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

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.