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For a second order reaction, if the conc...

For a second order reaction, if the concentration of a reactant decreases from 0.08 M to 0.04 M in ten minutes, what would be the time taken for the concentration to decreases to 0.01 M from 0.08 M.

Text Solution

Verified by Experts

The correct Answer is:
`70.00`
For second reaction,
`K = (1)/(t)xx(x)/(a(a-x))`
`K = (1)/(10)xx(0.04)/(0.08xx0.04)`
`K=(1)/(0.8)`
Now
`t=(1)/(K)(x)/(a(a-x))`
`= 0.8xx(0.07)/(0.08xx0.01)=(8xx7xx10^(-3))/(8xx10^(-4))`
`t = 70` min
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