Question

At 1000°C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 68 1/s, to two molecules of ethylene (C2H4). The initial cyclobutane concentration is 1.84. How long will it take for 52% of the cyclobutane to decompose? Enter to 4 decimal places.

Answer 1

Answer:

$t=0.0114 s$

Explanation:

For first order reaction we are going to use the following formula:

$t=\frac{2.73}{k} \log\frac{x}{x-a}$

where:

k is the constant

x is the initial concentration

a is the final concentration

x-a is the amount left

IN our Case:

k=68 (1/s), x=1.84. Calculating x-a:

% of cyclobutane left after time t=100-52=48%

Final Concentration=1.84*48%=0.8832

x-a=Initial concentration - Final Concentration

x-a=1.84-0.8832

x-a=0.9568

Now:

$t=\frac{2.73}{68} \log\frac{1.84}{0.9568}$

$t=0.0114 s$

In order,for cyclobutane to decompose to 52%, time taken is 0.0114 sec.