Q6.
Ans:AC.
Explain how?

Dear student, 
The graph of second order concentration is given below:

Thus the above plot represents 2nd order reaction. 
Initial concentration of 2nd order reaction is given by (for above graph) [Ao ] =1/y-intercept= 1/2 = 0.5.
Half life of 2nd order reaction  = 1/(k*[Ao]) 
 k →slope = 0.5
Thus half-life = 1/(0.5*0.5) 
Therefore half-life = 4 min. 
Hence options A and C are correct. 
Regards
 

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