# Gaseous cyclobutene isomerizes to butadiene in a first order process which has a

Question:

Gaseous cyclobutene isomerizes to butadiene in a first order process which has a ' $\mathrm{k}$ ' value of $3.3 \times 10^{-4} \mathrm{~s}^{-1}$ at $153^{\circ} \mathrm{C}$. The time in minutes it takes for the isomerization to proceed $40 \%$ to completion at this temperature is___________ . (Rounded off to the nearest integer)

Solution:

(26)

For firdst order $\operatorname{Rxn}: t=\frac{2.303}{k} \log \left[\frac{100}{100-x}\right]$

$\mathrm{X}=40, \mathrm{k}=3.3 \times 10^{-4}$

$\mathrm{t}=\frac{2.303}{3.3 \times 10^{-4}} \log \left[\frac{100}{60}\right]$

For firdst order $\mathrm{Rxn}: \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \left[\frac{100}{100-\mathrm{x}}\right]$

$X=40, k=3.3 \times 10^{-4}$

$t=\frac{2.303}{3.3 \times 10^{-4}} \log \left[\frac{100}{60}\right]$

$\mathrm{t}=\frac{2.303}{3.3 \times 10^{-4}} \times 0.22$

$\mathrm{t}=0.1535 .3 \times 10^{4}$

$\mathrm{t}=1535 \mathrm{sec}$

$\mathrm{t}=0.1535 .3 \times 10^{4}$

$\mathrm{t}=1535 \mathrm{sec}=25.6 \mathrm{Min}$