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Question

t1/2 of first order reaction is 10 min. Starting with 10 molL1, rate after 20 min is :

A
0.0693 molL1min1
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B
0.0693×2.5 molL1min1
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C
0.0693×5 molL1min1
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D
0.0693×10 molL1min1
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Solution

The correct option is B 0.0693×2.5 molL1min1
For a first order reaction, we have

Initial concentration =10 mol L1

Concentration after 20 min (two half lives) =2.5 mol L1

Now, k=0.693t1/2=0.69310 or 0.0693 min1

Rate=k×[reactant] =0.0693×2.5 mol L1min1

Option (B) is correct.

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