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The following results were obtained during kinetic studies of the reaction:

2A + B $\to$ Products

Experiment[A]
(in mol L-1)
[B]
(in mol L-1)
Initial rate of reaction
(in mol L-1 min-1)
(I)0.100.206.93 x 10-3
(II)0.100.256.93 x 10-3
(III)0.200.301.386 x 10-2

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

(1) 10 

(2) 5 

(3) 100 

(4) 1

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2 Answers

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Best answer

On increasing [B] no effect on rate of rxn

Thus order with respect to B is ZERO.

Now on doubling [A] doubles the rate of rxn

​​​​​​​thus order with respect to A is 1

Now finding rate constant k.  

Rate =k[A]

6.93 * 10-3=k (0.1)

K=6.93*10-2

t1/2=ln2/2k      

(2k because we have been asked for halftime of A)​​​​​​​​​​​​​​​​​​​​

t1/2=5min   ​​​​​​​ans

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It has silly answer that is high/low calculation 2/1 and 3/1 get answer correctly if not get answer I will prove and written as a comment.
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Can you explain?
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