What is δg′° for the reaction ( k ′ eq measured at 25 °c)? (b) if the concentration of fructose 6-phosphate is adjusted to 1.5 m and that of glucose 6-phosphate is adjusted to 0.50 m, what is δg? (c) why are δg′° and δg different?

a.

b.

c. Standard conditions provide 1 M molarity for each component, while non-standard conditions provide specific concentrations

Explanation:

a. The original problem states that:

For an equilibrium:

This means:

Solving for the standard Gibbs free energy change:

b. Now we will solve for the non-standard Gibbs free energy change given by the equation:

Here:

We obtain:

c. The standard Gibbs free energy change is measured for 1 M concentration of each molarity, however, the non-standard Gibbs free energy change is measured for the given molarities of 1.5 M and 0.50 M. Due to different conditions, we obtain different values.

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.