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Calculate the change in entropy (ΔS˚) of reaction (J/K) for the equation as written.

BaCO3(s) → BaO(s) + CO2(g)


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?

  • chemistry -

    It looks ok to me.

  • chemistry -

    its products - reactants not reactants - products

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