± The Arrhenius Equation Part A The activation energy of a certain reaction is 3

Question

± The Arrhenius Equation

Part A

The activation energy of a certain reaction is 38.6 kJ/mol . At 24 C , the rate constant is 0.0160s1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0160s1 at an initial temperature of 24  C , what would the rate constant be at a temperature of 140.  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Explanation / Answer

the Arrhenius equation is as follow:

K = A exp(-Ea/RT)

We want the temperature where K2 = 2K1 so:

K1/K2 = exp(-Ea/RT1 + Ea/RT2)
ln(K1/K2) = Ea/R(1/T2 - 1/T1) ---> T1 = 24+273 = 297 K
ln(1/2) = 38600/8.3144 (1/T2 - 1/297)
-0.693 = 4642.55(1/T2 - 3.37x10-3)
-1.49x10-4 = 1/T2 - 3.37x10-3
1/T2 = 3.221x10-3
T2 = 310.46 K ---> 310.46-273 = 37.46 °C

For the second part, we just solve for K2:
ln(0.0160/K2) = 38600/8.3144(1/413 - 1/297)
ln(0.0160/K2) = -4.39
0.0160/K2 = exp(-4.39)
K2 = 0.0160/0.0124
K2 = 1.2908 s-1

Hope this helps

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