Question

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) if the initial concentration of cyclopropane was 0.25M , what is the concentration after 8.8min?b) how long (in min) will it take for the concentration of cyclopropane to decrease from 0.25M to 0.15M?

Answer 1

Answer: a)  The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$