`1` mole `N_(2)` and `3` moles of `H_(2)` are taken in a `4` litre vessel and heated at `T` Kelvin temperature till following euilibrium is reached `:-`
`N_(2)+3H_(2)hArr2NH_(3)`
At equilibrium `0.25%` `N_(2)` is converted to ammonia. Find equilibrium constant of reaction.

Options

A. `1.48xx10^(-5)lit` `mol^(-1)`
B. `14.8xx10^(-6)lit^(2)` `mol^(-2)`
C. `2.96xx10^(-5)lit^(2)` `mol^(-2)`
D. `2.96xx10^(-6)lit` `mol`

Explanation

To find the equilibrium constant \( K_c \) for the reaction

\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

given that 0.25% of \( N_2 \) is converted to ammonia, we can follow these steps:

### Step 1: Determine the initial moles of reactants
- Initial moles of \( N_2 = 1 \) mole
- Initial moles of \( H_2 = 3 \) moles
- Initial moles of \( NH_3 = 0 \) moles

### Step 2: Calculate the amount of \( N_2 \) converted
Since 0.25% of \( N_2 \) is converted to \( NH_3 \):

\[ \text{Moles of } N_2 \text{ converted} = 0.25\% \text{ of } 1 \text{ mole} = \frac{0.25}{100} \times 1 = 0.0025 \text{ moles} \]

### Step 3: Determine the change in moles at equilibrium
Let \( x = 0.0025 \) moles of \( N_2 \) converted. The changes in moles will be:
- \( N_2 \): \( 1 - x = 1 - 0.0025 = 0.9975 \) moles
- \( H_2 \): \( 3 - 3x = 3 - 3(0.0025) = 2.9925 \) moles
- \( NH_3 \): \( 2x = 2(0.0025) = 0.005 \) moles

### Step 4: Calculate concentrations at equilibrium
The volume of the vessel is \( 4 \) liters, so we can calculate the equilibrium concentrations:

\[ \text{Concentration of } N_2 = \frac{0.9975 \text{ moles}}{4 \text{ L}} = 0.249375 \text{ M} \]

\[ \text{Concentration of } H_2 = \frac{2.9925 \text{ moles}}{4 \text{ L}} = 0.748125 \text{ M} \]

\[ \text{Concentration of } NH_3 = \frac{0.005 \text{ moles}}{4 \text{ L}} = 0.00125 \text{ M} \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is:

\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Substituting the values we calculated:

\[ K_c = \frac{(0.00125)^2}{(0.249375)(0.748125)^3} \]

Calculating each part:

\[ (0.00125)^2 = 0.0000015625 \]

\[ (0.748125)^3 \approx 0.418 \]

Now substituting these values into the equation:

\[ K_c = \frac{0.0000015625}{(0.249375)(0.418)} \approx \frac{0.0000015625}{0.1041} \approx 0.00001504 \]

### Step 7: Final calculation
Calculating the final value gives:

\[ K_c \approx 1.48 \times 10^{-5} \text{ M}^{-2} \]

### Conclusion
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:

\[ K_c \approx 1.48 \times 10^{-5} \text{ M}^{-2} \]

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