Questions 7-9 refer to the following information. At 27°C, five identical rigid 2.0 L vessels are fi — Stoichiometry Chemistry Question

STEPS:

1. Understand the initial conditions of the vessels: At $27^\circ\text{C}$, both Vessel 1 and Vessel 2 are filled with the exact same volume of inert $\text{N}_2(g)$ ($2.0\text{ L}$) at the exact same initial pressure ($720\text{ mm Hg}$). Additionally, Vessel 1 contains $0.050\text{ mol}$ of solid sodium bicarbonate ($\text{NaHCO}_3(s)$), and Vessel 2 contains $0.050\text{ mol}$ of solid sodium bromide ($\text{NaBr}(s)$).
2. Account for the temperature change: When both vessels are heated from $27^\circ\text{C}$ ($300\text{ K}$) to $127^\circ\text{C}$ ($400\text{ K}$), the average kinetic energy of the gas molecules increases. According to Gay^-Lussac's Law ($P \propto T$ when volume and moles of gas are constant), this temperature increase causes the pressure of the $\text{N}_2(g)$ in both vessels to rise proportionally (to approximately $960\text{ mm Hg}$).
3. Analyze the chemical behavior of the solid in Vessel 2: Vessel 2 contains sodium bromide ($\text{NaBr}(s)$), which is a highly stable ionic salt with an extremely high melting point. It does not decompose, melt, or vaporize at $127^\circ\text{C}$. Therefore, the number of moles of gas ($n$) in Vessel 2 remains completely constant, and the pressure inside is due solely to the heated $\text{N}_2(g)$.
4. Relate the pressure difference to the number of gas molecules: The problem states that at $127^\circ\text{C}$, the pressure in Vessel 1 is higher than that in Vessel 2. According to the Ideal Gas Law ($PV = nRT$), because volume ($V$) and temperature ($T$) are identical in both rigid containers, the higher pressure in Vessel 1 can only be explained by a larger number of moles of gas ($n$) in the container.
5. Evaluate the decomposition of sodium bicarbonate: Heating sodium bicarbonate ($\text{NaHCO}_3(s)$) causes it to undergo thermal decomposition. The balanced chemical equation in Option C shows that solid sodium bicarbonate decomposes to form a solid carbonate and two gaseous products: water vapor and carbon dioxide gas:
$$2\text{ NaNaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \mathbf{\text{H}_2\text{O}(g)} + \mathbf{\text{CO}_2(g)}$$
The generation of these extra gas molecules increases the total moles of gas ($n$) in Vessel 1, resulting in a higher measured pressure compared to Vessel 2.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This reaction proposes that solid sodium bicarbonate decomposes into sodium metal and a solid bicarbonate radical. Because both of the proposed products are in the solid phase ($s$), no new gas molecules would be generated. As a result, the pressure in Vessel 1 would remain identical to Vessel 2.
• Option B is incorrect: This reaction suggests the formation of solid sodium hydride ($\text{NaH}(s)$) and neutral carbonate gas ($\text{CO}_3(g)$). Carbonate ($\text{CO}_3$) exists as a polyatomic ion in solid lattices and solutions, but it is not a stable, independent gas molecule that forms during the thermal decomposition of sodium bicarbonate.
• Option D is incorrect: This option represents a reaction where sodium bicarbonate reacts with the nitrogen gas ($\text{N}_2(g)$) in the vessel. However, nitrogen gas is extremely stable and unreactive under these mild heating conditions ($127^\circ\text{C}$) due to its strong triple covalent bond. Furthermore, even if this reaction did occur, it consumes 1 mole of reactant gas ($\text{N}_2(g)$) to produce 1 mole of product gas ($\text{H}_2(g)$), meaning there would be no net increase in the moles of gas to explain the higher pressure.