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

STEPS:

1. Determine the concentration change of $\text{H}_2\text{O}_2$ at 400. s:
* From the provided data table, the initial concentration of $\text{H}_2\text{O}_2$ at $t = 0\text{ s}$ is $2.7\text{ M}$.
* The concentration of $\text{H}_2\text{O}_2$ at $t = 400.\text{ s}$ is $1.7\text{ M}$.
* The decrease in concentration (the amount reacted) is:
$$\Delta[\text{H}_2\text{O}_2] = 2.7\text{ M} - 1.7\text{ M} = \mathbf{1.0\text{ M}}$$
2. Calculate the moles of $\text{H}_2\text{O}_2$ that decomposed:
* The volume of the aqueous $\text{H}_2\text{O}_2$ solution is $1.0\text{ L}$.
* Using the definition of molarity ($\text{moles} = \text{molarity} \times \text{volume}$):
$$\text{Moles of }\text{H}_2\text{O}_2\text{ reacted} = 1.0\text{ mol/L} \times 1.0\text{ L} = \mathbf{1.0\text{ mol}}$$
3. Use stoichiometry to determine the moles of $\text{O}_2(g)$ produced:
* The balanced chemical equation is:
$$2\ \text{H}_2\text{O}_2(aq) \rightarrow 2\ \text{H}_2\text{O}(l) + \text{O}_2(g) \quad \text{}$$
* The stoichiometric ratio shows that $2\text{ moles}$ of $\text{H}_2\text{O}_2$ decompose to produce $1\text{ mole}$ of $\text{O}_2(g)$.
* Therefore, the moles of oxygen gas produced is:
$$1.0\text{ mol }\text{H}_2\text{O}_2\text{ reacted} \times \frac{1\text{ mol }\text{O}_2}{2\text{ mol }\text{H}_2\text{O}_2} = \mathbf{0.50\text{ mol }\text{O}_2}$$
4. Calculate the concentration of $\text{O}_2(g)$ in the collection flask:
* The produced oxygen gas is collected in a previously evacuated $10.0\text{ L}$ flask.
* The concentration of $\text{O}_2$ in this gas flask is:
$$\text{Concentration} = \frac{\text{moles of gas}}{\text{volume of flask}} = \frac{0.50\text{ mol}}{10.0\text{ L}} = \mathbf{0.050\text{ mol/L}}$$
5. Estimate the final pressure using the provided conversion factor:
* The problem states that $1.0\text{ mole}$ of gas in $1.0\text{ L}$ (a concentration of $1.0\text{ mol/L}$) exerts a pressure of $24\text{ atm}$ at $300.\text{ K}$.
* Because pressure is directly proportional to concentration at a constant temperature ($P = \left(\frac{n}{V}\right)RT$), we can use a direct proportion to find the pressure of our sample:
$$P = 0.050\text{ mol/L} \times \frac{24\text{ atm}}{1.0\text{ mol/L}} = \mathbf{1.2\text{ atm}}$$
* This matches Option A.

*

WHY_OTHERS_WRONG:

• Option B is incorrect (2.4 atm): This value is obtained if a student forgets to apply the 1:2 stoichiometric ratio between $\text{O}_2$ produced and $\text{H}_2\text{O}_2$ consumed. Assuming $1.0\text{ mole}$ of $\text{O}_2$ is produced yields a gas concentration of $0.10\text{ mol/L}$, resulting in an incorrect pressure of $2.4\text{ atm}$.
• Option C is incorrect (12 atm): This value is off by a factor of 10. This mistake occurs if a student calculates the pressure of the gas as if it were still constrained to the original $1.0\text{ L}$ solution volume instead of the $10.0\text{ L}$ collection flask volume.
• Option D is incorrect (24 atm): This is the pressure that $1.0\text{ mole}$ of gas would exert in a $1.0\text{ L}$ container. This completely neglects both the chemical reaction stoichiometry and the actual volume of the collection flask.