The volumes of two vessels at same temperature are in the ratio of `2 : 3`. One vessel contains `H_(2)` and other `N_(2)` at 600 mm and 900 mm respectively. The final pressure when they are connected together is : (Assume that `N_(2) and H_(2)` react to form `NH_(3)`)

Options

A. 620 mm
B. 760 mm
C. 780 mm
D. 800 mm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
We have two vessels with volumes in the ratio of 2:3. One vessel contains hydrogen (H₂) at a pressure of 600 mm, and the other contains nitrogen (N₂) at a pressure of 900 mm.

### Step 2: Assign Variables to the Volumes
Let the volume of the first vessel (H₂) be \( V_1 = 2V \) and the volume of the second vessel (N₂) be \( V_2 = 3V \).

### Step 3: Calculate the Number of Moles of Each Gas
Using the ideal gas equation \( PV = nRT \), we can find the number of moles of each gas.

For H₂:
\[ n_{H_2} = \frac{P_1 V_1}{RT} = \frac{600 \, \text{mm} \times 2V}{RT} = \frac{1200V}{RT} \]

For N₂:
\[ n_{N_2} = \frac{P_2 V_2}{RT} = \frac{900 \, \text{mm} \times 3V}{RT} = \frac{2700V}{RT} \]

### Step 4: Write the Balanced Reaction Equation
The reaction between hydrogen and nitrogen to form ammonia is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 5: Determine the Limiting Reagent
From the stoichiometry of the reaction:
- 1 mole of N₂ reacts with 3 moles of H₂.
- We have \( \frac{2700V}{RT} \) moles of N₂ and \( \frac{1200V}{RT} \) moles of H₂.

To find the limiting reagent, we calculate the required amount of H₂ for the available N₂:
\[ \text{Required } H_2 = 3 \times n_{N_2} = 3 \times \frac{2700V}{RT} = \frac{8100V}{RT} \]
Since we only have \( \frac{1200V}{RT} \) moles of H₂, H₂ is the limiting reagent.

### Step 6: Calculate Remaining Moles of N₂
The amount of N₂ that reacts with \( \frac{1200V}{RT} \) moles of H₂:
\[ \text{Moles of } N_2 \text{ reacted} = \frac{1200V}{RT} \times \frac{1}{3} = \frac{400V}{RT} \]
Remaining moles of N₂:
\[ n_{N_2, \text{remaining}} = n_{N_2} - \text{Moles of } N_2 \text{ reacted} = \frac{2700V}{RT} - \frac{400V}{RT} = \frac{2300V}{RT} \]

### Step 7: Calculate Moles of Ammonia Formed
From the reaction, \( 3H_2 \) produces \( 2NH_3 \):
\[ \text{Moles of } NH_3 = \frac{2}{3} \times \frac{1200V}{RT} = \frac{800V}{RT} \]

### Step 8: Total Moles After Reaction
Total moles after the reaction:
\[ n_{\text{total}} = n_{N_2, \text{remaining}} + n_{NH_3} = \frac{2300V}{RT} + \frac{800V}{RT} = \frac{3100V}{RT} \]

### Step 9: Calculate Final Pressure
The total volume when the vessels are connected is \( V_{total} = V_1 + V_2 = 2V + 3V = 5V \).
Using the ideal gas equation again for the final pressure:
\[ P = \frac{nRT}{V} = \frac{\frac{3100V}{RT} \cdot RT}{5V} = \frac{3100}{5} = 620 \, \text{mm} \]

### Final Answer
The final pressure when the two vessels are connected is **620 mm**.

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