The total pressure of a mixture of oxygen and hydrogen is 1.0 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.40 atm when measured at the same values of T and V as the original mixture. What was the composition of the original mixture in mole fraction ?

Options

A. `x_(o_(2))=0.2,x_(H_(2))=0.8`
B. `x_(o_(2))=0.4,x_(H_(2))=0.6`
C. `x_(o_(2))=0.6,x_(H_(2))=0.4`
D. `x_(o_(2))=0.8,x_(H_(2))=0.2`

Explanation

To find the composition of the original mixture of oxygen (O₂) and hydrogen (H₂) in terms of mole fraction, we can follow these steps:

### Step 1: Understand the given information
- The total pressure of the mixture (P_total) is 1.0 atm.
- After ignition, the remaining gas is pure hydrogen (H₂) with a pressure of 0.40 atm.
- The reaction that occurs is:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]

### Step 2: Determine the amount of gas consumed
- The initial total pressure of the mixture is 1.0 atm.
- After the reaction, the pressure of the remaining hydrogen is 0.40 atm.
- Therefore, the pressure of the gases consumed (which is the pressure of oxygen and the pressure of hydrogen that reacted) is:
\[ P_{consumed} = P_{total} - P_{H_2 \, remaining} = 1.0 \, atm - 0.40 \, atm = 0.60 \, atm \]

### Step 3: Relate the pressures to the mole fractions
- The reaction shows that 2 moles of H₂ react with 1 mole of O₂. This means the ratio of moles of H₂ to O₂ is 2:1.
- Let the partial pressure of hydrogen in the original mixture be \( P_{H_2} \) and the partial pressure of oxygen be \( P_{O_2} \).
- According to Dalton's Law of Partial Pressures:
\[ P_{total} = P_{H_2} + P_{O_2} \]
\[ 1.0 \, atm = P_{H_2} + P_{O_2} \]

### Step 4: Express the pressures in terms of the mole fractions
- From the stoichiometry of the reaction, we know that for every 2 moles of H₂, 1 mole of O₂ is consumed. Therefore, if \( P_{H_2} \) is the initial pressure of hydrogen, then the pressure of oxygen that reacts can be expressed as:
\[ P_{O_2} = \frac{1}{2} P_{H_2} \]
- The total pressure consumed (0.60 atm) is due to the consumption of both gases. Since 2 parts of H₂ react with 1 part of O₂, we can express the consumed pressures as:
\[ P_{consumed} = P_{H_2 \, consumed} + P_{O_2 \, consumed} = P_{H_2} - P_{H_2 \, remaining} + P_{O_2} \]
\[ 0.60 \, atm = (P_{H_2} - 0.40 \, atm) + P_{O_2} \]

### Step 5: Solve the equations
1. Substitute \( P_{O_2} = \frac{1}{2} P_{H_2} \) into the total pressure equation:
\[ 1.0 = P_{H_2} + \frac{1}{2} P_{H_2} \]
\[ 1.0 = \frac{3}{2} P_{H_2} \]
\[ P_{H_2} = \frac{2}{3} \, atm \]

2. Now substitute \( P_{H_2} \) back to find \( P_{O_2} \):
\[ P_{O_2} = \frac{1}{2} P_{H_2} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \, atm \]

### Step 6: Find the mole fractions
- The mole fraction of hydrogen (\( X_{H_2} \)) and oxygen (\( X_{O_2} \)) can be calculated as:
\[ X_{H_2} = \frac{P_{H_2}}{P_{total}} = \frac{\frac{2}{3}}{1.0} = \frac{2}{3} \]
\[ X_{O_2} = \frac{P_{O_2}}{P_{total}} = \frac{\frac{1}{3}}{1.0} = \frac{1}{3} \]

### Final Answer
The composition of the original mixture in mole fraction is:
- Mole fraction of H₂: \( \frac{2}{3} \)
- Mole fraction of O₂: \( \frac{1}{3} \)