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AVprozaik [17]
3 years ago
13

Predict which member of each pair will be more acidic. Explain your answers. (a) methanol or tert-butyl alcohol (b) 2-chloroprop

an-1-ol
Chemistry
1 answer:
Art [367]3 years ago
6 0

Answer:

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Explanation:

Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.

This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).

By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.

By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.

Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.

Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.

From above,

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

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myrzilka [38]

Answer: A. an electron

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2 years ago
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The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for
jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)      

KP = 0.13 = \frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

  • With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
  • With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

  • XSO₂ = 0.58/3.29 = 0.176
  • XO₂ = 1.29/3.29 = 0.392
  • XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

  • p(SO₂) = 0.176 * PT
  • p(O₂) = 0.392 * PT
  • p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm

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Answer:

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