`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 need to follow the process of establishing the equilibrium for the reaction and applying the equilibrium constant expression.

### Step 1: Write the balanced chemical equation.
The reaction between nitrogen and hydrogen to form ammonia can be represented as:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Set up the initial concentrations.
We have:
- Initial moles of \( N_2 = 1 \) mole
- Initial moles of \( H_2 = 3 \) moles
- Initial moles of \( NH_3 = 0 \) moles

Since the volume of the vessel is 1 L, the initial concentrations are:
- \([N_2] = 1 \, \text{M}\)
- \([H_2] = 3 \, \text{M}\)
- \([NH_3] = 0 \, \text{M}\)

### Step 3: Define the change in concentrations at equilibrium.
Let \( x \) be the amount of \( N_2 \) that reacts at equilibrium. The changes in concentrations will be:
- For \( N_2 \): \( 1 - x \)
- For \( H_2 \): \( 3 - 3x \) (since 3 moles of \( H_2 \) react for every mole of \( N_2 \))
- For \( NH_3 \): \( 2x \) (since 2 moles of \( NH_3 \) are produced for every mole of \( N_2 \))

### Step 4: Write the equilibrium concentrations.
At equilibrium, the concentrations will be:
- \([N_2] = 1 - x\)
- \([H_2] = 3 - 3x\)
- \([NH_3] = 2x\)

### Step 5: Write the expression for the equilibrium constant \( K_c \).
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(1 - x)(3 - 3x)^3} \]

### Step 6: Substitute the value of \( K_c \) and solve for \( x \).
Given \( K_c = \frac{4}{27} \), we have:
\[ \frac{4}{27} = \frac{4x^2}{(1 - x)(3 - 3x)^3} \]
Cross-multiplying gives:
\[ 4(1 - x)(3 - 3x)^3 = 4x^2 \cdot 27 \]
Simplifying this leads to:
\[ (1 - x)(3 - 3x)^3 = 27x^2 \]

### Step 7: Solve the equation for \( x \).
This is a cubic equation in \( x \). After simplifying and solving, we find:
\[ x \approx 0.381 \]

### Step 8: Calculate the concentration of \( NH_3 \) at equilibrium.
The concentration of \( NH_3 \) at equilibrium is:
\[ [NH_3] = 2x = 2(0.381) \approx 0.762 \, \text{M} \]

### Final Answer:
The concentration of \( NH_3 \) at equilibrium is approximately **0.762 M**.

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