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Chemical Kinetics and Nuclear Chemistry question

2018 · 16 Apr · Shift 1 · Q8
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Chemical Kinetics and Nuclear Chemistry question

2018 · 16 Apr · Shift 1 · Q8

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 secod, the order of this reaction is :
  1. A
    Zero
  2. B
    1
  3. C
    2
  4. D
    3
View written solutionFree

Correct answer: B

  1. Let the initial concentration of reactant be aaa.

  2. After 100 s100\ \text{s}100 s, 50%50\%50% reaction has occurred.

    So, concentration left is: [A]100=a−0.5a=a2[A]_{100} = a - 0.5a = \frac{a}{2}[A]100​=a−0.5a=2a​

  3. After 200 s200\ \text{s}200 s, 75%75\%75% reaction has occurred.

    So, concentration left is: [A]200=a−0.75a=a4[A]_{200} = a - 0.75a = \frac{a}{4}[A]200​=a−0.75a=4a​

  4. Compare the two times:

    • At t=100 st=100\ \text{s}t=100 s, reactant left =a2= \frac{a}{2}=2a​
    • At t=200 st=200\ \text{s}t=200 s, reactant left =a4= \frac{a}{4}=4a​

    This means in the next 100 s100\ \text{s}100 s, the concentration again becomes half: a2→a4\frac{a}{2} \to \frac{a}{4}2a​→4a​

  5. Thus, the reaction takes equal time intervals (100 s100\ \text{s}100 s) for the concentration to halve.

    Equal half-life is the characteristic of a first-order reaction.

  6. Therefore, the order of the reaction is: 1\boxed{1}1​

  7. Checking options:

    • A: Zero →\rightarrow→ Incorrect
    • B: 111 →\rightarrow→ Correct
    • C: 222 →\rightarrow→ Incorrect
    • D: 333 →\rightarrow→ Incorrect
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