Question

A reaction that is second-order in one reactant has a rate constant of 1 × 10–2 L/mol · s. If the initial concentration of the reactant is 0.360 mol/L, how long will it take for the concentration to become 0.180 mol/L?

Answer 1

Answer : It take time for the concentration to become 0.180 mol/L will be, 277.8 s

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 = $1\times 10^{-2}mol/L.s$

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.180 mol/L

$[A]_o$ = Initial concentration = 0.360 mol/L

Putting values in above equation, we get:

$1\times 10^{-2}=\frac{1}{t}\left (\frac{1}{0.180}-\frac{1}{0.360}\right)$

$t=277.8s$

Hence, it take time for the concentration to become 0.180 mol/L will be, 277.8 s