menu search
brightness_auto
Feel free to answer or ask any questions.
If you are new here please see how to use or FAQ.
more_vert

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

2 Answers

more_vert
 
verified
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

thumb_up_off_alt 2 like thumb_down_off_alt 0 dislike
more_vert
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.
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
more_vert
Can you explain?

Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.


Join our Telegram group for live discussion.

Telegram Group

Subscribe our YouTube channel for video solutions with explanation.

YouTube Channel

Download Jee Neet QnA Books in PDF for offline learning.

Jee Neet QnA Books

1.2k questions

844 answers

385 comments

139 users

...