ATP hydrolysis & ΔG’°

[luckynumber23]

Experimentally, the standard free energy change associated with the hydrolysis of ATP can be determined by measuring the equilibrium constants for the following two reactions:
i. ATP + glucose --> ADP + glucose 6-phosphate K’eq = 890
ii. glucose + inorganic phosphate --> G6P + H2O K’eq = 0.0037

1) Based on these data, what is the ΔG’° for the hydrolysis of ATP?
Draw a reaction coordinate that demonstrates your solution.

Should I just find the ΔG’° of the 2 rxns and add them? Then I get 705.44 cal/mol.
or
should I invert the second rxn and cancel out the glucose and G6P? Then I get the ΔG’° = 7336.6 cal/mol.

2) Explain why this indirect method is experimentally more feasible than a direct measurement of the equilibrium constant for the hydrolysis of ATP.

If I calculate ΔG’° the first way, the ΔG’° is much less than the ΔG’° for reaction (i), which is the direct hydrolysis of ATP. So is it because the indirect method would lower the free energy needed to hydrolyze ATP?

PLEASE HELP!!