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Chemical Kinetics and Nuclear Chemistry question

2019 · 9 Jan · Shift 1 · Q13
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Chemical Kinetics and Nuclear Chemistry question

2019 · 9 Jan · Shift 1 · Q13

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The following results were obtained during kinetic studies of the reaction ;

2A + B →\to→ Products

Experiment [A] (in mol L−-− 1) [b] (in mol L−-− 1) Initial Rate of reaction
(In mol L−-− 1 min−-− 1)
I 0.10 0.20 6.93 G 10−-− 3
II 0.10 0.25 6.93 G 10−-− 3
III 0.20 0.30 1.386 G 10−-− 2


The time (in minutes) required to consume half of A is :
  1. A
    5
  2. B
    10
  3. C
    1
  4. D
    100
View written solutionFree

Correct answer: B

  1. Write the rate law

For the reaction 2A+B→Products2A + B \to \text{Products}2A+B→Products assume r=k[A]m[B]nr = k[A]^m[B]^nr=k[A]m[B]n

We determine the orders mmm and nnn from the data.

  1. Find order with respect to BBB

Compare Experiments I and II:

  • [A][A][A] is constant at 0.100.100.10
  • [B][B][B] changes from 0.200.200.20 to 0.250.250.25
  • Rate remains same: 6.93×10−36.93 \times 10^{-3}6.93×10−3

So rate is independent of [B][B][B]. Hence, n=0n=0n=0

  1. Find order with respect to AAA

Now compare Experiments I and III. Since we already know n=0n=0n=0, dependence on BBB does not matter.

From I to III:

  • [A][A][A] doubles: 0.10→0.200.10 \to 0.200.10→0.20
  • Rate doubles: 6.93×10−3→1.386×10−26.93 \times 10^{-3} \to 1.386 \times 10^{-2}6.93×10−3→1.386×10−2

Thus, 2m=2  ⟹  m=12^m = 2 \implies m=12m=2⟹m=1

So the rate law is r=k[A]r = k[A]r=k[A] This is a first-order reaction with respect to AAA.

  1. Calculate the rate constant kkk

Using Experiment I: r=k[A]r = k[A]r=k[A] 6.93×10−3=k(0.10)6.93 \times 10^{-3} = k(0.10)6.93×10−3=k(0.10) k=6.93×10−30.10=6.93×10−2 min−1k = \frac{6.93 \times 10^{-3}}{0.10} = 6.93 \times 10^{-2}\,\text{min}^{-1}k=0.106.93×10−3​=6.93×10−2min−1

So, k=0.0693 min−1k = 0.0693\,\text{min}^{-1}k=0.0693min−1

  1. Use first-order half-life formula

For a first-order reaction, t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

Substitute k=0.0693k=0.0693k=0.0693: t1/2=0.6930.0693=10 mint_{1/2} = \frac{0.693}{0.0693} = 10 \text{ min}t1/2​=0.06930.693​=10 min

  1. Final answer

The time required to consume half of AAA is 10 min\boxed{10\text{ min}}10 min​

So the correct option is B.

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