`2 "mole" N_(2)` and `3 "mole" H_(2)` gas are allowed to react in a `20 L` flask at `400 K` and after complete conversion of `H_(2)` into `NH_(3)`. `10 L H_(2)O` was added and temperature reduced to `300 K`. Pressure of the gas after reaction is :
`N_(2)+3H_(2)to2NH_(3)`

Options

A. `3Rxx300//20`
B. `3Rxx300//10`
C. `Rxx300//20`
D. `Rxx300//10`

Answer

To solve the problem step by step, we will follow the stoichiometry of the reaction and the changes in conditions after the reaction is complete.

### Step 1: Identify the Reaction and Initial Moles
The reaction given is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

Initially, we have:
- 2 moles of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 2: Determine Limiting Reactant
From the stoichiometry of the reaction, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \). Therefore, for 2 moles of \( N_2 \), we would need:
\[ 2 \text{ moles of } N_2 \times 3 \text{ moles of } H_2 = 6 \text{ moles of } H_2 \]

However, we only have 3 moles of \( H_2 \). Thus, \( H_2 \) is the limiting reactant.

### Step 3: Calculate Moles After Reaction
Since \( H_2 \) is the limiting reactant, it will completely react:
- 3 moles of \( H_2 \) will react with \( 1 \) mole of \( N_2 \) (as per the stoichiometric ratio).
- Therefore, \( 2 \text{ moles of } N_2 - 1 \text{ mole of } N_2 = 1 \text{ mole of } N_2 \) remains unreacted.
- The reaction produces \( 2 \text{ moles of } NH_3 \) from \( 3 \text{ moles of } H_2 \).

Final moles after the reaction:
- \( N_2 \): 1 mole
- \( NH_3 \): 2 moles
- \( H_2 \): 0 moles

### Step 4: Calculate Total Moles of Gas
After the reaction, the total moles of gas in the flask are:
\[ 1 \text{ mole of } N_2 + 2 \text{ moles of } NH_3 = 3 \text{ moles of gas} \]

### Step 5: Consider the Addition of Water
After the reaction, 10 L of water is added. Since ammonia (\( NH_3 \)) is highly soluble in water, it will dissolve completely. Therefore, the only gas remaining in the flask is:
- \( N_2 \): 1 mole

### Step 6: Calculate the Volume of Gas
The total volume of the flask is 20 L, but since 10 L of water is added, the volume available for the gas becomes:
\[ 20 \text{ L} - 10 \text{ L} = 10 \text{ L} \]

### Step 7: Use the Ideal Gas Law to Calculate Pressure
Using the ideal gas law:
\[ P = \frac{nRT}{V} \]
Where:
- \( n = 1 \) mole of \( N_2 \)
- \( R \) is the universal gas constant
- \( T = 300 \text{ K} \)
- \( V = 10 \text{ L} \)

Substituting the values:
\[ P = \frac{1 \times R \times 300}{10} = 30R \]

### Final Answer
The pressure of the gas after the reaction is:
\[ P = 30R \]

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