# chemistry

posted by
**Carter** on
.

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?