Calculate Delta G

Question

A key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P) as shown below.

G6P  F6P

At 298 K, the equilibrium constant for the isomerization is 0.510.

(a) Calculate G° at 298 K.

  1.67  kJ/mol

(b) Calculate G when Q, the [F6P]/[G6P] ratio, equals 12.0.

??

(c) Calculate G when Q = 0.300.

??  kJ/mol

(d) Calculate Q in the cell if G = -4.10 kJ/mol.

??

Answer 1

(b) Calculate G when Q, the [F6P]/[G6P] ratio, equals 12.0.

G = G⁰ + RTlnQ

G = 1.67 kj/mole + 8.314 x 298C x ln 12.0 = 7.83 kj/mole

(c) Calculate G when Q = 0.300.

G = G⁰ + RTlnQ = 1.67 kj/mole + 8.314 x 298 x ln 0.300 = - 1.31 kj/mole

(d) Calculate Q in the cell if G = -4.10 kJ/mol.

G = G⁰ + RTlnQ , -4.10 kj = 1.67 kj/mole + 8.314 x 298 x lnQ

lnQ = (-4.10 kj/mole -1.67) kj/mole / (8.314 x 298) x 10^-3 kj/mole = - 2.33

Q = anti ln 2.33 = 9.74

Check all math!

Answer 2

These 2 equations can help you solve these problems:

1) ∆G° = -RTlnK

2) ∆G = RTln[Q/K]

a) Where R is gas constant 8.314JK^-1mol^-1,. T is temperature, and K is equilibrium constant.

Since this is ∆G° it refers to standard conditions. Standard conditions are 1M, 1atm, For both sides products and reactants, 25° or 298 K. The ln is the natural log. So evaluating the equation with these conditions:

∆G° = -(8.314)298ln[1/1]

Since the natural log or ln of 1 is 0, everything is 0 based on the scenario in part a. So ∆G° = 0 at equilibrium neither product nor reactants are favored.

b) ∆G = RTln[Q/K]

Since this is not indicating standard conditions you can use the temperature in the problem, 298K.

(8.314)298ln[12/1]. When ∆G° = 0,. K= 1. So evaluating this equation gives:

(8.314)298ln[12/1] = 6156.5J. In the units of Gibbs free in kJ we just divide this number by 1000 or multiply by the equivalency 1000J = 1kJ

Dividing by 1000 gives ∆G = 6.1565 kJ/mol

c) use the same equation:

∆G = RTln[Q/K]. = (8.314)298ln[0.300/1] = -2982.9 J or 2.9829 kJ/mol

d) Now given ∆G, solve for Q:

∆G = RTln[Q/K = -4.10kJ/mol= (8.314)298ln[Q/1]

-4.10/(8.314)298 = ln[Q]

e^(-4.10/(8.314)298) = e^Q

Q = 0.99835. or 0.998

This makes sense because it is similar to the numerical expression, Q value in part C of this problem.