To calculate $\Delta H^\circ_{\text{rxn}}$ for the reaction:

$2\text{H}_2\text{O}_2(l) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g),$

we use the enthalpy of formation ($\Delta H^\circ_f$) values given in the problem. The equation for $\Delta H^\circ_{\text{rxn}}$ is:

$\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants}).$

Step 1: List the $\Delta H^\circ_f$ values

• $\Delta H^\circ_f (\text{H}_2\text{O}_2(l)) = -187.6 \, \text{kJ/mol}$,
• $\Delta H^\circ_f (\text{H}_2\text{O}(l)) = -285.8 \, \text{kJ/mol}$,
• $\Delta H^\circ_f (\text{O}_2(g)) = 0 \, \text{kJ/mol}$ (by definition).

Step 2: Apply the equation

For the products: $\text{Products: } 2 \text{H}_2\text{O}(l) + \text{O}_2(g),$ $\sum \Delta H^\circ_f (\text{products}) = 2(-285.8) + 1(0) = -571.6 \, \text{kJ}.$

For the reactants: $\text{Reactants: } 2 \text{H}_2\text{O}_2(l),$ $\sum \Delta H^\circ_f (\text{reactants}) = 2(-187.6) = -375.2 \, \text{kJ}.$

Step 3: Calculate $\Delta H^\circ_{\text{rxn}}$

$\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants}),$ $\Delta H^\circ_{\text{rxn}} = -571.6 - (-375.2),$ $\Delta H^\circ_{\text{rxn}} = -571.6 + 375.2 = -196.4 \, \text{kJ}.$

Final Answer:

$\Delta H^\circ_{\text{rxn}} = -196.4 \, \text{kJ}.$

Let me know if you need further details or have any questions! Here are some additional related questions:

1. What is the significance of a negative $\Delta H^\circ_{\text{rxn}}$?
2. How would the enthalpy change if the reaction produced gaseous water instead of liquid water?
3. How can Hess's Law help calculate enthalpies for more complex reactions?
4. Why is the standard enthalpy of formation for oxygen gas ($\text{O}_2(g)$) zero?
5. What assumptions are made when using enthalpy of formation values?

Tip: Always ensure you balance the chemical equation correctly before applying enthalpy calculations!