Can someone help me with this entropy problem?

Question

for the hydrolysis of ATP at 25 degree C (298k) and pH7 the standard free energy of hydrolysis is -30.5 KJ/mol (-7.3 kal/mol) and the standard enthalpy change is -20.1 KJ/mol (-4.8 kcal/mol). calculate the standard entropy change for the reaction in both joules and calories. why is the positive sign of the answer expected in view of the nature of the reaction

Answer 1

dG = dH - TdS

-30.5kJ = -20.1kJ - (298K)(dS)

why is the positive sign of the answer expected in view of the nature of the reaction,... your answer is: no way you could get a dG = -30.5 with only a dH = -20.1 unless dS is a "+",  multiplying "-TdS" will make  it the amount of extra "- kJ's" that you need

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-30.5kJ = -20.1kJ - (298K)(dS)

298(dS) = 10.4

dS =0.0349 kilojoules

your answer is: 34.9 joules

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dG = dH - TdS

-7.3kcal= -4.8kcal - (298K)(dS)

298 (dS) = 2.5

dS = 0.00839 kcal

your answer = 8.4 cal