Question

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 1

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