Question

If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it will take ________ for the concentration to drop to 0.150 . 1.01 5.19 0.299 0.0880 3.34

Answer 1

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

$NO_2\rightarrow NO+\frac{1}{2}O_2$

The reaction is second order for $NO_2$ with a rate constant of $0.543M^{-1}s^{-1}$ at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

Answer: The time taken is 5.19 seconds

Explanation:

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.543M^{-1}s^{-1}$

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

$[A]_o$ = Initial concentration = 0.260 M

Putting values in above equation, we get:

$0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s$

Hence, the time taken is 5.19 seconds