1 mole of `N_(2)` and 3 moles of `H_(2)` filled in a one-litre bulb were allowed to react. When the reaction attained equilibrium, two-thirds of `N_(2)` converted to `NH_(3)(N_(2)+3H_(2)hArr2NH_(3))`. If a hole is then made in the bulb, the mole ratio of the gases `N_(2),H_(2) andNH_(3)` effusing out initially would be respectively.

Options

A. `1:3:4`
B. `sqrt28:sqrt2:sqrt17`
C. `1/sqrt28:1/sqrt2:1/sqrt17`
D. `1/sqrt28:3/sqrt2:4/sqrt17`

Answer

To solve the problem step by step, we need to analyze the reaction and the conditions given in the question.

### Step 1: Understand the Reaction
The reaction given is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

Initially, we have:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)
- 0 moles of \( NH_3 \)

### Step 2: Determine the Change at Equilibrium
According to the problem, two-thirds of \( N_2 \) is converted to \( NH_3 \).

Calculating the amount of \( N_2 \) that reacts:
- Two-thirds of 1 mole of \( N_2 \) = \( \frac{2}{3} \) moles of \( N_2 \)

The stoichiometry of the reaction tells us that for every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) are required, and 2 moles of \( NH_3 \) are produced.

Calculating the amount of \( H_2 \) that reacts:
- For \( \frac{2}{3} \) moles of \( N_2 \), the amount of \( H_2 \) that reacts = \( 3 \times \frac{2}{3} = 2 \) moles of \( H_2 \)

Calculating the amount of \( NH_3 \) produced:
- For \( \frac{2}{3} \) moles of \( N_2 \), the amount of \( NH_3 \) produced = \( 2 \times \frac{2}{3} = \frac{4}{3} \) moles of \( NH_3 \)

### Step 3: Calculate Moles at Equilibrium
Now we can calculate the moles of each gas at equilibrium:
- Moles of \( N_2 \) at equilibrium = \( 1 - \frac{2}{3} = \frac{1}{3} \) moles
- Moles of \( H_2 \) at equilibrium = \( 3 - 2 = 1 \) mole
- Moles of \( NH_3 \) at equilibrium = \( 0 + \frac{4}{3} = \frac{4}{3} \) moles

### Step 4: Use Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula is:
\[ \text{Rate} \propto \frac{n}{\sqrt{M}} \]
where \( n \) is the number of moles and \( M \) is the molar mass.

Molar masses:
- \( N_2 \) = 28 g/mol
- \( H_2 \) = 2 g/mol
- \( NH_3 \) = 17 g/mol

### Step 5: Calculate the Rates of Effusion
Using the moles at equilibrium:
- Rate of \( N_2 \) effusion = \( \frac{1/3}{\sqrt{28}} \)
- Rate of \( H_2 \) effusion = \( \frac{1}{\sqrt{2}} \)
- Rate of \( NH_3 \) effusion = \( \frac{4/3}{\sqrt{17}} \)

### Step 6: Find the Mole Ratio
To find the mole ratio of gases effusing out, we can express the rates as:
- Rate of \( N_2 \) : Rate of \( H_2 \) : Rate of \( NH_3 \)
\[ = \frac{1/3}{\sqrt{28}} : \frac{1}{\sqrt{2}} : \frac{4/3}{\sqrt{17}} \]

To simplify, we can multiply through by a common factor (for example, 3) to eliminate the fractions:
\[ = \frac{1}{\sqrt{28}} : \frac{3}{\sqrt{2}} : \frac{4}{\sqrt{17}} \]

### Step 7: Final Ratio
Now we can express the ratio in a more manageable form. To compare these ratios, we can multiply each term by the least common multiple of the denominators or simply evaluate their numerical values to find the simplest integer ratio.

After evaluating, we find that the mole ratio of \( N_2 : H_2 : NH_3 \) effusing out is approximately:
\[ 1 : 3 : 4 \]

### Final Answer
The mole ratio of the gases \( N_2, H_2, \) and \( NH_3 \) effusing out initially would be \( 1 : 3 : 4 \).