`1 "mole of" N_(2) ` and 3 `"moles of " H_(2)` are placed in `1L` vessel. Find the concentration of `NH_(3)` at equilibrium, if the equilibrium constant `(K_(c))` at `400K "is" (4)/(27)`

Answer

To solve the problem step-by-step, we will follow the process of establishing the reaction, setting up the equilibrium expression, and calculating the concentration of NH₃ at equilibrium.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]

### Step 2: Write down the initial concentrations
Given:
- Initial moles of N₂ = 1 mole
- Initial moles of H₂ = 3 moles
- Volume of the vessel = 1 L

Thus, the initial concentrations are:
- \([\text{N}_2] = \frac{1 \text{ mole}}{1 \text{ L}} = 1 \text{ M}\)
- \([\text{H}_2] = \frac{3 \text{ moles}}{1 \text{ L}} = 3 \text{ M}\)
- \([\text{NH}_3] = 0 \text{ M}\)

### Step 3: Define the change in concentrations at equilibrium
Let \( x \) be the amount of N₂ that reacts at equilibrium. The changes in concentrations will be:
- \([\text{N}_2] = 1 - x\)
- \([\text{H}_2] = 3 - 3x\)
- \([\text{NH}_3] = 2x\)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(1 - x)(3 - 3x)^3} \]

### Step 5: Substitute the given value of \( K_c \)
We know that \( K_c = \frac{4}{27} \). Therefore, we can set up the equation:
\[ \frac{(2x)^2}{(1 - x)(3 - 3x)^3} = \frac{4}{27} \]

### Step 6: Simplify the equation
This simplifies to:
\[ \frac{4x^2}{(1 - x)(3 - 3x)^3} = \frac{4}{27} \]

Cross-multiplying gives:
\[ 4x^2 \cdot 27 = 4(1 - x)(3 - 3x)^3 \]

### Step 7: Cancel out the common factor
Dividing both sides by 4:
\[ 27x^2 = (1 - x)(3 - 3x)^3 \]

### Step 8: Expand and solve for \( x \)
Expanding the right side and solving for \( x \) can be complex, but we can use numerical or graphical methods to find \( x \). After solving, we find:
\[ x \approx 0.381 \text{ M} \]

### Step 9: Calculate the concentration of NH₃ at equilibrium
Now, we can find the concentration of NH₃ at equilibrium:
\[ [\text{NH}_3] = 2x = 2 \times 0.381 = 0.762 \text{ M} \]

### Final Answer
The concentration of NH₃ at equilibrium is:
\[ \boxed{0.762 \text{ M}} \]