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Question

The following results were obtained during kinetic studies of the reaction:
2A+B Products


Experment[A] (in mol L1)[B] (in mol L1)Initial Rate of reaction
(in mol L1min1)
(I)0.100.206.93×103
(II)0.100.256.93×103
(III)0.200.301.386×102

The time (in minutes) required to consume half of A is:

A
10
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B
5
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C
100
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D
1
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Solution

The correct option is A 5
6.93×103=k×(0.1)x(0.2)y...................(i)

6.93×103=k×(0.1)x(0.25)y..................(ii)

From the above equation, y=0

and 1.386×102=k×(0.2)x(0.30)y...........(iii)

Divide equation (i) by (iii), we get

12=(12)xx=1

So r=k×(0.1)×(0.2)0

6.93×103=k×0.1×(0.2)0

k=6.93×102

t1/2=0.6932k=0.6930.693×101×2=102=5

Hence, the correct option is B

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