At 125 °C, 1 mol O, and 2 mol H2 in a chamber have a total pressure of 3 atm. A current is run through the chamber, combining all the molecules into H,0 at constant volume. The pressure of the chamber, in atmospheres, is now:

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**Combining Gases into Water in a Closed System**

**Scenario:**
At 125°C, a chamber contains 1 mole of O₂ and 2 moles of H₂, with a total initial pressure of 3 atm. An electric current is applied, causing the O₂ and H₂ to react and form water (H₂O) while the chamber's volume remains constant.

**Question:**
After the reaction is complete, what is the new pressure inside the chamber measured in atmospheres?

**Analysis:**
The reaction between hydrogen and oxygen to form water can be represented by the chemical equation:

\[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \]

Initially, we have 1 mol of O₂ and 2 mol of H₂, which perfectly matches the stoichiometry of the reaction, meaning all reactants are fully consumed in forming water.

Since water is in a liquid state under these conditions at constant volume, and given that liquid does not exert significant pressure compared to gases, the resulting pressure in the chamber is effectively reduced to zero as no gaseous moles remain.

**Conclusion:**
The pressure in the chamber after the reaction is complete is effectively 0 atm, assuming the water remains in a liquid state and does not contribute to the gaseous pressure.
Transcribed Image Text:**Combining Gases into Water in a Closed System** **Scenario:** At 125°C, a chamber contains 1 mole of O₂ and 2 moles of H₂, with a total initial pressure of 3 atm. An electric current is applied, causing the O₂ and H₂ to react and form water (H₂O) while the chamber's volume remains constant. **Question:** After the reaction is complete, what is the new pressure inside the chamber measured in atmospheres? **Analysis:** The reaction between hydrogen and oxygen to form water can be represented by the chemical equation: \[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \] Initially, we have 1 mol of O₂ and 2 mol of H₂, which perfectly matches the stoichiometry of the reaction, meaning all reactants are fully consumed in forming water. Since water is in a liquid state under these conditions at constant volume, and given that liquid does not exert significant pressure compared to gases, the resulting pressure in the chamber is effectively reduced to zero as no gaseous moles remain. **Conclusion:** The pressure in the chamber after the reaction is complete is effectively 0 atm, assuming the water remains in a liquid state and does not contribute to the gaseous pressure.
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