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:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
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:
[tex]8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}[/tex]
[tex]\log\frac{0.25}{a-x}=0.15[/tex]
[tex]\frac{0.25}{a-x}=1.41[/tex]
[tex](a-x)=0.17M[/tex]
b) for concentration to decrease from 0.25M to 0.15M
[tex]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[/tex]
[tex]t=687s[/tex]
