Calculate the change in entropy (ΔS˚) of reaction (J/K) for the equation as written.
BaCO3(s) → BaO(s) + CO2(g)
Table:
S° (J/(K mol))
BaCO3(s) = 112
BaO(s) = 70.3
CO2(g) = 213.7
So the answer is -41.7
I thought to find the entropy of reaction you subract the entropy of formation of the products from those of the reactants.
So I did:
(213.7+70.3)-112 = 172
What did I do wrong?
3 answers
It looks ok to me.
its products - reactants not reactants - products
second way is correct dude.