Sure! Let's solve this step by step:
### 1. Write down the equilibrium expression:
For the reaction:
[tex]\[ 3 H_2(g) + N_2(g) \rightleftharpoons 2 NH_3(g) \][/tex]
The equilibrium constant expression ([tex]\( K_{eq} \)[/tex]) is given by:
[tex]\[ K_{eq} = \frac{[NH_3]^2}{[H_2]^3 \cdot [N_2]} \][/tex]
### 2. Insert the given equilibrium values:
- [tex]\( K_{eq} = 1.7 \times 10^2 \)[/tex]
- [tex]\( [H_2] = 0.13 \)[/tex] M
- [tex]\( [N_2] = 0.025 \)[/tex] M
We need to find the concentration of [tex]\( NH_3 \)[/tex], denoted as [tex]\( [NH_3] \)[/tex].
### 3. Calculate the concentration of [tex]\( [H_2]^3 \)[/tex]:
[tex]\[ [H_2]^3 = (0.13)^3 = 0.002197 \][/tex]
### 4. Use the equilibrium constant to solve for [tex]\( [NH_3]^2 \)[/tex]:
[tex]\[ K_{eq} = \frac{[NH_3]^2}{[H_2]^3 \cdot [N_2]} \][/tex]
Rearranging for [tex]\( [NH_3]^2 \)[/tex]:
[tex]\[ [NH_3]^2 = K_{eq} \cdot [H_2]^3 \cdot [N_2] \][/tex]
[tex]\[ [NH_3]^2 = 1.7 \times 10^2 \cdot 0.002197 \cdot 0.025 \][/tex]
[tex]\[ [NH_3]^2 = 0.00933725 \][/tex]
### 5. Solve for [tex]\( [NH_3] \)[/tex]:
[tex]\[ [NH_3] = \sqrt{0.00933725} \][/tex]
[tex]\[ [NH_3] \approx 0.09663 \][/tex]
### 6. Expressing the concentration to two significant figures:
[tex]\[ [NH_3] = 0.097 \, \text{M} \][/tex]
### Final Answer:
[tex]\[ \left[ NH_3 \right] = 0.097 \, \text{M} \][/tex]
Therefore, the molar concentration of [tex]\( NH_3 \)[/tex] at equilibrium is [tex]\( 0.097 \)[/tex] M.
