900 ml of pure & dry O2 is subjected to O2 seilent electric discharge, so that after a time 10 min. volume of ozonized oxygen is found to be 870 ml. Now, average rate of reaction in this interval is (in ml/min)
(1) 3 (2) 9 (3) 90 (4) 60
Dear student,
Rate of reaction = change in volume of O2 / change in time
= 870-900 / 10
= -30 / 10
= -3 ml min-1
It is negative because volume is decreasing with time.
So the answer is ( 1 ) i.e 3 when only magnitude is considered
Regards
Rate of reaction = change in volume of O2 / change in time
= 870-900 / 10
= -30 / 10
= -3 ml min-1
It is negative because volume is decreasing with time.
So the answer is ( 1 ) i.e 3 when only magnitude is considered
Regards