The decarboxylation of lysine catalyzed by lysine decarboxylase has a kcat value of 500 s-1 at 298K, and loss of CO2 is the rate-determining step. What is the free energy of activation for the CO2 loss step? The half-life for the uncatalyzed reaction under the same conditions is 4 billion years (1017 seconds). How much does the enzyme lower the free energy of activation for this reaction? Show your work.

The decrease in free energy is 113.299kJ

Explanation:

K for enzyme catalyzed reaction = 500s^-1

Temperature (T) =298k

ΔG =?

ΔG = - 2.303 RT log k

ΔG = (-2.303)(8.314)(298) log 500

ΔG = - 15399.9 J

ΔG catalyzed = - 15. 399kJ

The first order reaction is given as:

t1/2= 0.693/k

or k= 0.693/t1/2

0.693/10^17

Therefore,

K= 0.693 × 10^-17

Now,

K= 0.693 × 10^-17

T= 298k

ΔG uncatalyzed =?

ΔG uncatalyzed = - 2.303 RT log k

ΔG uncatalyzed = (-2.303)(8.314)(298) log0.693 × 10^-17

= 97908.1J

ΔG uncatalyzed = 97.9081kJ

Therefore,

The decrease in free energy is:

ΔG uncatalyzed - ΔG catalyzed

97.908 - (-15.399)

= 113.299KJ

The decrease in free energy is 113.299kJ

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"