The initial concentrations of both the reactants of a second order reaction are equal and 60% of the reaction gets completed in 30 s. How much time will be taken in 20% completion of the reaction?
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answer is 5.
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Detailed Solution
For second order:k2=1t⋅xa(a−x) [Let a = 1]=130s×0.61(1−0.6)=130×0.60.4Now for 20% completion,k2=1t⋅xa(1−x)(Since k2 is constant)130×0.60.4=1t×14t=300.6×0.44=5s