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
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