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Sagot :
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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