Question

Cyclopropane is converted to propene in a first–order process with a rate constant of 5.4 × 10–2 hour–1 . The initial concentration of cyclopropane is 0.150 M. Estimate the concentration of cyclopropane after 22.0 hours

Answer 1

Answer:

The concentration of cyclopropane after 22.0 hour is 0.0457 M.

Explanation:

Conversion of cyclopropane into propene follows first order kinetics.

The integrated rate of first order kinetic is given by :

$[A]=[A_o]\times e^{-kt}$

$[A_o]$ = Initial concentration of reactant

$[A]$ = final concentration of reactant after time t

k = rate constant of the reaction

We have :

Rate constant of the reaction = k = $5.4\times 10^{-2} hour^{-1}$

$[A_o]=0.150 M$

t = 22.0 hour

[A] =?

$[A]=0.150 M\times e^{-5.4\times 10^{-2} hour^{-1}\times 22.hour}$

$[A]=0.0457 M$

The concentration of cyclopropane after 22.0 hour is 0.0457 M.