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The decomposition of azomethane, at a certain temperature according to the equation CH 3 2 N 2 → C 2 H 6 + N 2 is a first-order reaction. After 20 minutes from the start, the total pressure developed is found to be 350   mm   Hg in place of initial pressure 200   mm   of   Hg of azomethane. The value of rate constant k is:

Options

  1. A2.303 × 1 0 - 2 min -1
  2. B0.693 × 1 0 - 5 min -1
  3. C6.93 × 1 0 - 2 min -1
  4. D13.8 × 1 0 - 7 min -1

Correct answer

C. 6.93 × 1 0 - 2 min -1

Step-by-step solution

This problem is based on conceptual mixing of determination of partial pressure and rate constant, which can be solved as follows. k = 2.303 t log Pi Pt CH 3 2 N 2 → C 2 H 6 + N 2 t = 0 2 0 0 0 0 t = t 200-x x x Given, 2 0 0 - x + x + x = 3 5 0 x = 1 5 0 k = 2.303 t log 2 0 0 2 0 0 - 1 5 0 k = 2.303 2 0 log 2 0 0 5 0 k = 2.303 2 0 × 2 log 2 k = 2.303 2 0 × 2 × 0.3010 = 0.693 1 0 = 6.93 × 1 0 - 2 min -1

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