If the initial number of moles/L of `N_2, H_2, and NH_3, ` are 1, 2 and 3 respectively, their concentrations at equilibrium will be `N_2+3H_2 Leftrightarrow 2NH_3`

Options

A. `"(1-x) (2-3x) 2x`
B. `"(1-x/3) (2-x) 2x/3"`
C. `"(1-x) (2-x) (3+x)"`
D. `"(1-x) (2-3x) (3+2x)"`

Answer

To solve the problem of determining the concentrations of \( N_2 \), \( H_2 \), and \( NH_3 \) at equilibrium for the reaction:

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

given the initial concentrations of \( N_2 \), \( H_2 \), and \( NH_3 \) as 1 mol/L, 2 mol/L, and 3 mol/L respectively, we can follow these steps:

### Step 1: Write the Initial Concentrations
The initial concentrations of the reactants and products are:
- \( [N_2] = 1 \, \text{mol/L} \)
- \( [H_2] = 2 \, \text{mol/L} \)
- \( [NH_3] = 3 \, \text{mol/L} \)

### Step 2: Set Up the Change in Concentration
Let \( x \) be the amount of \( N_2 \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- For every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) react and 2 moles of \( NH_3 \) are produced.

Thus, the changes in concentration will be:
- \( N_2 \): decreases by \( x \)
- \( H_2 \): decreases by \( 3x \)
- \( NH_3 \): increases by \( 2x \)

### Step 3: Write the Equilibrium Concentrations
The equilibrium concentrations can be expressed as:
- \( [N_2]_{eq} = 1 - x \)
- \( [H_2]_{eq} = 2 - 3x \)
- \( [NH_3]_{eq} = 3 + 2x \)

### Step 4: Establish the Equilibrium Condition
At equilibrium, we need to ensure that the concentrations satisfy the equilibrium expression. However, we don't have a specific equilibrium constant given, so we can analyze the system qualitatively or assume \( x \) is small enough that it doesn't significantly change the initial concentrations.

### Step 5: Solve for \( x \)
To find the value of \( x \), we can assume that the reaction goes to completion or reaches a certain extent. In this case, we can analyze the system based on the limiting reactant.

Since \( N_2 \) is present in the least amount relative to the stoichiometric coefficients, it will be the limiting reactant. Therefore, we can set \( x = 1 \) (the initial concentration of \( N_2 \)):

- \( [N_2]_{eq} = 1 - 1 = 0 \)
- \( [H_2]_{eq} = 2 - 3(1) = -1 \) (not possible, indicating that \( x \) must be less than 1)
- \( [NH_3]_{eq} = 3 + 2(1) = 5 \)

### Step 6: Adjust \( x \)
Since \( H_2 \) cannot have a negative concentration, we need to find a suitable \( x \) such that \( 2 - 3x \geq 0 \).

Setting \( 2 - 3x = 0 \):
\[ 3x = 2 \implies x = \frac{2}{3} \]

### Step 7: Calculate Equilibrium Concentrations
Now substituting \( x = \frac{2}{3} \):
- \( [N_2]_{eq} = 1 - \frac{2}{3} = \frac{1}{3} \)
- \( [H_2]_{eq} = 2 - 3 \left(\frac{2}{3}\right) = 0 \)
- \( [NH_3]_{eq} = 3 + 2 \left(\frac{2}{3}\right) = 3 + \frac{4}{3} = \frac{13}{3} \)

### Final Answer
Thus, the equilibrium concentrations are:
- \( [N_2]_{eq} = \frac{1}{3} \, \text{mol/L} \)
- \( [H_2]_{eq} = 0 \, \text{mol/L} \)
- \( [NH_3]_{eq} = \frac{13}{3} \, \text{mol/L} \)