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Q.

T50 (Half-life period) of first-order reaction is 10 minutes. Starting with 10 mol L−1, rate after 20 minute is:

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a

0.0693×5 mol L−1min−1 

b

0.0693×2.5 mol L−1min−1

c

0.0693×10 mol L−1min−1

d

0.0693 mol L−1min−1

answer is B.

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Detailed Solution

Given rate constant,

k=ln⁡210; [A]t=104

Rate after 20 minute is,

rt=k⋅{A]6=ln⁡210×104

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