Answered

For the reaction

[tex]\[ 3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g) \][/tex]

at [tex]\( 225^{\circ}C \)[/tex], the equilibrium constant is [tex]\( 1.7 \times 10^2 \)[/tex]. The equilibrium mixture contains [tex]\( 0.13 \, M \, H_2 \)[/tex] and [tex]\( 0.025 \, M \, N_2 \)[/tex]. What is the molar concentration of [tex]\( NH_3 \)[/tex]?

Express your answer to two significant figures and include the appropriate units.

[tex]\[ \left[ NH_3 \right] = \, \boxed{\hspace{2cm}} \, \text{Units} \][/tex]



Answer :

GinnyAnswer
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.