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jasenka [17]
3 years ago
8

When propyl alcohol is treated with acid, the initially formed intermediate is known as an oxonium ion. There is a scheme of a r

eversible chemical reaction. The substrates are CH3CH2CH2OH molecule and H with a charge of 1 plus ion. The product is CH3CH2CH2OH2 with a charge of 1 plus ion. Oxygen atom in CH3CH2CH2OH molecule has 2 lone pairs. Oxygen atom in CH3CH2CH2OH2 with a charge of 1 plus ion has a lone pair. All bonds are single.
Using the curved arrow formalism, show how this process most likely occurs.
Chemistry
1 answer:
MAVERICK [17]3 years ago
4 0

Answer:

See explanation and image attached

Explanation:

The question has to do with the protonation of an alcohol. We know that oxygen has two lone pairs of electrons. Now a lone pair of electrons can pick up a proton leading to the formation of a coordinate bond between the proton and the oxygen atom.

The oxygen atom in the alcohol now has a positive charge as shown in the image attached to this answer. All the bonds remain single.

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Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
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Explanation:

1) 2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)

Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091

2) H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)

Kc=\frac{[H_2SO_4]}{[SO_3]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3

3)  P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)

Kc=\frac{1}{[O_2]^3}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641

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4 years ago
Thermoplastics can be made to act more like thermosets by a process called cross-linking
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hat is the pressure of CO(g) in equilibrium with the CO2(g) and O2(g) in the atmosphere at 25 ????C? The partial pressure of O2(
Lubov Fominskaja [6]

Answer:

The partial pressure of CO is 5.54x10⁻⁴⁹atm. You shouldn't worry because it is very low pressure

Explanation:

First, the balanced reaction is:

CO + 1/2O₂ → CO₂

The energies of formation are:

ΔG(CO)=-137.168kJ/mol

ΔG(O₂)=0

ΔG(CO₂)=-394.359kJ/mol

The energy of the reaction is:

delta-G_{reaction} =delta-G_{CO_{2} } -(delta-G_{CO} +1/2delta-G_{O_{2} } )\\delta-G_{reaction}=-394.359-(-137.168+0)=-257.191kJ/mol

The expression for calculate the partial pressure of CO is:

p_{CO} =\frac{p_{CO2} }{p_{O_{2} }^{1/2}*exp^{-\frac{delta-G}{RT} }   } \\p_{CO}=\frac{3x10^{-4} }{0.2^{1/2}*exp(-\frac{-257.191*1000}{8.314*298} ) } \\p_{CO}=5.54x10^{-49} atm

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