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alexdok [17]
3 years ago
7

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
Chemistry
1 answer:
Fantom [35]3 years ago
7 0

Answer:

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

<em>ΔH°rxn = -9.2kJ/mol</em>

<em />

b) Using the second equation:

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

ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ

<em>ΔH°rxn = -9.2kJ/mol</em>

<em />

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