standard entropy change for the following reaction,?

Question

Calculate the standard entropy change for the following reaction,

2 HgO(s) → 2 Hg(λ) + O2(g)

given S°[HgO] = 70.3 J/K·mol, S°[ Hg(λ)] =76.0 J/K·mol, and S°[O2(g)] = 205.1 J/K·mol.

a.+210.8 J/K
b.-216.5 J/K
c.+351.4 J/K
d.+497.7 J/K
e.+216.5 J/K

Answer 1

ΔS° = S°products - S°reactants
ΔS° = 205.1+2*76.0-2*70.3=+216.5 J/K
e

Answer 2

Just use the idea that the sum of the entropies of the products minus the sum of the entropies of the reactants will give you the entropy change for the process.

You should be able to do the calculation yourself.

Answer 3

ask your teacher
no one here is that serious