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?

It looks ok to me.

its products - reactants not reactants - products

second way is correct dude.

To calculate the change in entropy (ΔS˚) of a reaction, you need to consider the stoichiometric coefficients of the compounds involved.

In this case, the stoichiometry of the reaction is:
BaCO3(s) → BaO(s) + CO2(g)

Given:
S° (J/(K mol))
BaCO3(s) = 112
BaO(s) = 70.3
CO2(g) = 213.7

To calculate ΔS˚ for the reaction, you need to consider the change in entropy between the reactants and the products. The change in entropy can be calculated by considering the stoichiometric coefficients of the compounds.

For BaCO3(s), you have a coefficient of 1.
So the change in entropy for BaCO3(s) is: 112 J/(K mol) x 1 = 112 J/(K mol)

For BaO(s), you have a coefficient of 1.
So the change in entropy for BaO(s) is: 70.3 J/(K mol) x 1 = 70.3 J/(K mol)

For CO2(g), you have a coefficient of 1.
So the change in entropy for CO2(g) is: 213.7 J/(K mol) x 1 = 213.7 J/(K mol)

Now, add up the change in entropy values for the products and subtract the sum of the change in entropy values for the reactants to get the total change in entropy (ΔS˚):

ΔS˚ = (70.3 J/(K mol) + 213.7 J/(K mol)) - 112 J/(K mol)
ΔS˚ = 283 J/(K mol) - 112 J/(K mol)
ΔS˚ = 171 J/(K mol)

Therefore, the correct value for the change in entropy (ΔS˚) of the reaction BaCO3(s) → BaO(s) + CO2(g) is 171 J/(K mol), not -41.7 J/(K mol) as you mentioned in your calculation.

The negative sign might indicate an error in sign convention or direction of the reaction, so please double-check the instructions or information provided to ensure the correct calculation.