Question

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

Answer 1

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

`C_(2) H_(4)` ΔS° 219.4 (J/K mol)

REACTANT ENTHALPIES


`H^(2)` ΔS° 130.58 (J/K mol)


`C_(2) H_(2)` 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

Answer: d. - 112 J/K mol

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