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enyata [817]
3 years ago
15

Enthalpy of combustion of ethyl alcohol, C2 H5 OH, is – 950 kJ mol–1. How much heat is evolved when one gram of ethyl alcohol bu

rns?​
Chemistry
1 answer:
melomori [17]3 years ago
7 0
<h3>Answer:</h3>

20.62 Kilo-joules

<h3>Explanation:</h3>
  • The Enthalpy of combustion of ethyl alcohol is -950 kJ/mol.
  • This means that 1 mole of ethyl alcohol evolves a quantity of heat of 950 Joules when burned.

Molar mass of ethyl ethanol = 46.08 g/mol

Therefore;

46.08 g of  C₂H₅OH evolves heat equivalent to 950 kilojoules

We can calculate the amount of heat evolved by 1 g of C₂H₅OH

Heat evolved by 1 g of C₂H₅OH  = Molar enthalpy of combustion ÷ Molar mass

                                      = 950 kJ/mol ÷ 46.08 g/mol

                                      = 20.62 Kj/g

Therefore, a gram of C₂H₅OH  will evolve 20.62 kilo-joules of heat

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A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
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pH = 3.70

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0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

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C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

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HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

7 0
3 years ago
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