`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. `3Rxx(300)/(20)`
B. `3Rxx(300)/(10)`
C. `Rxx(300)/(20)`
D. `Rxx(300)/(10)`

Explanation

To solve the problem step by step, we will follow the stoichiometry of the reaction and apply the ideal gas law.

### 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 the limiting reactant
From the stoichiometry of the reaction:
- 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \).
- Therefore, 2 moles of \( N_2 \) would require \( 2 \times 3 = 6 \) moles of \( H_2 \).

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

### Step 3: Calculate the moles of products formed
Using the stoichiometry:
- 3 moles of \( H_2 \) will produce \( \frac{2}{3} \times 3 = 2 \) moles of \( NH_3 \).
- The remaining moles of \( N_2 \) after the reaction will be:
\[ 2 - 1 = 1 \text{ mole of } N_2 \]
(since 1 mole of \( N_2 \) is consumed for 3 moles of \( H_2 \)).

### Step 4: Calculate the total moles after the reaction
After the reaction, the total moles of gas in the flask will be:
- Moles of \( N_2 \): 1
- Moles of \( NH_3 \): 2

Total moles of gas = \( 1 + 2 = 3 \) moles.

### Step 5: Consider the addition of water
After the reaction, 10 L of water is added to the system. Since \( NH_3 \) is soluble in water, it will dissolve in the added water. However, we will consider the volume of gas remaining in the flask.

### Step 6: Calculate the volume of gas remaining
Initially, the volume of the flask is 20 L. After adding 10 L of water, the volume available for gas becomes:
\[ 20 \text{ L} - 10 \text{ L} = 10 \text{ L} \]

### Step 7: Apply the ideal gas law to find pressure
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in L)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in K)

Substituting the values:
- \( n = 3 \) moles (total moles of gas)
- \( V = 10 \) L (volume after adding water)
- \( T = 300 \) K (final temperature)

The equation becomes:
\[ P \cdot 10 = 3 \cdot 0.0821 \cdot 300 \]

### Step 8: Solve for pressure \( P \)
\[ P = \frac{3 \cdot 0.0821 \cdot 300}{10} \]
\[ P = \frac{73.89}{10} \]
\[ P = 7.389 \text{ atm} \]

### Final Answer
The pressure of the gas after the reaction is approximately **7.39 atm**.

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