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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?

Answer :

Answer: 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:

[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]

a= concentration left after time t = [tex]\frac{28.2}{100}\times 1.77=0.50M[/tex]

k = rate constant =[tex]2.85M^{-1}min^{-1}[/tex]

[tex]{a_0}[/tex]  = initial concentration = 1.77

t= time taken for decomposition

[tex]\frac{1}{0.50}=2.85\times t+\frac{1}{1.77}[/tex]

[tex]t=0.50min[/tex]

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

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