Question

UL. Use the following data to determine AHºfor P.(s) + 5 CIF3(g) ----> 4 PF3 (g) + 3 CIF(g) + Cl2(g) Data Substance AH°(kJ.mole) Substance AH°(kJ/mole) H2O2(1) H2O2(aq) NO2(g) CO2(g) CO(g) H2O(g) H2O(1) C3H2(g) C3H18(1) Fe3O4(s) C6H129) CIF3(g) HCN(g) CH (g) -187.78 -191.17 33.18 -393.51 -110.53 -241.82 -285.83 -3537 -5471 -1118.4 -3953 -158.87 135.1 -75.0 PH3(g) 5.40 KNO3(s) -492.70 HNO3(1) 63.01 POC13(g) -542.2 PC13(1) -287.0 POC13(1) -597.0 KCl(s) -435.9 KCIO3(s) -391.5 KCIO3(aq) -349.5 Fe2O3(s) -824.2 NH3(g) -46.19 CIF(g) -50.29 NO(g) 90.37 PF3(g) -958.44 AH(elements in reference form) = 0 (kJ/mole)

Answer 1

Given:

Hof(P4(s)) = 0.0 KJ/mol

Hof(Cl2(g)) = 0.0 KJ/mol

Hof(ClF3(g)) = -158.87 KJ/mol

Hof(PF3(g)) = -958.44 KJ/mol

Hof(ClF(g)) = -50.29 KJ/mol

Balanced chemical equation is:

P4(s) + 5 ClF3(g) ---> 4 PF3(g) + 3 ClF(g) + Cl2(g)

ΔHo rxn = 4*Hof(PF3(g)) + 3*Hof(ClF(g)) + Hof(Cl2(g)) - 1*Hof( P4(s)) - 5*Hof(ClF3(g))

ΔHo rxn = 4*(-958.44) + 3*(-50.29) + 1*(0.0) - 1*(0.0) - 5*(-158.87)

ΔHo rxn = -3190.28 KJ

Answer: -3190.28 KJ