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solong [7]
3 years ago
7

If ∆G◦ = 27.1 kJ at 25◦C for the reaction CH3COOH (aq) + H2O (l) → CH3COO− (aq) + H3O+ (aq), calculate Ka for this reaction at 2

98 K.1. 1.012. 5.63 × 1043. 1.15 × 10−114. 9.89 × 10−15. 1.78 × 10−5
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

Ka = 1.78 × 10⁻⁵

Explanation:

Let's consider the following thermochemical equation.

CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)   ∆G° = 27.1 kJ/mol

At 25°C (298 K), we can find the equilibrium constant (Ka) using the following expression.

∆G° = - R × T × lnKa

where,

R: ideal gas constant

T: absolute temperature

27.1 × 10³ J/mol = - (8.314 J/K.mol) × 298 K × lnKa

Ka = 1.78 × 10⁻⁵

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A 104 m3 thoroughly mixed pond has a water inflow and outflow of 5 m3/h. The inflow water contains 0.01 mol/m3 of chemical. Chem
Naily [24]

Answer:

C ≈ 1.44 × 10⁻³ mol/m³

Explanation:

The given information are;

The liquid volume the pond can hold = 104 m³

The volume of inflow into the pond = 5 m³/h

The volume of outflow into the pond = 5 m³/h

The concentration of the chemical in the inflow water = 0.01 mol/m³

The concentration of the chemical discharged directly into the water = 0.1 mol/h

The concentration, c_{(inflow)}, of chemical that enters the water through inflow per hour is given as follows;

c_{(inflow)} = 0.01 mol/m³ × 5 m³/h = 0.05 mol/h

The concentration, c_{(discharge)}, of chemical that enters the water through direct discharge per hour is given as follows;

c_{(discharge)} = 0.1 mol/h

The total concentration that enters the pond per hour is given as follows;

c_{(inflow)} + c_{(discharge)} = 0.1 mol/h + 0.05 mol/h = 0.15 mol/h

Whereby the water in the pond properly mixes with the pond, we have;

The concentration of chemicals (C) in the outflow water = 0.15 mol/(104 m³) ≈ 0.00144 mol/m³

C ≈ 1.44 × 10⁻³ mol/m³.

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3 years ago
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IceJOKER [234]

Answer:

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Explanation:

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Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.
il63 [147K]

Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O  ⇄  C₆H₅-COO⁻  +  H₃O⁺    (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺        (2)

K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}      (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

6 0
3 years ago
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