Question

Question #2 Consider the following reaction: 1. If a solution of fructose-1-phosphate is incubated with a catalytic amount of enzyme X, the fructose-1-phosphate is transformed to fructose-6-phosphate. At equilibrium, the concentrations of the reaction components are fructose-1-phosphate <-> fructose-6-phosphate 10 mM 40 mM a. Calculate K'eq for this reaction at 25°C. b. Calculate AG" for this reaction at 25°C. (use Gas constant-R=8J/molK) c. If the concentration of fructose-1-phosphate is adjusted to 20 M and the concentration of fructose-6-phosphate is adjusted to 2 M, what is AG?

a. through c. please!

Answer 1

Answer:

1. Assuming the equilibrium rate constant, K_eq = [F-6-P]/[F-1-P] = 40/10 = 4.0

2. Now the relation between G⁰ and K_eq is follows as;

G⁰ = -RTlnK_eq = - 8 J/mok.K x 298.16 K x ln(4.0) = -3306.7002 J/mol

3. Now the relation between the G⁰ and G is as follows;

G = G⁰ - RTlnQ

Here Q = [P]/[R] = 2/20 = 0.1

G = -3306.7002 + 8 J/mok.K x 298.16 K x ln(0.1) = -3306.7002 - 5492.310 = -8799.0103 J/mol

As G is positive. The reaction is spontaneous.  

Thanks