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DedPeter [7]
3 years ago
5

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

c acid (HCl) is 3.32 × 10−10 s−1 at 25°C. Calculate the rate constant at 48.5°C if the activation energy is 116 kJ/mol.
Chemistry
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

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The I.C.E table can be represented as:

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However, Since the amount of sulfur trioxide gas to be 1.6 mol.

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For 2SO₂; we have 14 - 2x

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= 14 - 1.6

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Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

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= 0.8325

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If we want to find:

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Then:

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