Question

Use the data from table to estimate the enthalpy of formation of \(CH_3CHO\).
\[ \begin{array}{|c|c|c|} \hline \text{Bond Enthalpy} & \text{Bond} & \text{Enthalpy of formation} \\ \hline 400\ \text{kJ mol}^{-1} & C-H & C(g)=700\ \text{kJ mol}^{-1} \\ \hline 350\ \text{kJ mol}^{-1} & C-C & H(g)=200\ \text{kJ mol}^{-1} \\ \hline 700\ \text{kJ mol}^{-1} & C=O & O(g)=250\ \text{kJ mol}^{-1} \\ \hline \end{array} \]

• A. \(-200\ \text{kJ mol}^{-1}\)
• B. \(-400\ \text{kJ mol}^{-1}\)
• C. \(-350\ \text{kJ mol}^{-1}\)
• D. \(-150\ \text{kJ mol}^{-1}\)

Hint:

Enthalpy of formation using bond enthalpy is calculated by: \[ \Delta H= \sum(\text{bond breaking energies}) - \sum(\text{bond formation energies}) \] Bond formation releases energy, so it is subtracted.

Answer 1

The Correct Option is A

Step 1: Identify bonds in acetaldehyde $CH_3CHO$.
Acetaldehyde contains: 4 C-H bonds, 1 C-C bond, 1 C=O bond. Formation from elements requires 2 C(g) + 4 H(g) + 1 O(g).
Step 2: Calculate energy released when bonds form in $CH_3CHO$.
Given bond enthalpies: C-H = 400, C-C = 350, C=O = 700 kJ/mol. \[ E_{\text{bonds}} = 4(400) + 1(350) + 1(700) = 1600 + 350 + 700 = 2650\ \text{kJ mol}^{-1} \]
Step 3: Calculate energy needed to produce gaseous atoms from elements.
Given atomisation enthalpies: $C(g) = 700$, $H(g) = 200$, $O(g) = 250$ kJ/mol. For 2 C, 4 H, 1 O: \[ E_{\text{atoms}} = 2(700) + 4(200) + 1(250) = 1400 + 800 + 250 = 2450\ \text{kJ mol}^{-1} \]
Step 4: Apply Hess's Law to find enthalpy of formation.
\[ \Delta H_f = E_{\text{atoms}} - E_{\text{bonds}} = 2450 - 2650 = -200\ \text{kJ mol}^{-1} \]
Step 5: Interpret the sign of the result.
A negative value means formation of $CH_3CHO$ from its elements is exothermic. More energy is released forming bonds in the product than is absorbed creating gaseous atoms.
Step 6: State the final answer.
\[ \boxed{-200\ \text{kJ mol}^{-1}} \]