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 | k_{1} = 0,5 | 40 | T_{1} = 313,15 |
2 | k_{2} = 5 | 70 | T_{2} = 343,15 |
Partial results: ln (k_{1} / k_{2}) = -2.3026; (1 / T_{2} - 1 / T_{1}) = -2,7918*10^{-4};
E._{A.}= ln (k_{1 }/ k_{2}) * R / {(1 / T_{2}) - (1 / T_{1})} = 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: T_{1} = 40 ° C = 313.15 K: k_{1} searched; T_{2} = 60 ° C = 333.15 K; k_{2} = 15.
Note in the equation that E_{A.} is used in J / mol!
(E._{A. }/ R) * (1 / T_{2} - 1 / T_{1}) = (69.800 / 8,314) * ... = 8.395,48 * (-1,917*10^{-4}) = -1,6095
ln (k_{1} / k_{2}) = -1.6095; so (k_{1} / k_{2}) = exp (-1.6095) = 0.1999;
k_{1} = 0.1999 * k_{2} = 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: T_{1} = -5 ° C = 268.15 K: k_{1} = 4; T_{2} searched; k_{2} = 800; E._{A.} = 70,000 J / mol.
When resolving, be careful not to confuse "1" and "2".
Rearranging the equation: (1 / T_{2} - 1 / T_{1}) = ln (k_{1} / k_{2}) * R / E_{A.};
(1 / T_{2} - 1 / 268.15) = ln (4/800) * 8.314 / 70,000 = -6.2929 * 10^{-4};
1 / T_{2} = -6,2929*10^{-4} + 3,7293*10^{-3} = 3,0999*10^{-3}; T_{2} = 322.58 K = 49.43 ~ 49.4 ° C.
Back to theory (Arrhenius equation) Additional exercise material: Downloads No. 3
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