Question

Hydrolysis reactions are commonly used inside the cell to split high-energy covalent bonds. For each of the three reactions below, use the ΔG° for each reaction to determine the equilibrium constants (K). Assume that each reaction occurs independently of the other two.

ΔG° (kcal/mole)
Reaction 1: acetyl-P --> actetate + P
-10.3
Reaction 2: ATP --> ADP + P
-7.3
Reaction 3: glucose 6-P ! glucose + P
-3.3

Answer 1

The equilibrium constants (K) for the given hydrolysis reactions can be determined using the equation ΔG° = -RT ln(K) and the provided ΔG° values. The calculated equilibrium constants are approximately 3.5 x 10^7 for Reaction 1, 2.2 x 10^5 for Reaction 2, and 2.5 x 10^2 for Reaction 3.

Step-by-step explanation:

To determine the equilibrium constants (K) from the given ΔG° values for hydrolysis reactions, we use the equation ΔG° = -RT ln(K), where R is the gas constant (1.987 cal/(mol·K)) and T is the temperature in Kelvin (assumed to be 298K for standard conditions).

For Reaction 1 (acetyl-P --> acetate + P), with ΔG° = -10.3 kcal/mol:

K = exp(-ΔG° / (RT)) = exp(10300 cal/mol / (1.987 cal/(mol·K) * 298K))

K ≈ exp(10300 / 592.706) ≈ exp(17.38) ≈ 3.5 x 107

For Reaction 2 (ATP --> ADP + P), with ΔG° = -7.3 kcal/mol:

K = exp(-ΔG° / (RT)) = exp(7300 cal/mol / (1.987 cal/(mol·K) * 298K))

K ≈ exp(12.3) ≈ 2.2 x 105

For Reaction 3 (glucose 6-P --> glucose + P), with ΔG° = -3.3 kcal/mol:

K = exp(-ΔG° / (RT)) = exp(3300 cal/mol / (1.987 cal/(mol·K) * 298K))

K ≈ exp(5.54) ≈ 2.5 x 102