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

Question

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)	0.10	0.20	6.93 x 10-3
(II)	0.10	0.25	6.93 x 10-3
(III)	0.20	0.30	1.386 x 10-2

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

(1) 10 

(2) 5 

(3) 100 

(4) 1

Answer 1

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

^t_1/2=ln2/2k      

^{(2k because we have been asked for halftime of A)}​​​​​_{^{​​​​​​}​​​​​​​​​}

t_1/2=5min   ​​​​​​​ans

Answer 2

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.

Comments

Can you explain?