$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago

The oxygen atom of a ketone (such as cyclohexanone) contains 2 lone pairs of electrons. These pairs of electrons most likely res

skelet666 [1.2K]

Answer: The given pairs of electrons most likely reside in $sp^{2}$ type of orbital.

Explanation:

As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).

Also, there will be one bond pair between carbon and oxygen atom.

Hence, total electrons present in the domain are as follows.

2 lone pairs + 1 bon pair of electron = 3 electron domains

This means that there will be $sp^{2}$ type of orbital present.

Thus, we can conclude that given pairs of electrons most likely reside in $sp^{2}$ type of orbital.

5 0

3 years ago