If 1 mole of `H_(2)`, 2 moles of `O_(2)` and 3 moles of `N_(2)` are mixed in a vessel and total pressure was found to be 12 atm then the partial pressure exerted by `N_(2)` in the vessel will be

Options

A. 2 atm
B. 4 atm
C. 6 atm
D. 12 atm

Answer

To find the partial pressure exerted by \( N_2 \) in the vessel, we can follow these steps:

### Step 1: Determine the total number of moles of gas in the mixture.
We have:
- Moles of \( H_2 = 1 \)
- Moles of \( O_2 = 2 \)
- Moles of \( N_2 = 3 \)

Total moles = \( 1 + 2 + 3 = 6 \) moles.

### Step 2: Calculate the mole fraction of \( N_2 \).
The mole fraction (\( \chi \)) of a gas is calculated using the formula:
\[ \chi_{N_2} = \frac{\text{Number of moles of } N_2}{\text{Total number of moles}} \]
Substituting the values:
\[ \chi_{N_2} = \frac{3}{6} = 0.5 \]

### Step 3: Use the mole fraction to find the partial pressure of \( N_2 \).
The partial pressure (\( P_{N_2} \)) can be calculated using the formula:
\[ P_{N_2} = \chi_{N_2} \times P_{\text{total}} \]
Where \( P_{\text{total}} \) is the total pressure, which is given as 12 atm. Therefore:
\[ P_{N_2} = 0.5 \times 12 \text{ atm} = 6 \text{ atm} \]

### Final Answer:
The partial pressure exerted by \( N_2 \) in the vessel is \( 6 \text{ atm} \).

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