Be sure to answer all parts. Calculate ΔH o for the following reaction in two ways, using the data given below. H2 (g) + I2 (g) → 2HI (g) (a) Using the equation ΔH o rxn = ∑ (BE (reactants)) - ∑ (BE (products)) kJ/mol (b) Using the equation ΔH o rxn = ∑ (nΔH o f (products)) - ∑ (mΔH o f (reactants)) kJ/mol Bond BE kJ/mol Substance ΔH o f kJ/mol H-H 436.4 H2 (g) 0 I-I 151 I2 (g) 61.0 H-I 298.3 HI (g) 25.9

a) ΔH°rxn = - 9.2kJ/mol

b) ΔH°rxn = - 9.2kJ/mol

Explanation:

Using Hess's law, you can find ΔH of a reaction from ΔH of formation of the substances involved in the reaction, thus:

ΔH°rxn = ∑ (BE (reactants)) - ∑ (BE (products))

Or:

ΔH°rxn = ∑ (nΔH°f (products)) - ∑ (mΔH°f (reactants))

For the reaction:

H₂ (g) + I₂ (g) → 2HI (g)

a) Using the first equation:

ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)

ΔH°rxn = 436.4kJ + 151kJ - 2*298.3kJ

ΔH°rxn = - 9.2kJ/mol

b) Using the second equation:

ΔH°rxn = 2Δ°f (HI) - ΔH°f (H₂) - ΔH°f (I₂)

ΔH°rxn = 2*25.9kJ - 0kJ - 61.0kJ

ΔH°rxn = - 9.2kJ/mol