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Chemical Kinetics — Chemistry STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryChemical Kinetics2 Marks
Question
A reaction is first order in A and second order in B.
  1. Write the differential rate equation.
  2. How is the rate affected on increasing the concentration of B three times?
  3. How is the rate affected when the concentrations of both A and B are doubled?
✓

Answer

  1. Differential rate equation of reaction is
$\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^1[\text{B}]^2=\text{k}[\text{A}]\ [\text{B}]^2$
  1. When cone. of B is tripled, it means cone. of B becomes (3 × B)
$\therefore$ New rate of reaction,
$\frac{\text{dx'}}{\text{dt}}=\text{k}[\text{A}]\ [\text{3B}]^2=9\text{k}[\text{A}]\ [\text{B}]^2=\bigg(\frac{\text{dx}}{\text{dt}}\bigg)$
i.e., rate of reaction will become 9 times.
  1. When cone. of A is doubled and that of B is also doubled, then cone. of A becomes [2A] and that of B becomes [2B] rate of reaction,
$\frac{\text{dx''}}{\text{dt}}=\text{k}[\text{2A}]\ [\text{3B}]^2=8\text{k}[\text{A}]\ [\text{B}]^2$
i.e., the rate of reaction will become 8 times the rate as in (1).

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