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: Write the structure of acetaldehyde.
The structure of \[ CH_3CHO \] is \[ H_3C-CHO \] It contains: \[ 4\ \text{C-H bonds} \] \[ 1\ \text{C-C bond} \] and \[ 1\ \text{C=O bond} \]

Step 2: Calculate energy required to form gaseous atoms.
Formation of gaseous atoms from elements requires: \[ 2C(g)+4H(g)+O(g) \] Using the given atomic formation enthalpies: \[ 2\times700=1400\ \text{kJ mol}^{-1} \] \[ 4\times200=800\ \text{kJ mol}^{-1} \] \[ 1\times250=250\ \text{kJ mol}^{-1} \] Total energy required: \[ 1400+800+250 \] \[ =2450\ \text{kJ mol}^{-1} \]

Step 3: Calculate energy released during bond formation.
Energy released due to bond formation is: For \(4\) C-H bonds: \[ 4\times400=1600\ \text{kJ mol}^{-1} \] For \(1\) C-C bond: \[ 1\times350=350\ \text{kJ mol}^{-1} \] For \(1\) C=O bond: \[ 1\times700=700\ \text{kJ mol}^{-1} \] Total energy released: \[ 1600+350+700 \] \[ =2650\ \text{kJ mol}^{-1} \]

Step 4: Calculate enthalpy of formation.
\[ \Delta H_f = \text{Energy absorbed} - \text{Energy released} \] \[ \Delta H_f = 2450-2650 \] \[ \Delta H_f=-200\ \text{kJ mol}^{-1} \]

Step 5: Final conclusion.
Hence, the enthalpy of formation of \[ CH_3CHO \] is \[ \boxed{-200\ \text{kJ mol}^{-1}} \]