Answer to Question #321358 in General Chemistry for TJW

Question

H2 and F2 react to form HF:

H2 (g) + F2 (g) → 2 HF (g)

the relevant bond dissociation energies, in kJ/mol, are: 

H—H: 436 F—F: 155 H—F: 567

Showing your work, calculate the value of ∆H for this reaction. Is this exothermic or endothermic?

Answer 1

Solution:

Balanced chemical equation:

H₂(g) + F₂(g) → 2HF(g)

To find enthalpy of the reaction, use the following formula:

ΔH_rxn = ∑(bonds broken) − ∑(bonds formed)

 

In this reaction, 1 H–H bond and 1 F–F bond must be broken. Also, 2 H–F bonds are formed.

Bond energies of these bonds:

• H–H bond: 436 kJ/mol
• F–F bond: 155 kJ/mol
• H–F bond: 567 kJ/mol

Bonds broken = 1 × (H–H) + 1 × (F–F) = 1 × (436 kJ/mol) + 1 × (155 kJ/mol) = 591 kJ/mol

Bonds broken = 591 kJ/mol

Bonds formed = 2 × (H–F) = 2 × (567 kJ/mol) = 1134 kJ/mol

Bonds formed = 1134 kJ/mol

ΔH = Bonds broken − Bonds formed = 591 kJ/mol − 1134 kJ/mol = −543 kJ/mol

ΔH = −543 kJ/mol

ΔH < 0, thus this is an exothermic reaction

Answer:

ΔH = −543 kJ/mol

Exothermic reaction