Question

At 1000°C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 79 1/s, to two molecules of ethylene (C2H4). The initial cyclobutane concentration is 1.68. How long will it take for 52% of the cyclobutane to decompose?

Answer 1

Explanation:

It is known that for first order reaction, the equation is as follows.

           t = $\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}$

    t = ?,        K = rate constant = 79 1/s

Initial conc. of $C_{4}H_{8}$ = 1.68

Decompose amount of $C_{4}H_{8}$ = 52% of 1.68

                                                   = $\frac{52}{100} \times 1.68$

                                                   = 0.8736

                                                   = 0.87

Now, $[C_{4}H_{8}]_{t}$ = (1.68 - 0.87)

                         = 0.81

Therefore, calculate the value of t as follows.

               t = $\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}$

                  = $\frac{2.303}{79 s^{-1}} log \frac{1.68}{0.81}$

                  = $\frac{2.303}{79 s^{-1}} \times 0.316$

                  = $9.212 \times 10^{-3}$ s

                  = $0.00921 s$

Thus, we can conclude that 0.00921 s will be taken for 52% of the cyclobutane to decompose.