Gaseous cyclobutene isomerizes to butadiene in a first order process which has a 'K' value of $$3.3 \times 10^{-4}$$ s$$^{-1}$$ at 153°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)

We are given that gaseous cyclobutene isomerizes to butadiene in a first order process with rate constant $$k = 3.3 \times 10^{-4} \text{ s}^{-1}$$ at 153°C. We need to find the time for 40% completion.

For a first order reaction, the integrated rate law is:

$$t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right)$$

If the reaction is 40% complete, then 40% of the reactant has been consumed and 60% remains. So $$[A] = 0.60 \, [A]_0$$.

$$t = \frac{1}{k} \ln\left(\frac{[A]_0}{0.60 \, [A]_0}\right) = \frac{1}{k} \ln\left(\frac{1}{0.60}\right) = \frac{1}{k} \ln\left(\frac{5}{3}\right)$$

Now, $$\ln\left(\frac{5}{3}\right) = \ln 5 - \ln 3 = 1.6094 - 1.0986 = 0.5108$$

Substituting:

$$t = \frac{0.5108}{3.3 \times 10^{-4}} = 1548 \text{ s}$$

Converting to minutes:

$$t = \frac{1548}{60} = 25.8 \text{ min} \approx 26 \text{ min}$$

So, the answer is $$26$$.