Question

Observe the following equation
$H_2O_2(g) \rightarrow 2H(g) + 2O(g)$; $\Delta_a H^\ominus = 1066 \, kJ mol^{-1}$.
If $\Delta_{O-H} H^\ominus$ is $464 \, kJ mol^{-1}$, then $\Delta_{O-O} H^\ominus$ (in $kJ mol^{-1}$) value will be

• A. 602
• B. 138
• C. 690
• D. 276

Hint:

Always draw the Lewis structure to ensure you count the number of bonds correctly (e.g., $H_2O_2$ is not just HO + OH, it involves the central O-O bond).

Answer 1

The Correct Option is B

Step 1: Understand the Reaction Enthalpy:
The equation represents the atomization of Hydrogen Peroxide ($H_2O_2$). Structure of $H_2O_2$: $H - O - O - H$. To break this molecule into atoms ($2H + 2O$), we need to break all the chemical bonds. Total Enthalpy ($\Delta_a H^\ominus$) = Sum of Bond Dissociation Enthalpies.
Step 2: Identify Bonds Broken:
The molecule contains:

• Two $O-H$ bonds.
• One $O-O$ bond (peroxide linkage).

Step 3: Calculation:
\[ \Delta_a H^\ominus = 2 \times (\Delta_{O-H} H^\ominus) + 1 \times (\Delta_{O-O} H^\ominus) \] Given: $\Delta_a H^\ominus = 1066$ kJ/mol. $\Delta_{O-H} H^\ominus = 464$ kJ/mol. Substitute values: \[ 1066 = 2(464) + \Delta_{O-O} H^\ominus \] \[ 1066 = 928 + \Delta_{O-O} H^\ominus \] \[ \Delta_{O-O} H^\ominus = 1066 - 928 \] \[ \Delta_{O-O} H^\ominus = 138 \, kJ mol^{-1} \]
Step 4: Final Answer:
The bond enthalpy of the O-O bond is 138 kJ/mol.