Answer:
To determine the equilibrium constant \( K_{eq} \) for the reaction using an ICE table, we need the balanced chemical equation. For the synthesis of ammonia from nitrogen and hydrogen, the balanced equation is:
\[ \text{N}_2(g) + 3\text{H}_2(g) \leftrightarrow 2\text{NH}_3(g) \]
We will use an ICE (Initial, Change, Equilibrium) table to find the equilibrium constant.
**Initial Conditions:**
- \([ \text{N}_2 ]_0 = 0.450 \, \text{M}\)
- \([ \text{H}_2 ]_0 = 0.600 \, \text{M}\)
- \([ \text{NH}_3 ]_0 = 0.0250 \, \text{M}\)
**Change:**
Let \( x \) be the change in concentration of \(\text{N}_2\) and \(\text{H}_2\) that reacts to reach equilibrium. According to the balanced equation, 1 mole of \(\text{N}_2\) reacts with 3 moles of \(\text{H}_2\) to produce 2 moles of \(\text{NH}_3\).
Therefore, at equilibrium:
- \([\text{N}_2] = 0.450 - x\)
- \([\text{H}_2] = 0.600 - 3x\)
- \([\text{NH}_3] = 0.0250 + 2x\)
Given that \([\text{NH}_3]\) at equilibrium is 0.0250 M, and assuming this is the concentration of \(\text{NH}_3\) formed, we can find \(x\) as follows:
At equilibrium:
\[ 0.0250 = 2x \]
\[ x = \frac{0.0250}{2} = 0.0125 \, \text{M} \]
**Equilibrium Concentrations:**
- \([\text{N}_2] = 0.450 - 0.0125 = 0.4375 \, \text{M}\)
- \([\text{H}_2] = 0.600 - 3(0.0125) = 0.600 - 0.0375 = 0.5625 \, \text{M}\)
- \([\text{NH}_3] = 0.0250 \, \text{M}\)
**Equilibrium Constant Calculation:**
The equilibrium constant \( K_{eq} \) is given by:
\[ K_{eq} = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Substitute the equilibrium concentrations:
\[ K_{eq} = \frac{(0.0250)^2}{(0.4375) \times (0.5625)^3} \]
First, calculate the denominator:
\[ (0.5625)^3 = 0.178 \]
\[ 0.4375 \times 0.178 = 0.078 \]
Now calculate \( K_{eq} \):
\[ K_{eq} = \frac{0.000625}{0.078} \approx 0.008 \]
So, the equilibrium constant \( K_{eq} \) is approximately 0.008.