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3 years ago

Benzene is the simplest example of _____ compound.

1 answer:

4 0

Benzene is the simplest example of the aromatic compound
Firstly, we want to explain what is the meaning of aromaticity?
the aromaticity is called on any compound which has cyclic shape and has a resonance. Resonance is delocalization of the electrons (which form the double bond) in the molecules, in other words, the double bond moves through the compound by the movement of electrons which leads to increasing the stability of the compound by this process which called resonance. The resonance may occur in open or cyclic compounds, and if it occurs in cyclic compounds, these compounds are called aromatic compounds. So, the benzene considered to be the simplest aromatic compound where it is not only has a resonance in a cyclic ring but also it was the first aromatic compound discovered and also, the most common aromatic compounds are derivatives of it (benzene).

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deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

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Jlenok [28]

Answer:

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Explanation:

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