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Mrrafil [7]
3 years ago
6

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?
Chemistry
1 answer:
leva [86]3 years ago
7 0

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.

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