Calcuating Gibbs Energy...?

Question

I need some chemistry help. :-/

I know how to calculate Gibbs energy, ∆G=∆H-T∆S. However, I don't know how to solve it without the temperature given in the problem.

The question is: For the reaction H2O --> H2 + ½O2, the value of ∆G is
a. positive
b. negative
c. positive at low temperatures and negative at high temperatures
d. negative at low temperatures and positive at high temperatures.

The standard enthalpy of H2O is -242. The standard entropy of H2O is 188.7. The standard entropies of H2 and O2 are 130.7 and 205.1.

Could anyone please explain to me how you would go about calculating the Gibbs energy of this problem? Thank you very much.

Answer 1

The change in entropy is + 44.55 J/molK.
The enthalpy change is +242 kJ/mol, at standard conditions (SATP).
You can use standard ambient temperature (298 K) for T.

Gibbs = +242.2 kJ/mol - (13.27 kJ/mol)
= +228.9 kJ/mol

If the temperature increased by more than a factor of about 20, the second value would increase, and the Gibb's energy would be a negative quantitity.