Food Chemistry

132 2 Enzymes

For the reaction rate follows:

k=M·

A

A^ =

=M·

k 1

k− 1

=M·K^ = (2.86)

with

M=

KB·T

h

=

R·T

NA·h

(2.87)

(K#equilibrium constant, kBBoltzmannconstant,
h:Planckconstant, NA:Avogadronumber).
For the equilibrium constant follows:

K^ ==e−G

=/RT

(2.88)

Resulting for the equilibrium constant in:

k=

kB·T

h

e−ΔG

=/RT

(2.89)

and for the free activation enthalpy:

ΔG^ ==−RT ln

k·h

kB·T

(2.90)

If k is known for any temperature,ΔG^ =can be
calculated according to equation 2.90. Further-
more, the following is valid:

ΔG^ ==ΔH^ =−TΔS^ = (2.91)

A combination with equation 2.90 results in:

−RT ln

k·h

kB·T

=ΔH^ =−TΔS^ = (2.92)

and

log

k

T

=−log

h

kB

−

ΔH^ =

2 .3RT

+

TΔS^ =

2 .3R

(2.93)

It is possible to determineΔH^ =graphically based
on the above equation if k is known for several
temperatures and logk/T is plotted against 1/T.

IfΔG^ =andΔH^ =are known,ΔS^ =can be calcu-
lated from equation 2.91.
The activation entropy is contained in theAr-
rheniusfactor A as can be seen by comparing
the empiricalArrheniusequation 2.84 with equa-
tion 2.89 which is based on the transition state
hypothesis:

k=A·e−Ea/RT (2.94a)

k=

kB

h

·e−S

=/R

·T·e−H

=/RT

(2.94b)

Activation energy Ea and activation en-

thalpy H^ = are linked with each other as

follows:

dlnk

dT

=

Ea

RT^2

(2.95)

dlnk

dT

=

1

T

+

H^ =

RT^2

=

RT+H^ =

RT^2

(2.96)

Ea=H^ =+RT (2.97)

Using plots of logk against 1/T, the activation en-

ergy of theArrheniusequation can be determined.

For enzyme catalyzed reactions, Eais 10–60, for

chemical reactions this value is 50–150 and for

the inactivation of enzymes, the unfolding of pro-

teins, and the killing of microorganisms, 250–

350 kJ/mol are required.

For enzymes which are able to convert more

than one substrate or compound into product,

the activation energy may be dependent on the

substrate. One example is alcohol dehydroge-

nase, an important enzyme for aroma formation

in semiripened peas (Table 2.13). In this case the

activation energy for the reverse reaction is only

slightly influenced by substrate.

Under consideration of the temperature depend-

ence of the rate constant k in equation 2.80, the

implementation of the expression fromArrhenius

equation 2.84 leads to:

c 1 =c 0 ·e−k^0 ·t·e

−Ea/RT

(2.98)

For a constant effect follows:

ct

c 0

=const.=e−k^0 ·t·e

−Ea/RT

(2.99)

Table 2.13.Alcohol dehydrogenase from pea seeds: ac-

tivation energy of alcohol dehydrogenation and alde-

hyde reduction

Alcohol Ea Aldehyde Ea

(kJ·mole−^1 )(kJ·mole−^1 )

Ethanol 20

n-Propanol 37 n-Propanal 20

2-Propenol 18

n-Butanol 40 n-Butanal 21

n-Hexanol 37 n-Hexanal 18

2-trans- 2-trans-

hexenol 15 Hexenal 19

2-trans-

Heptenal 18