Using the table provided, which of the following answer choices represents δs for the reaction in the image? standard thermodynamic values select one: a. 18.6 j/k ·mol b. -18.6 j/k ·mol c. 550.8 j/k ·mol d. -112 j/k ·mol

You are decreasing "disorder" by building one molecule out of two. The entropy will be negative. So, we know "a" and "c" are incorrect.

ΔS°reaction = (∑"n" moles of products × the ΔS° of the products) - (∑ "n" moles of reactants × the ΔS° of the reactants) 

 (I looked these up. I hope they are the same as your table.)

PRODUCT ENTHALPIES
      ΔS° 219.4 (J/K mol)

REACTANT ENTHALPIES

      ΔS° 130.58 (J/K mol)

    ΔS° 200.8(J/K mol)

Now, we solve:
 ΔS°reaction = (∑"n" moles of products × the ΔS° of the products) - (∑ "n" moles of reactants × the ΔS° of the reactants) 

 ΔS°= ((n* 219.4))-((n*200.8)+(n*130.58))
(because all the molec. have a coefficient of 1 in front of them, "n"= 1 for all of them.)

 ΔS°= (219.4)-((200.8)+(130.58))
 ΔS°= (219.4)-(331.38)
 ΔS°= (-111.98) 
rounded to -112

d. - 112 J/K mol

Comment with any further questions! Hope it all made sense!

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