Question

Part A. Find the average enthalpy of the PF bond in PF5(g) from the following standard molar enthalpies of formation (in kJ/mol): PF5(g) (–1595), P(g) (315), F(g) (79).

Part B.

Using the bond energies from the table below estimate AH for the following reaction C2H2(g) + H2(g) → C2H4(g) Bond Bond energy [kJ/mol] [C-H 413 H-H 436 |C-C 348 C=C 614 csc 839

Answer 1

A)

Given:

Hof(P(g)) = 315 KJ/mol

Hof(F(g)) = 79 KJ/mol

Hof(PF5(g)) = -1595 KJ/mol

Balanced chemical equation is:

P(g) + 5 F(g) ---> PF5(g)

ΔHo rxn = 1*Hof(PF5(g)) - 1*Hof( P(g)) - 5*Hof(F(g))

ΔHo rxn = 1*(-1595.0) - 1*(315.0) - 5*(79.0)

ΔHo rxn = -2305 KJ

There are no bonds in reactant and there are 5 P-F bonds in product

ΔHo rxn can also be written as:

ΔHo rxn = BE(reactant) - BE(products)

-2305 = 0 - 5*BE(P-F)

5*BE(P-F) = 2305

BE(P-F) = 461 KJ/mol

Answer: 461 KJ/mol

B)

On reactant side we have:

1 C≡C bond

2 C-H bond

1 H-H bond

On product side we have:

1 C=C bond

4 C-H bond

Use:

Delta H = BE(reactant) - BE(product)

=1*BE(C≡C) + 2*BE(C-H) + 1*BE(H-H) - 1*BE(C=C) - 4*BE(C-H)

=1*BE(C≡C) + 1*BE(H-H) - 1*BE(C=C) - 2*BE(C-H)

= 839 + 436 - 614 - 2*413

= -165 KJ

Answer: -165 KJ