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.

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

b) nuclear ive learned that in a pedtrodic table

c. electrical charges of ionic bonds.