Question

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.

Answer 1

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

Answer 2

The rate constant at 48.5°C if the activation energy is 116 kJ/mol is $1.017 E-8 s^{-1}$

Chemical reaction:

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 = ?

According to Arrhenius equation:

$K(T) = A*e^{(-Ea/RT)}\\\\ln K = ln(A) - [(Ea/R)(1/T)]$

where,

A = frequency factor

R = 8.314 E-3 KJ/K.mol

$ln K_1 = ln(A) - [Ea/R)*(1/T_1)]$..........(1)

$ln K_2 = ln(A) - [(Ea/R)*(1/T_2)]$...........(2)

On dividing equation 1 by 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 = $1.017 E-8 s^{-1}$

Thus, the rate constant at 48.5°C if the activation energy is 116 kJ/mol is 1$0.017 E-8 s^{-1}$

Find more information about Activation energy here:

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