`N_(2)+3H_(2)rarr2NH_(3)`. 1 mol `N_(2)` and 4 mol `H_(2)` are taken in 15L flask at `27^(@)C` After complete conversion of `N_(2)` into `NH_(3),5L` of `H_(2)O` is added. pressure set up in the flask is:

Options

A. `(3xx0.0821xx300)/15`atm
B. `(2xx0.0821xx300)/10` atm
C. `(1xx0.0821xx300)/15` atm
D. `(1xx0.0821xx300)/10`atm

Answer

To solve the problem step-by-step, we will follow the reaction and the changes in the system as described in the question.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine the initial moles of reactants
From the problem, we have:
- 1 mole of \( N_2 \)
- 4 moles of \( H_2 \)

### Step 3: Identify the limiting reactant
According to the stoichiometry of the reaction:
- 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \).
- For 1 mole of \( N_2 \), we need 3 moles of \( H_2 \).

Since we have 4 moles of \( H_2 \), \( N_2 \) is the limiting reactant and will be completely consumed.

### Step 4: Calculate the moles of products formed
From the balanced equation:
- 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \).
Thus, when 1 mole of \( N_2 \) is completely consumed, it produces:
- 2 moles of \( NH_3 \)

### Step 5: Calculate the remaining moles of \( H_2 \)
Initially, we had 4 moles of \( H_2 \). Since 3 moles of \( H_2 \) are needed to react with 1 mole of \( N_2 \):
- Remaining \( H_2 = 4 - 3 = 1 \) mole

### Step 6: Determine the total moles of gas after the reaction
After the reaction, we have:
- 2 moles of \( NH_3 \) (which will dissolve in water)
- 1 mole of unreacted \( H_2 \)

Since \( NH_3 \) is soluble in water, it will not contribute to the pressure in the flask. Therefore, the total moles of gas remaining in the flask is:
- Total moles of gas = 1 mole of \( H_2 \)

### Step 7: Calculate the volume of the gas after adding water
The initial volume of the flask is 15 L. After adding 5 L of water, the volume available for the gas becomes:
- Volume available for gas = \( 15 L - 5 L = 10 L \)

### Step 8: Use the ideal gas law to calculate the pressure
We will use the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in L)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = \( 27°C + 273 = 300 K \)

Rearranging the ideal gas law to find pressure:
\[ P = \frac{nRT}{V} \]

Substituting the values:
- \( n = 1 \) mole (of \( H_2 \))
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 300 \) K
- \( V = 10 \) L

Calculating the pressure:
\[ P = \frac{1 \times 0.0821 \times 300}{10} \]
\[ P = \frac{24.63}{10} \]
\[ P = 2.463 \, \text{atm} \]

### Final Answer
The pressure set up in the flask after the addition of water is approximately **2.46 atm**.

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