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Chemical Kinetics and Nuclear Chemistry question

2016 · 10 Apr · Shift 1 · Q14
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Chemical Kinetics and Nuclear Chemistry question

2016 · 10 Apr · Shift 1 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate law for the reaction below is given by the expression k [A] [B] A + B →\to→ Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :
  1. A
    k
  2. B
    k/3
  3. C
    3k
  4. D
    9k
View written solutionFree

Correct answer: A

  1. Given rate law

    The reaction is: A+B→ProductA + B \to \text{Product}A+B→Product

    with rate law: Rate=k[A][B]\text{Rate} = k[A][B]Rate=k[A][B]

  2. What is being changed?

    • Initial concentration of B=0.1B = 0.1B=0.1 mole
    • Final concentration of B=0.3B = 0.3B=0.3 mole
    • Concentration of AAA remains constant at 0.10.10.1 mole

  3. Meaning of rate constant kkk

    The rate constant kkk depends only on factors like:

    • temperature
    • catalyst
    • nature of reactants

    It does not depend on the concentrations of reactants.

  4. Effect of increasing [B][B][B]

    Since Rate=k[A][B],\text{Rate} = k[A][B],Rate=k[A][B], increasing [B][B][B] from 0.10.10.1 to 0.30.30.3 will increase the rate by a factor of 333 (because AAA is constant), but the rate constant kkk remains unchanged.

  5. Option check

    • A: kkk ✅ Correct
    • B: k/3k/3k/3 ❌ Incorrect
    • C: 3k3k3k ❌ Incorrect
    • D: 9k9k9k ❌ Incorrect

  6. Final answer

    The rate constant remains: k\boxed{k}k​

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