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Use data in Appendix C in the textbook to calculate ΔH∘ in (kJ/mol) , ΔS∘ in...

Use data in Appendix C in the textbook to calculate ΔH∘ in (kJ/mol) , ΔS∘ in (j/mol-K) , and ΔG∘ in (kJ/mol) at 25 ∘C for each of the following reactions: 2P(g)+10HF(g)→2PF5(g)+5H2(g)

Appendix C:

P(g): dH:316.4; dG: 280.0; S:163.2

HF(g): dH: -268.61; dG: -270.70; S:173.51

PF5(g): dH:-1594.4; dG: -1520.7; S: 300.8

H2(g): dH:217.94; dG: 203.26; S: 114.60

Solutions

Expert Solution

Given:

Hof(P(g)) = 316.4 KJ/mol

Hof(HF(g)) = -268.61 KJ/mol

Hof(PF5(g)) = -1594.4 KJ/mol

Hof(H2(g)) = 217.94 KJ/mol

Balanced chemical equation is:

2 P(g) + 10 HF(g) ---> 2 PF5(g) + 5 H2(g)

ΔHo rxn = 2*Hof(PF5(g)) + 5*Hof(H2(g)) - 2*Hof( P(g)) - 10*Hof(HF(g))

ΔHo rxn = 2*(-1594.4) + 5*(217.94) - 2*(316.4) - 10*(-268.61)

ΔHo rxn = -45.8 KJ/mol

Answer: -45.8 KJ/mol

Given:

Sof(P(g)) = 163.2 J/mol.K

Sof(HF(g)) = 173.51 J/mol.K

Sof(PF5(g)) = 300.8 J/mol.K

Sof(H2(g)) = 114.6 J/mol.K

Balanced chemical equation is:

2 P(g) + 10 HF(g) ---> 2 PF5(g) + 5 H2(g)

ΔSo rxn = 2*Sof(PF5(g)) + 5*Sof(H2(g)) - 2*Sof( P(g)) - 10*Sof(HF(g))

ΔSo rxn = 2*(300.8) + 5*(114.6) - 2*(163.2) - 10*(173.51)

ΔSo rxn = -886.9 J/mol.K

Answer: -886.9 J/mol.K

Given:

Gof(P(g)) = 280.0 KJ/mol

Gof(HF(g)) = -270.7 KJ/mol

Gof(PF5(g)) = -1520.7 KJ/mol

Gof(H2(g)) = 203.26 KJ/mol

Balanced chemical equation is:

2 P(g) + 10 HF(g) ---> 2 PF5(g) + 5 H2(g)

ΔGo rxn = 2*Gof(PF5(g)) + 5*Gof(H2(g)) - 2*Gof( P(g)) - 10*Gof(HF(g))

ΔGo rxn = 2*(-1520.7) + 5*(203.26) - 2*(280.0) - 10*(-270.7)

ΔGo rxn = 121.9 KJ/mol

Answer: 121.9 KJ/mol

queen_honey_blossom answered 1 year ago

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