A certain reaction has an activation energy of 43.41 kj/mol. at what kelvin temperature will the reaction proceed 7.50 times faster than it did at 317 k?

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06-09-2022
Physics

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A certain reaction has an activation energy of 43.41 kj/mol. at what kelvin temperature will the reaction proceed 7.50 times faster than it did at 317 k?

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smecloudbird18 smecloudbird18 · Answer 1 · 2022-09-20T06:41:30-04:00

The temperature required to increase the reaction rate by 7.50 times will be 332.63 K .

The speed at which a chemical reaction proceeds is known as the reaction rate.

We are given that ,

The activation energy = E = 43.41 kj/mol

Initial temperature = T₁ = 317 k

Therefore we can write as, the temperature at which reaction has 7.50 times faster than at 317 K be T₂ . Then the rate of a reaction increases because the rate constant increases with temperature.

The relation between temperatures, rate constants, and activation energy is given as,

(k₁/k₂) = E/R [(1/T₁ ) - (1/T₂)]

Where, k₁ is rate constant at temperature T₁ and k₂ is rate constant at temperature T₂ & R is gas constant having value 8.314 j/kmol .

Thus putting all the given values in above equation then we get,

7.50k/k = (434100 j/mol)/(8.314j/kmol)[(1/317k) - (1/T₂)]

7.50 = 52213.1344 k [ 0.00315 k⁻¹ - 1/T₂]

1/T₂ = 0.00315 K⁻¹ - (7.50/52213.1344)

T₂ = 332.63 K

Therefore the temperature required to increase the reaction rate by 7.50 times is given as T₂ = 332.63 K

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