Answer:
−153.1 J / (K mol)
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Hg(liq) + Cl2(g) → HgCl2(s)
Given that;
The standard molar entropies of the species at that temperature are:
Sºm (Hg,liq) = 76.02 J / (K mol) ;
Sºm (Cl2,g) = 223.07 J / (K mol) ;
Sºm (HgCl2,s) = 146.0 J / (K mol)
The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]
= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]
= -153.09 J / (K mol)
= or -153.1 J / (K mol)
Hence the answer is −153.1 J / (K mol)
Answer:
Rubidium Acetate
Explanation:
Rb is Rubidium C2H3O2 is acetate
