If a reaction occurs at 293K that has a change in enthalpy of +125 kJ and a change in entropy of +35 J/K, then

Question

is the reaction spontaneous? If so, why? If not, why not?

Use Hess’s Law to find the change in entropy for the following reaction: S8(s) + 12O2(g) à 8SO3(g). Given the following entropy changes,

S8(s) + 8O2(g) à 8SO2(g),

DS = 89 J/K

2SO2(g) + O2(g) à2SO3(g)

DS = -188 J/K

Answer 1

ΔG = ΔH - TΔS

ΔG = 125kJ - (293)(35/1000kJ) = 114.745kJ

Since G >0 (Free energy), the reaction is non-spontaneous.

For the other question, try multiplying everything by 4 to get

8SO2(g) + 4O2(g) --> 8SO3(g)

DS = -188*4 J/K = -752 J

Combining the 2nd and third equations, we end up with our main equation

S8(s) + 12O2(g) --> 8SO3(g)

Thus, adding the ΔS together, we get 89+-752 = -663J = answer

[Answer: see above]