For the equilibrium `N_(2)+3H_(2)hArr2NH_(3),K_(c)` at 1000K is `2.73xx10^(-3)` if at equlibrium `[N_(2)]=2M,[H_(2)]=3M,` the concentraion of `NH_(3)` is

Options

A. `0.00358` M
B. `0.0358` M
C. `0.358` M
D. `3.58` M

Explanation

To find the concentration of \( NH_3 \) at equilibrium for the reaction

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

given that \( K_c = 2.73 \times 10^{-3} \), \( [N_2] = 2 \, M \), and \( [H_2] = 3 \, M \), we can use the equilibrium constant expression.

### Step-by-Step Solution:

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

2. **Substitute the known values**:
We know that \( K_c = 2.73 \times 10^{-3} \), \( [N_2] = 2 \, M \), and \( [H_2] = 3 \, M \). We need to find \( [NH_3] \).
\[ 2.73 \times 10^{-3} = \frac{[NH_3]^2}{(2)(3^3)} \]

3. **Calculate \( [H_2]^3 \)**:
Calculate \( 3^3 \):
\[ 3^3 = 27 \]

4. **Substitute \( [H_2]^3 \) into the equation**:
Substitute \( 27 \) into the equilibrium expression:
\[ 2.73 \times 10^{-3} = \frac{[NH_3]^2}{(2)(27)} \]

5. **Simplify the equation**:
Calculate \( (2)(27) = 54 \):
\[ 2.73 \times 10^{-3} = \frac{[NH_3]^2}{54} \]

6. **Rearrange to solve for \( [NH_3]^2 \)**:
Multiply both sides by 54:
\[ [NH_3]^2 = 54 \times 2.73 \times 10^{-3} \]

7. **Calculate \( [NH_3]^2 \)**:
\[ [NH_3]^2 = 0.14622 \]

8. **Take the square root to find \( [NH_3] \)**:
\[ [NH_3] = \sqrt{0.14622} \approx 0.3825 \, M \]

9. **Round to three significant figures**:
The concentration of \( NH_3 \) at equilibrium is approximately:
\[ [NH_3] \approx 0.383 \, M \]

### Final Answer:
The concentration of \( NH_3 \) at equilibrium is approximately **0.383 M**.