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Question

For a consecutive reaction R1k1R2  R1k2R3 If initial concentration of R1 is 100 M and k1:k2=1:0.15 the value of [R2]max is   [Given:k1=4.0×102min1]


  1. 55.71 M

  2. 71.55 M

  3. 7.155 M

  4. 5.571 M


Solution

The correct option is B

71.55 M


The values of [R2]max is given by the relation
[R2]max=[R1]0[k2k1]k2(k2k1)
[R2]max=100[0.15×1021.0×102]0.015×102(10.15)×102=100[0.151]0.150.85=71.55M.

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