At a certain temperature, the decomposition of NO2 is second order with a rate constant of 2.85 M-1min-1. If one starts with 1.77 M of NO2, how many minutes will it take for it to decompose to 28.2% of its initial value?

It will take 0.50 minutes for it to decompose to 28.2% of its initial value

Explanation:

Integrated rate law for second order kinetics is given by:

a= concentration left after time t =

k = rate constant =

 = initial concentration = 1.77

t= time taken for decomposition

Thus it will take 0.50 minutes for it to decompose to 28.2% of its initial value

they are all solids.

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