Question

Determine AS for the reaction to form 45.92 g of, Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g), given (answer in J) Substance Sº (J/mol · K) Zn(s) 60.9 HCl(aq) 56.5 H2(g) 130.58 Zn2+ (aq) -106.5 cl(aq) 55.10 Answer: Check

Answer 1

Zn(s) + 2HCl(aq) -----------> ZnCl2(aq) + H2(g)

Zn(s) + 2HCl(aq) -----------> Zn^2+(aq) + 2Cl^-(aq) + H2(g)

$\Delta$ S⁰    = S⁰f products - S⁰f reactants

           = 130.58 -106.5 +2*55.10-(60.9 +2*56.5)

            = -39.62J/mole-K

no of moles of Zn   = W/G.A.Wt

                                = 45.92/65.4   = 0.702moles

1 moles of Zn react with excess of HCl to gives -39.62J/K

0.702 moles of Zn react with excess of HCl to gives = -39.62*0.702/1   = -27.8J/K   >>>>answer