4FeS (s) + 7O2 (g) --> 2Fe2O3 (s) + 4SO2 (g)

Entropy value for iron (II) sulfide is 67 J/K mol
I don't understand how to find entropy

Do you want the entropy of the reaction?

dSrxn = (n*dSo products) - (n*dSo reactants).
Look up the standard S values in your text/notes and plug into the above equation and solve for dSrxn.

To find the entropy for a chemical reaction, you need to consider the entropy values of each of the reactants and products involved. The entropy (S) of a substance is a measure of the disorder or randomness of its particles.

In this case, the given entropy value of iron (II) sulfide (FeS) is 67 J/K mol. This value represents the entropy per mole of FeS.

To find the overall entropy change for the chemical reaction, you need to calculate the difference between the sum of the entropy values of the products and the sum of the entropy values of the reactants.

In the given balanced equation:
Reactants:
4FeS (s) - This means you have 4 moles of FeS

Products:
2Fe2O3 (s) - This means you have 2 moles of Fe2O3
4SO2 (g) - This means you have 4 moles of SO2

To calculate the entropy change, you need to multiply each product and reactant quantity by its respective entropy value:

Entropy change (∆S) = Σ(S_products) - Σ(S_reactants)

Σ(S_products) = (2 moles of Fe2O3) x (entropy value of Fe2O3) + (4 moles of SO2) x (entropy value of SO2)

Σ(S_reactants) = (4 moles of FeS) x (entropy value of FeS)

Therefore, the entropy change for the given reaction is:

∆S = [(2 x entropy value of Fe2O3) + (4 x entropy value of SO2)] - (4 x entropy value of FeS)

You can substitute the given entropy value of FeS (67 J/K mol) into the equation and use the entropy values for Fe2O3 and SO2 to calculate the entropy change for the reaction.