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NemiM [27]
3 years ago
11

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 1.7

7 M of NO2, how many minutes will it take for it to decompose to 28.2% of its initial value?
Chemistry
1 answer:
alex41 [277]3 years ago
8 0

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:

\frac{1}{a}=kt+\frac{1}{a_0}

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

k = rate constant =2.85M^{-1}min^{-1}

{a_0}  = initial concentration = 1.77

t= time taken for decomposition

\frac{1}{0.50}=2.85\times t+\frac{1}{1.77}

t=0.50min

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

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