Question

2. The figure below shows two possible methods for the production of acetyl CoA from acetate as part of the biosynthesis of fatty acids in the body. The first method is a direct reaction, and the second method involves coupled reactions.

Direct Reaction

Coupled Reactions

Step 1

Step 2

a. The standard Gibbs free energy for the direct reaction, ΔG°', is 32 kJ/mol. (Note that the prime symbol is used for biological systems.) Calculate the equilibrium constant for the reaction at a temperature of 298 K. Is the reaction favorable or unfavorable? Explain. (1 point)

b. Calculate the equilibrium constant for each step of the coupled reactions. ΔG₁°' is about 11 kJ/mol for the first reaction, and ΔG₂°' is about –10 kJ/mol for the second reaction. Is each step favorable or unfavorable? Explain. (1 point)

c. Determine the net reaction for the coupled reaction, its Gibbs free energy value, and its overall equilibrium constant. (1 point)

d. Predict whether the direct method or the coupled reaction method would be more likely to occur in the body to produce acetyl CoA from acetate. What might happen to make the reaction even more likely to occur? (1 point)

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For a) I got K = 0.00000258 or 2.58 * 10^-6

For b) I got

K₁ = exp[ (-11000J mol­­^-­) / (8.314 J K^-1 mol^-1)(298.15 K)] = 0.01182

K₂ = exp[ -(-10000J mol^-) / ((8.314 J K^-1 mol^-1)(298.15 K)] = 56.4965

ΔG₁^o is unfavourable as it is greater than 0

ΔG₂^o is favorable as it is less than 0     

I can't figure out C and D at all.   

  

Answer 1

Given (a) Standard Gibb's free energy (Delta G degree) = 32 kJ (mol) = 32000J/mod Temperature of reaction = 298K we know that Delta G or delta G degree = -RT lnk where R rightarrow gas constant = 8.314 JK^-3 mol^-1 K - equilibrium constant Now, lnk = -delta G degree /RT K = exp (-32000 Jmol /8.314 J mol k times 298 K) K = exp (-12.915) K = 2.458 times 10^-6 Hence, equilibrium constant for direct reaction (k) is 2.458 times 10^-6 Here, delta G degree is positive i.e > 0 so reaction is unfavourable. (b) Given, Delta G_i for first reaction in complied reaction is 11kJ/mol Delta G degree = 11Kj/mol = 11000 J/mol Here, delta G_i value is positive i.e > 0 so first reaction is not Favourable \r\n we know that delta G_i = -RT ln k_1 K_1 = exp (-delta G_i/RT) = exp (-11060 J mole/ 8.314 J mol times 29.8 k) k_1 = exp (-11000/8.314 times 298) = exp (-4.439)