2 H2O2(aq) → 2 H2O(l) + O2(g) ΔH° = −196 kJ/mol_rxn The decomposition of H2O2(aq) is represented by — Thermodynamics Chemistry Question

STEPS:

1. Analyze the molecular structures and bonds of reactants and products:
* Reactants: The reaction begins with 2 moles of hydrogen peroxide ($\text{H}_2\text{O}_2$). The molecular structure of $\text{H}_2\text{O}_2$ is $\text{H}-\text{O}-\text{O}-\text{H}$. Therefore, 2 moles of reactant contain:
* 4 moles of $\text{O}-\text{H}$ single bonds
* 2 moles of $\text{O}-\text{O}$ single bonds
* Products: The reaction yields 2 moles of water ($\text{H}_2\text{O}$) and 1 mole of oxygen gas ($\text{O}_2$).
* The molecular structure of water is $\text{H}-\text{O}-\text{H}$. Thus, 2 moles of water contain 4 moles of $\text{O}-\text{H}$ single bonds.
* The molecular structure of oxygen gas is $\text{O}=\text{O}$. Thus, 1 mole of oxygen contains 1 mole of $\text{O}=\text{O}$ double bonds.

2. Set up the relationship between bond enthalpies and reaction enthalpy ($\Delta H^\circ$):
* The enthalpy of a reaction can be estimated by subtracting the total bond energy of the bonds formed (products) from the total bond energy of the bonds broken (reactants):
$$\Delta H^\circ \approx \sum D_{\text{broken}} - \sum D_{\text{formed}}$$
$$\Delta H^\circ \approx [4 \cdot D(\text{O}-\text{H}) + 2 \cdot D(\text{O}-\text{O})] - [4 \cdot D(\text{O}-\text{H}) + 1 \cdot D(\text{O}=\text{O})]$$

3. Simplify the mathematical expression using the given assumption:
* The question states to assume that the bond enthalpies of the $\text{O}-\text{H}$ bonds in $\text{H}_2\text{O}$ are not significantly different from those in $\text{H}_2\text{O}_2$.
* Because of this, the $4 \cdot D(\text{O}-\text{H})$ term on the reactant side and the $4 \cdot D(\text{O}-\text{H})$ term on the product side are essentially equal and cancel each other out.
* This simplifies our $\Delta H^\circ$ expression to:
$$\Delta H^\circ \approx 2 \cdot D(\text{O}-\text{O}) - D(\text{O}=\text{O})$$

4. Substitute the experimental $\Delta H^\circ$ value to find the numerical target:
* We are given that $\Delta H^\circ = -196\text{ kJ/mol}_{\text{rxn}}$. Substituting this value:
$$-196\text{ kJ/mol} \approx 2 \cdot D(\text{O}-\text{O}) - D(\text{O}=\text{O})$$
* This tells us that the bond energy of 1 mole of $\text{O}=\text{O}$ double bonds must be approximately $200\text{ kJ/mol}$ larger than the energy required to break 2 moles of $\text{O}-\text{O}$ single bonds to make the reaction exothermic.

5. Test the given options using the simplified relationship:
* Let's substitute the values from Option B into our simplified equation ($D(\text{O}-\text{O}) = 150\text{ kJ/mol}$ and $D(\text{O}=\text{O}) = 500\text{ kJ/mol}$):
$$2(150) - 500 = 300 - 500 = \mathbf{-200\text{ kJ/mol}}$$
* This result is extremely close to the experimental value of $-196\text{ kJ/mol}$, identifying Option B as the correct choice.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Substituting these values yields $2(300) - 500 = +100\text{ kJ/mol}$. This would represent an endothermic process, which contradicts the known exothermic nature of the decomposition reaction ($\Delta H^\circ = -196\text{ kJ/mol}$).
• Option C is incorrect: Substituting these values yields $2(500) - 300 = +700\text{ kJ/mol}$, which describes a highly endothermic reaction that is nowhere near the target value of $-196\text{ kJ/mol}$.
• Option D is incorrect: Substituting these values yields $2(250) - 300 = +200\text{ kJ/mol}$, which is also endothermic rather than exothermic.