Question

using average bond enthalpies (linked above), estimate the enthalpy change for the following reaction: ch₃oh(g) + hi(g) → ch₃i(g) + h₂o(g) kj

Explanation:

To solve this, we use the formula for enthalpy change (\(\Delta H\)) using bond enthalpies: \(\Delta H=\sum \text{Bond enthalpies of bonds broken}-\sum \text{Bond enthalpies of bonds formed}\)

Step 1: Identify bonds in reactants and products

• Reactants: \(\ce{CH3OH(g)}\) and \(\ce{HI(g)}\)
• \(\ce{CH3OH}\) has: \(3\ \ce{C-H}\), \(1\ \ce{C-O}\), \(1\ \ce{O-H}\) bonds.
• \(\ce{HI}\) has: \(1\ \ce{H-I}\) bond.
• Products: \(\ce{CH3I(g)}\) and \(\ce{H2O(g)}\)
• \(\ce{CH3I}\) has: \(3\ \ce{C-H}\), \(1\ \ce{C-I}\) bonds.
• \(\ce{H2O}\) has: \(2\ \ce{O-H}\) bonds.

Step 2: List bond enthalpies (average values, in kJ/mol)

• \(\ce{C-H}\): \(413\)
• \(\ce{C-O}\): \(358\)
• \(\ce{O-H}\): \(463\)
• \(\ce{H-I}\): \(299\)
• \(\ce{C-I}\): \(240\)

Step 3: Calculate bonds broken (reactants)

Bonds broken: \(3\ \ce{C-H} + 1\ \ce{C-O} + 1\ \ce{O-H} + 1\ \ce{H-I}\)

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Step 4: Calculate bonds formed (products)

Bonds formed: \(3\ \ce{C-H} + 1\ \ce{C-I} + 2\ \ce{O-H}\)

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Step 5: Calculate \(\Delta H\)

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Answer:

\(\boldsymbol{-46}\) kJ (per mole of reaction)