Question

Question: Acetic acid is responsible for the sour taste of vinegar. It can be manufactured using the following reaction:CH₃OH _(g) + CO _(g) CH₃COOH _(l)Use tabulated values of bond energies (Table 13.6) to estimate 𝛥H for this reaction. Compare this result to the 𝛥H value calculated using standard enthalpies of formation in Appendix 4. Explain any discrepancies.

Short answer

The reaction is as follows -

CH₃OH _(g) + CO _(g) CH₃COOH _(l)

The enthalpy of reaction (𝛥H) can be found by two formulas/methods -

1. Bond Energy(BE)
2. Standard enthalpy of formation(𝛥H_f ^o)

Step-by-step solution

Determining the enthalpy of reaction (𝛥H) by bond energy

Given Data (From table 13.6)

Bond	CC	CH	OH	CO	CO	CO
Bond Energy(BE) (KJ/mol)	347	413	467	358	745	1072

For CH₃OH,

3(CH) = 3 * 413 KJ/mol = 1239

1(OH) = 1 * 467 KJ/mol = 467

1(CO) = 1 * 358 KJ/mol = 358

For CO,

1(CO) = 1 * 1072 KJ/mol =1072

For CH₃COOH,

3(CH) = 3 * 413 KJ/mol = 1239

1(CC) = 1 * 347 KJ/mol = 347

1(CO) = 1 * 745 KJ/mol = 745

1(OH) = 1 * 467 KJ/mol = 467

1(CO) = 1 * 358 KJ/mol = 358

The enthalpy of reaction (𝛥H) is given as

𝛥H = (BE)_reactants –𝞢(BE)_products

Therefore,

𝛥H = [3(CH) + 1(OH) + 1(CO) + 1(CO)] –[3(CH) + 1(CC) +

1(CO) + 1(OH) + 1(CO)]

= [1239 + 467 + 358 + 1072] – [1239 + 347 + 745 + 467 + 358]

= 3136 - 3156

= -20 KJ/mol

Therefore, the enthalpy of reaction (𝛥H) is-20 KJ/mol.

 Determining the enthalpy of reaction (𝛥H) by standard enthalpy of formation

Now using,

𝛥H_f ^o(CH₃OH) = - 201 KJ/mol

𝛥H_f ^o (CO) = - 110.5 KJ/mol

𝛥H_f ^o (CH₃COOH) = - 484 KJ/mol

The enthalpy of reaction (𝛥H) =𝞢𝛥H_f ^o_(products) - 𝛥H_f ^o_(reactants)

Hence,

𝛥H = [𝛥H_f ^o (CH₃COOH)] - [𝛥H_f ^o (CO) +𝛥H_f ^o(CH₃OH) ]

= [-484] – [(-110.5) + (-201)]

= (-484) – (-311.5)

= - 172.5 (KJ/mol)

The enthalpy of reaction (𝛥H) is-172.5 (KJ/mol).

Comparing the enthalpy of reaction (𝛥H)

Here, we can say that both the reactions are exothermic. Comparing both values from two different methods indicates that heat released in the second method is more than in the first. The energy released in step 1 is 20 KJ/mol of acetic acid formed. The concept of the use of bond energy is not very feasible as the reaction takes place in the gas phase which means intermolecular forces are less.