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BARSIC [14]
3 years ago
13

If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation

barrier for the reaction?
Chemistry
1 answer:
Airida [17]3 years ago
4 0

Answer:

59.077 kJ/mol.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

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