`N_(2) + 3H_(2)hArr2NH_(3)`
1mole `N_(2)` and 3 moles `H_(2)` are present at start in 1 L flask. At equilibrium `NH_(3)` formed required 100 mL of 5 M HC1 for neutralisation hence `K_(c)` is

Options

A. `((0.5)^(2))/((0.75)(2.25)^(3))`
B. `((0.5)^(2))/((0.5)(2.5)^(3))`
C. `((0.5)L)/((0.75)(2.5)^(3))`
D. none of these

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Determine initial moles of reactants.
Initially, we have:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 3: Define the change in moles at equilibrium.
Let \( \alpha \) be the number of moles of \( N_2 \) that reacts at equilibrium. Then:
- Moles of \( N_2 \) at equilibrium = \( 1 - \alpha \)
- Moles of \( H_2 \) at equilibrium = \( 3 - 3\alpha \)
- Moles of \( NH_3 \) formed = \( 2\alpha \)

### Step 4: Use the information about neutralization to find \( \alpha \).
We know that the amount of \( NH_3 \) formed can neutralize 100 mL of 5 M HCl.

First, calculate the moles of HCl:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 5 \, \text{mol/L} \times 0.1 \, \text{L} = 0.5 \, \text{moles} \]

Since \( NH_3 \) reacts with HCl in a 1:1 ratio, the moles of \( NH_3 \) formed is also 0.5 moles:
\[ 2\alpha = 0.5 \implies \alpha = 0.25 \]

### Step 5: Calculate moles of reactants at equilibrium.
Now we can substitute \( \alpha \) back into our expressions for the moles of reactants:
- Moles of \( N_2 \) at equilibrium = \( 1 - 0.25 = 0.75 \)
- Moles of \( H_2 \) at equilibrium = \( 3 - 3(0.25) = 2.25 \)
- Moles of \( NH_3 \) at equilibrium = \( 0.5 \)

### Step 6: Calculate the equilibrium concentrations.
Since the volume of the flask is 1 L, the concentrations are equal to the number of moles:
- \([N_2] = 0.75 \, \text{M}\)
- \([H_2] = 2.25 \, \text{M}\)
- \([NH_3] = 0.5 \, \text{M}\)

### Step 7: Write the expression for \( K_c \).
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression.
\[ K_c = \frac{(0.5)^2}{(0.75)(2.25)^3} \]

### Step 9: Calculate \( K_c \).
Calculating the denominator:
\[ (2.25)^3 = 11.390625 \]
Now substituting:
\[ K_c = \frac{0.25}{0.75 \times 11.390625} = \frac{0.25}{8.54296875} \approx 0.0293 \]

### Final Answer:
Thus, the equilibrium constant \( K_c \) is approximately \( 0.0293 \).

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