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

Question

Chemistry

Kingston Lutz

27 May, 12:39

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

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Gould · Answer 1

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