Question

Isomerisation of gaseous cyclobutane to butadiene is a first-order reaction. At \( T(K) \), the rate constant of the reaction is \( 3.3 \times 10^{-4} \, \text{s}^{-1} \). What is the time required (in min) to complete 90% of the reaction at the same temperature? (log 2 = 0.3)

• A. 116.67
• B. 233.34
• C. 58.34
• D. 350.0

Hint:

For first-order reactions, use the logarithmic formula to calculate the time for a given percentage completion.

Answer 1

The Correct Option is A

We use the first-order rate equation for this reaction: \[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \] where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant. For 90% completion of the reaction, the concentration left is 10% of the initial concentration, i.e., \([A] = 0.1 [A]_0\). Substituting into the equation, we get: \[ \ln \left( \frac{[A]_0}{0.1 [A]_0} \right) = k t \] \[ \ln (10) = k t \] Since \(\ln 10 = 2.3\), the equation becomes: \[ 2.3 = k t \] Substitute the value of \( k = 3.3 \times 10^{-4} \, \text{s}^{-1} \): \[ t = \frac{2.3}{3.3 \times 10^{-4}} = 6969.7 \, \text{s} \] Convert the time into minutes: \[ t = \frac{6969.7}{60} \approx 116.67 \, \text{min} \]