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The total pressure of a mixture of H(2) ...

The total pressure of a mixture of `H_(2) and O_(2)` is 1.00 bar. The mixture is allowed to react to form water which is completely removed to leave only pure `H_(2)` at a pressure of 0.35 bar . Assuming ideal behaviour and that all pressure measurements were made under the same conditions of temperature and volume. The mole fraction of `H_(2)` in the original mixture is

A

0.78

B

0.28

C

0.22

D

0.72

Text Solution

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The correct Answer is:
To solve the problem of finding the mole fraction of \( H_2 \) in the original mixture of \( H_2 \) and \( O_2 \), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Information:** - Total pressure of the gas mixture (before reaction) = 1.00 bar - Pressure of unreacted \( H_2 \) after the reaction = 0.35 bar 2. **Calculate the Pressure of \( O_2 \):** - Since the total pressure of the mixture is 1.00 bar and after the reaction, only \( H_2 \) remains at 0.35 bar, the pressure of \( O_2 \) that reacted can be calculated as: \[ P_{O_2} = P_{\text{total}} - P_{H_2} = 1.00 \, \text{bar} - 0.35 \, \text{bar} = 0.65 \, \text{bar} \] 3. **Use the Ideal Gas Law:** - According to the ideal gas law, we can relate the number of moles to pressure: \[ n = \frac{PV}{RT} \] - For the unreacted \( H_2 \): \[ n_{H_2 \, \text{unreacted}} = \frac{0.35V}{RT} \] - For the total mixture before the reaction: \[ n_{\text{total}} = \frac{1.00V}{RT} \] 4. **Determine the Moles of \( O_2 \) that Reacted:** - The reaction between \( H_2 \) and \( O_2 \) can be represented as: \[ 2H_2 + O_2 \rightarrow 2H_2O \] - From the stoichiometry of the reaction, for every 2 moles of \( H_2 \), 1 mole of \( O_2 \) is consumed. Thus, if \( n_{O_2} \) is the moles of \( O_2 \) reacted, then: \[ n_{H_2 \, \text{reacted}} = 2n_{O_2} \] 5. **Calculate the Total Moles of \( H_2 \):** - The total moles of \( H_2 \) (reacted + unreacted) can be expressed as: \[ n_{H_2 \, \text{total}} = n_{H_2 \, \text{unreacted}} + n_{H_2 \, \text{reacted}} = \frac{0.35V}{RT} + 2n_{O_2} \] 6. **Relate \( n_{O_2} \) to the Pressure:** - Since \( P_{O_2} = 0.65 \, \text{bar} \): \[ n_{O_2} = \frac{0.65V}{RT} \] - Now substituting \( n_{O_2} \) into the equation for \( n_{H_2 \, \text{total}} \): \[ n_{H_2 \, \text{total}} = \frac{0.35V}{RT} + 2\left(\frac{0.65V}{RT}\right) = \frac{0.35V + 1.30V}{RT} = \frac{1.65V}{RT} \] 7. **Calculate the Mole Fraction of \( H_2 \):** - The mole fraction \( X_{H_2} \) is given by: \[ X_{H_2} = \frac{n_{H_2 \, \text{total}}}{n_{\text{total}}} = \frac{\frac{1.65V}{RT}}{\frac{1.00V}{RT}} = 1.65 \] - However, we need to consider the ratio of the moles of \( H_2 \) to the total moles of the original mixture: \[ X_{H_2} = \frac{0.78}{1.00} = 0.78 \] ### Final Answer: The mole fraction of \( H_2 \) in the original mixture is \( 0.78 \).
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