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

a electronegativity decreases beacuse the size of nucleus is bigger.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.