = 0.6 ...(iii)

From (i) ÷\div (iii)

(12)x(1)y=0.30.6{\left(\dfrac{1}{2}\right)}^{x}{\left(1\right)}^{y} = \dfrac{0.3}{0.6}

(12)x=12{\left(\dfrac{1}{2}\right)}^{x} = \dfrac{1}{2}

x=1x = 1

(ii) ÷\div (iii)

r2r3=2.40.6=(2)y\dfrac{{r}_{2}}{{r}_{3}} = \dfrac{2.4}{0.6} = {\left(2\right)}^{y}

4=(2)y4 = {\left(2\right)}^{y}

y=2y = 2

The order of reaction w.r.t. A=1A = 1 and w.r.t. B=2B = 2

Total order of reaction =1+2=3= 1 + 2 = 3

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