Question

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4.31

Calculate the enthalpy change or the heat of the following reactions at 298^oC by using the values from Standard Enthalpies and Free Energies of formation at 298^oC. Please, if you must use values form another tabulation please specify and comment.

a) 2HI(g) → H₂(g) + I₂ (s)

b) HI(g)  → 1/2H₂(g) + 1/2I₂ (s)

c) 2HI(g) → H₂(g) + I₂ (g)

d) H₂(g) + I₂ (s) → 2HI(g)

e) 2HI(g)  → H₂(g) + 2I (g)

f) 2HI(g)  → 2H(g) + 2I (g)

g) 3HI(g) → H₂(g) + I₂ (s) + HI(g)

What is the heat sublimation of I₂?

If you could also explain the answer to this question, it would be of much help:

4. 47 Calculate the heat of the following reaction at 298^o K; ∆H_f^o for PbO₂ (s) is -66.12 kcal

4Al(s) + PbO₂(s) →3Pb(s) + 2Al₂O₃(s)

What does the algebraic sign in your answer indicate?

Answer 1

The standard enthalpy of formation values are listed below.

Element/Compound (State)	Standard Enthalpy of Formation, ΔfH0 (kJ/mol)
HI (g)	26.48
H2 (g)	0
I2 (s)	0
I2 (g)	62.44
I (g)	106.84
H (g)	217.97

The standard enthalpy of formation values are derived from

https://www.york.ac.uk/media/chemistry/teachinglabs/instruments/instructions/DATABOOK%20FINAL%20VERSION%20for%20web%202011-12.pdf

The standard enthalpy of reaction is defined as

Δ_rH⁰ = ∑n.Δ_fH⁰ (products) - ∑n.Δ_fH⁰ (reactants)

where ∑ denotes summation and n = number of moles of reactants/products.

a) Δ_rH⁰ = [(1 mole H₂)*Δ_fH⁰ (H₂, g) + (1 mole I₂)*Δ_fH⁰ (I₂, s)] – [(2 mole HI)*Δ_fH⁰ (HI, g)]

= [(1 mole)*(0 kJ/mol) + (1 mole)*(0 kJ/mol)] – [(2 mole)*(26.48 kJ/mol)]

= (0 kJ) – (52.96 kJ)

= -52.96 kJ (ans).

b) Δ_rH⁰ = [(1/2 mole H₂)*Δ_fH⁰ (H₂, g) + (1/2 mole I₂)*Δ_fH⁰ (I₂, s)] – [(1 mole HI)*Δ_fH⁰ (HI, g)]

= [(1/2 mole)*(0 kJ/mol) + (1/2 mole)*(0 kJ/mol)] – [(1 mole)*(26.48 kJ/mol)]

= (0 kJ) – (26.48 kJ)

= -26.48 kJ (ans).

c) Δ_rH⁰ = [(1 mole H₂)*Δ_fH⁰ (H₂, g) + (1 mole I₂)*Δ_fH⁰ (I₂, g)] – [(2 mole HI)*Δ_fH⁰ (HI, g)]

= [(1 mole)*(0 kJ/mol) + (1 mole)*(62.44 kJ/mol)] – [(2 mole)*(26.48 kJ/mol)]

= (62.44 kJ) – (52.96 kJ)

= 9.48 kJ (ans).

d) Δ_rH⁰ = [(2 mole HI)*Δ_fH⁰ (HI, g)] - [(1 mole H₂)*Δ_fH⁰ (H₂, g) + (1 mole I₂)*Δ_fH⁰ (I₂, s)]

= [(2 mole)*(26.48 kJ/mol)] – [(1 mole)*(0 kJ/mol) + (1 mole)*(0 kJ/mol)]

= (52.96 kJ) – (0 kJ)

= 52.96 kJ (ans).

e) Δ_rH⁰ = [(1 mole H₂)*Δ_fH⁰ (H₂, g) + (2 mole I)*Δ_fH⁰ (I, g)] – [(2 mole HI)*Δ_fH⁰ (HI, g)]

= [(1 mole)*(0 kJ/mol) + (2 mole)*(106.84 kJ/mol)] – [(2 mole)*(26.48 kJ/mol)]

= (213.68 kJ) – (52.96 kJ)

= 160.72 kJ (ans).

f) Δ_rH⁰ = [(2 mole H)*Δ_fH⁰ (H, g) + (2 mole I)*Δ_fH⁰ (I, g)] – [(2 mole HI)*Δ_fH⁰ (HI, g)]

= [(2 mole)*(217.97 kJ/mol) + (2 mole)*(106.84 kJ/mol)] – [(2 mole)*(26.48 kJ/mol)]

= (649.62 kJ) – (52.96 kJ)

= 596.66 kJ (ans).

g) Δ_rH⁰ = [(1 mole H₂)*Δ_fH⁰ (H₂, g) + (1 mole I₂)*Δ_fH⁰ (I₂, s) + (1 mole HI)*Δ_fH⁰( HI, g)] – [(3 mole HI)*Δ_fH⁰ (HI, g)]

= [(1 mole)*(0 kJ/mol) + (1 mole)*(0 kJ/mol) + (1 mole)*(26.48 kJ/mol)] – [(3 mole)*(26.48 kJ/mol)]

= (26.48 kJ) – (79.44 kJ)

= -52.96 kJ (ans).

The sublimation reaction is given as

I₂ (s) ---------> I₂ (g)

The enthalpy change for the reaction is given as

Δ_rH⁰ = [(1 mole I₂)*Δ_fH⁰(I₂, g)] - [(1 mole I₂)*Δ_fH⁰(I₂, s)]

= [(1 mole)*(62.44 kJ/mol)] – [(1 mole)*(0 kJ/mol)]

= (62.44 kJ) – (0 kJ)

= 62.44 kJ (ans).