[stmoney]

Given...
Fe₂O_3(s) + 3H_2(g) --> 2 Fe_(s) + 3H₂O_)(g)

Hº = - 98.8 kJ
Sº = 141.5 kJ/T
-----------------------------------
A. Prove this reaction is nonspontaneous under the standard-state conditions at 25ºC
B. At what temp will it become spontaneous?

So, to start this problem I know the rule that if Hº and Sº are opposite signs, G will be either spontaneous at all temps or nonspontaneous at all temps, depending on whether Hº or Sº is positive (or negative). I'll just plug numbers...

G = Hº - T*Sº
   = (-98.8 kJ)- (298 K*141.5 kJ/T)
   = -98.9 kJ - 42167 kJ
   = -42265.8 kJ

Since G < 0 the reaction is going to be spontaneous, so it's like I've disproved question A. I'm curious if I've done something wrong with my math. And part B would be 'it's spontaneous at any temperature,' correct?

[enahs]

Entropy is usually given in units of Joules, not Kilojoules as Entropy is. Check your units and convert.

[stmoney]

I understand entropy is given in joules. But you have to convert the joules to kJ because standard free energy change (G) is given in units of kJ. So that's why the problem gave me entropy in kJ; it saved me a step.

Anyone else?

[enahs]

Well then either you are reading the problem wrong, or there is an error in it; as that Entropy just way way off. It is closer to 112 J/ mol^-1 K^-1.