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.

Explanation:

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

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:

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

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i think it is true since you can test the answer and can use observations to complete it.