What is the Arrhenius equation 2

First order reactions are always assumed, so the unit of k is always time-1.
The following is used for the relationship between ° C and K: 0 ° C = 273.15 K.

1) At 40 ° C the rate constant is 0.5 and at 70 ° C it is 5. Calculate E.A. in kJ / mol.
For the solution, a tabular arrangement is useful in order to insert the numbers correctly into the equation!

 

k

T (° C)

T (K)

1

k1 = 0,5

40

T1 = 313,15

2

k2 = 5

70

T2 = 343,15

Partial results: ln (k1 / k2) = -2.3026; (1 / T2 - 1 / T1) = -2,7918*10-4;
E.A.= ln (k1 / k2) * R / {(1 / T2) - (1 / T1)} = 68,571.06 J / mol (because R also contains J) ~ 68.6 kJ / mol.

2) At 60 ° C the rate constant is 15; the activation energy is 69.8 kJ / mol.
What is the rate constant at 40 ° C?

Approach: T1 = 40 ° C = 313.15 K: k1 searched; T2 = 60 ° C = 333.15 K; k2 = 15.
Note in the equation that EA. is used in J / mol!
(E.A. / R) * (1 / T2 - 1 / T1) = (69.800 / 8,314) * ... = 8.395,48 * (-1,917*10-4) = -1,6095
ln (k1 / k2) = -1.6095; so (k1 / k2) = exp (-1.6095) = 0.1999;
k1 = 0.1999 * k2 = 2,999 ~ 3.

3) At -5 ° C the rate constant is 4; the activation energy is 70 kJ / mol.
At what temperature, in ° C, is the rate constant 800?

Approach: T1 = -5 ° C = 268.15 K: k1 = 4; T2 searched; k2 = 800; E.A. = 70,000 J / mol.
When resolving, be careful not to confuse "1" and "2".
Rearranging the equation: (1 / T2 - 1 / T1) = ln (k1 / k2) * R / EA.;
(1 / T2 - 1 / 268.15) = ln (4/800) * 8.314 / 70,000 = -6.2929 * 10-4;
1 / T2 = -6,2929*10-4 + 3,7293*10-3 = 3,0999*10-3; T2 = 322.58 K = 49.43 ~ 49.4 ° C.

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