You are studying an enzyme catalyzed chemical reaction in which reactants X & Y

Question

You are studying an enzyme catalyzed chemical reaction in which reactants X & Y are combined to form Z (X +Y ® Z). Answer the questions below and show your work. All reactions are performed at room temperature (298 K). Hint: The gas constant (R) is equal to 8.314 J K-1 mol-1.

A.) Starting from the biochemical standard state, you measure the concentrations of products and reactants at equilibrium to find the results below. What is the ?G°’?

[X] = 0.5 M

[Y] = 0.1 M

[Z] = 2.3 M

B.) You alter the initial concentrations of products and reactants to observe the effect. What do you expect the new ?G to be starting from the initial concentrations below?

[X] = 0.2 M

[Y] = 0.7 M

[Z] = 5.0 M

C.) The reaction in part B was performed with 500 nM enzyme. How will the ?G change if the enzyme concentration is increased 100-fold to 50 ?M?

Explanation / Answer

Ans. Part A: Equilibrium constant, Keq = [Z] / ([X] [Y])

            Or, Keq = 2.3 / (0.5 x 0.1) = 46.0

Using the equation dG0’ = - RT ln Keq                - equation 1

Where, dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = equilibrium constant under standard condition

R = (0.001987 kcal mol-1K-1 or 0.008314 kJ mol-1 K-1)

Putting the values in equation 1-

            dG0’ = - (0.008314 kJ mol-1K-) x 298.0 K x ln 46.0

            Or, dG0’ = -2.477572 kJ mol-1 x 2.303 log 46.0 = -2.477572 kJ mol-1 x 3.8286

            Hence, dG0’ = -9.486 kJ mol-1

# Part B: Keq under given condition = 5.0 / (0.2 x 0.7) = 35.71

Using, dG = dG0 + RT (ln Keq)

Where, dG = free energy change of the reaction

dG0’ = standard free energy change of the reaction

R = universal gas constant = 0.008314 kJ mol-1K-1

T = temperature in kelvin

Keq = equilibrium constant

Putting the values in above equation –

dG = -9.486 kJ mol-1 + (0.008314 kJ mol-1K-1 x 298 K) x ln 35.71

Or, dG = -9.486 kJ mol-1 + (2.477572 kJ mol-1 x 3.5755)

            Or, dG = -9.486 kJ mol-1 + 8.859 kJ mol-1

            Hence, dG = -0.627 kJ mol-1

# Part C: Note the enzymes act as biocatalyst, they increase the rate of a reaction but does not affect the equilibrium.

Similarly, lowering [E] would lower the rate of catalysis but won’t change the equilibrium, that is Keq remains unaffected.

Note that the equation for dG calculation depends on Keq but not on [E].

So, dG remains unaffected when [E] is increased.

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