The activation energy of a certain reaction is 44.8 kj/mol. At 20 degree celcius

Question

The activation energy of a certain reaction is 44.8 kj/mol. At 20 degree celcius , the rate constant is0.0130s^-1 . At what temperature would this reaction go twice as fast?

Explanation / Answer

k1 = A · e^-(Ea / R·T1) ~(eq1) where, k1 = 0.0130 / sec A = frequency or pre-exponent factor Ea = 44.5 kJ/mol = 44,500 J/mol R = 8.314472 J/mol·K T1 = 20 ºC + 273.15 = 293.15 K At what temperature (T2), would this reaction go twice as fast? *Note below - k is proportional to rate: ==> k2 = 2k1 ~(eq2) Substituting k1 into ~(eq2): k2 = 2k1 = 2A · e^-(Ea / R·T1) ~(eq3) Also what k2 is by definition: k2 = A · e^-(Ea / R·T2) ~(eq4) Set ~(eq3) = ~(eq4) 2A · e^-(Ea / R·T1) = A · e^-(Ea / R·T2) Dividing both sides of previous equation by A: ==> 2·e^-(Ea / R·T1) = e^-(Ea / R·T2) ~(eq5) Substituting numbers into ~(eq5): 2·e^-{44,500 J/mol / [(8.314472 J/mol·K)·293.15 K]} = e^-{44,500 J/mol / [(8.314472 J/mol·K)·T2)} Evaluation the exponents on both sides: 2·e^-{18.257} = e^-{5352.1 K / T2} Solving for 2·e^-{18.257} on left: 2.3551 x 10^-8 = e^-{5352.1 K/ T2} Takinig ln of both sides: ln (2.3551 x 10^-8) = ln (e^-{5352.1 K/ T2}) Evaluating ln on left: Using rule ln (e^-x) = - x, on right: -17.564 = - 5352.1 K / T2 Multiplying both sides by T2: -17.564 ·T2 = - 5352.1 K Solving for T2: T2 = - 5352.1 K / -17.564 T2 = 304.72 K T2 = 304.72 K - 273.15 = 31.57 ºC

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