Isodesmic Equations

Question 1

a. SbCl₃ + F₂ → SbF₃ + Cl₂

Balanced Equation: 2SbCl₃ + 3F₂ → 2SbF₃ + 3Cl₂

ΔH = 6(Sb(III)-Cl) + 3(F-F) - 6(Sb(III)-F) - 3(Cl-Cl)

ΔH = 6(315) + 3(155) - 6(440) - 3(240) = -1005 kJ mol^-1

∴ The products are more thermodynamically stable than the reactants

b. P(CH₃)₃ + S₈ → S=P(CH₃)₃

Balanced Equation: 8P(CH₃)₃ + S₈ → 8S=P(CH₃)₃

ΔH = 24(P-C) + 72(C-H) + 8(S-S) - 24(P-C) - 72(C-H) - 8(P=S)

ΔH = 24(P-C) + 72(C-H) + 8(S-S) - 24(P-C) - 72(C-H) - 8(P=S)

ΔH = 8(S-S) - 8(P=S) = -872 kJ mol^-1

ΔH = 8(226) - 8(335) = -872 kJ mol^-1

∴ The products are more thermodynamically stable than the reactants

c. H₂CO₃ → CO₂ + H₂O

Balanced Equation: H₂CO₃ → CO₂ + H₂O

ΔH = 1(C=O) + 2(C-O) + 2(O-H) - 2(C=O) - 2(O-H)

ΔH = 1(C=O) + 2(C-O) + 2(O-H) - 2(C=O) - 2(O-H)

ΔH = 2(C-O) - 1(C=O)

ΔH = 2(358) - 1(799) = -83 kJ mol^-1

∴ The products are more thermodynamically stable than the reactants

Question 2

a. P₄ + Cl₂ → PCl₃ or PCl₅ ?

Balanced Equation 1: P₄ + 6Cl₂ → 4PCl₃

ΔH = 6(P-P) + 6(Cl-Cl) - 12(P(III)-Cl)

ΔH = 6(201) + 6(240) - 12(326) = -1266 kJ mol^-1

Balanced Equation 2: P₄ + 10Cl₂ → 4PCl₅

ΔH = 6(P-P) + 10(Cl-Cl) - 20(P(V)-Cl)

ΔH = 6(201) + 10(240) - 20(258) = -1554 kJ mol^-1

∴ PCl₅ is the expected product

b. HCCH + Br₂ → HBrCCBrH or HBr₂CCBr₂H ?

Balanced Equation 1: HCCH + Br₂ → HBrCCBrH

ΔH = 1(C≡C) + 2(C-H) + 1(Br-Br) - 1(C=C) - 2(C-H) - 2(C-Br)

ΔH = 1(C≡C) + 2(C-H) + 1(Br-Br) - 1(C=C) - 2(C-H) - 2(C-Br)

ΔH = 1(C≡C) + 1(Br-Br) - 1(C=C) - 2(C-Br)

ΔH = 1(835) + 1(190) - 1(602) - 2(285) = -147 kJ mol^-1

Balanced Equation 2: HCCH + Br₂ → HBr₂CCBr₂H

ΔH = 1(C≡C) + 2(C-H) + 2(Br-Br) - 1(C-C) - 2(C-H) - 4(C-Br)

ΔH = 1(C≡C) + 2(C-H) + 2(Br-Br) - 1(C-C) - 2(C-H) - 4(C-Br)

ΔH = 1(C≡C) + 2(Br-Br) - 1(C-C) - 4(C-Br)

ΔH = 1(835) + 2(190) - 1(346) - 4(285) = -271 kJ mol^-1

∴ HBr₂CCBr₂H is the expected product

c. CN^- + O₂ → OCN^- or CNO^- (linear) ?

Balanced Equation 1: CN^- + ½O₂ → OCN^-

ΔH = 1(C≡N) + ½(O=O) - 1(C=O) - 1(C=N)

ΔH = 1(887) + ½(494) - 1(799) - 1(615) = -280 kJ mol^-1

Balanced Equation 2: CN^- + ½O₂ → CNO^- (linear) ?

ΔH = 1(C≡N) + ½(O=O) - 1(C≡N) - 1(N-O)

ΔH = 1(C≡N) + ½(O=O) - 1(C≡N) - 1(N-O)

ΔH = ½(O=O) - 1(N-O)

ΔH = ½(494) - 1(222) = +25 kJ mol^-1

∴ OCN^- is the expected product

CGMT Theory

Question 1

a. H₂CCH₂

E_σ+π = 346 + 295 = 641 kJ mol^-1
ΣΔE_s→t = 2(-38) = -76 kJ mol^-1
2ΣΔE_s→t = -152 kJ mol^-1

Since E_σ+π > 2ΣΔE_s→t, a planar geometry would be expected.

b. H₂SiSiH₂

E_σ+π = 222 + 120 = 346 kJ mol^-1
ΣΔE_s→t = 2(88) = 176 kJ mol^-1
2ΣΔE_s→t = 352 kJ mol^-1

Since E_σ+π < 2ΣΔE_s→t, a trans-bent geometry would be expected.

c. H₂CSiH₂

E_σ+π = 300 + 139 = 439 kJ mol^-1
ΣΔE_s→t = -38 + 88 = 50 kJ mol^-1
2ΣΔE_s→t = 100 kJ mol^-1

Since E_σ+π > 2ΣΔE_s→t, a planar geometry would be expected.