J.R. i think i missed out something

Question

2 AB (g) + B2 (g) → 2AB2 (g); H° = +492.0 kJ/mol, S° = +236.9 J/Kmol

Use the data in the table below to calculate the value of the standard molar entropy S°, for B₂ (g) at 25°C.

Substance: AB (g) Standard Molar Entropy, S° (J/Kmol): 217.9

substance: AB₂ (g) Standard Molar Entropy, S° (J/Kmol): 238.1

Answer 1

Now the problem makes more sense, and can be solved as follows:

Sum of Sº of products - sum of Sº of reactants = ∆Sº for the reaction

Sum of products = 2 x 238.1 = 476.2

Sum of reactants = 2 x 217.9 + X = 435.8 + X (where X is Sº for B₂)

476.2 - 435.8 - X = 236.9

X = -196.5 kJ/mol = Sº for B₂