reneebrown017
11.01.2020 •
Chemistry
At 1000°c, cyclobutane (c4h8) decomposes in a first-order reaction, with the very high rate constant of 76 1/s, to two molecules of ethylene (c2h4). the initial cyclobutane concentration is 1.63. how long (in seconds) will it take for 79% of the cyclobutane to decompose? enter to 4 decimal places.
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Ответ:
It will take 0.0205s for 79% of cyclobutane to decompose
Explanation:
Step 1: Data given
Temperature = 1000 °C
Reaction is first order
rate constant = 76*1/s
The initial cyclobutane concentration is 1.63M
How long (in seconds) will it take for 79% of the cyclobutane to decompose?
When 79% decomposes, there will remain 21 %
ln ([A]0 / [A]t) * = k*t
There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.
⇒ with [A]0 = the original amount = 100 % =1
⇒ with [A]t = the amount after it's decomposed = 21 % =0.21
⇒ k = the rate constant = 76/s
⇒ t= the time needed = ?
ln (100/21) = 76t
t = 0.0205 s
It will take 0.0205s for 79% of cyclobutane to decompose
Ответ: