To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between nitrogen and hydrogen to form ammonia is given by the balanced equation:
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
### Step 2: Determine initial moles and changes
Initially, we have:
- Moles of \(N_2\) = 2
- Moles of \(H_2\) = 8
At equilibrium, we are told that there are 4 moles of \(N_2\). This indicates that some \(N_2\) has reacted.
### Step 3: Calculate the change in moles of \(N_2\)
Let \(x\) be the number of moles of \(N_2\) that reacted. Since we started with 2 moles of \(N_2\) and ended with 4 moles, we can set up the equation:
\[
2 - x = 4
\]
From this, we can solve for \(x\):
\[
-x = 4 - 2 \\
-x = 2 \\
x = -2
\]
This means that instead of reacting, \(N_2\) has increased, which suggests that there was no reaction and the initial moles were misinterpreted.
### Step 4: Reassess the equilibrium condition
Since we have 4 moles of \(N_2\) at equilibrium, it indicates that the reaction did not proceed as expected. The initial moles of \(N_2\) were 2, and since we have 4 moles at equilibrium, we need to consider the stoichiometry of the reaction.
### Step 5: Calculate the change in moles of \(H_2\)
Since the reaction consumes \(3\) moles of \(H_2\) for every \(1\) mole of \(N_2\), the change in \(H_2\) will be:
\[
\text{Change in } H_2 = 3 \times \text{Change in } N_2
\]
If \(N_2\) is increasing by 2 moles, then:
\[
\text{Change in } H_2 = 3 \times (-2) = -6
\]
Thus, the moles of \(H_2\) at equilibrium will be:
\[
8 - 6 = 2 \text{ moles}
\]
### Step 6: Calculate the equilibrium concentration of \(H_2\)
To find the concentration of \(H_2\) at equilibrium, we use the formula:
\[
\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}
\]
Given that the volume of the vessel is \(2 \, \text{liters}\):
\[
\text{Concentration of } H_2 = \frac{2 \, \text{moles}}{2 \, \text{liters}} = 1 \, \text{mol/L}
\]
### Final Answer
The equilibrium concentration of \(H_2\) is \(1 \, \text{mol/L}\).
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