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qwelly [4]
3 years ago
5

If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation

barrier for the reaction?
Chemistry
1 answer:
Alex73 [517]3 years ago
6 0

Answer: 48,501 J/mol


Explanation:


1) Action barrier = activation energy = Ea


2) Data:


i) T₁ = 12°C = 12 + 273.15 K = 285.15K

ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K

iii) rate constant = k: k₂ / k₁ = 2


iv) Ea = ?


3) Formula:


Arrhenius' law gives the relationship between the constant of reaction and the temperature:


k=Ae^{-\frac{Ea}{RT}}


4) Solution


By arranging the formula, you get:


㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]


Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15


ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K


Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol





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