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Z - Universiteit Utrecht

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Activeringsenergie<br />

Arrhenius vergelijking<br />

(heengaande reactie)<br />

k + = A + exp[ -E + a<br />

/ RT ]<br />

ln( k + ) = ln( A + ) - E a<br />

+<br />

/ RT<br />

ln( k + )<br />

a<br />

b<br />

-E<br />

+<br />

a<br />

/ Rc<br />

d<br />

e<br />

f<br />

1/T<br />

A + = constante<br />

R = 8,314 J K -1 mol -1<br />

Hoe verandert het met<br />

T in Kelvin<br />

een katalysator?<br />

(°C + 273,15)<br />

E a<br />

in J mol -1

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