Equilibrium constant, `K_(c)` for the reaction,
`N_(2(g))+3H_(2(g))hArr2NH_(3(g))`,
at `500 K` is `0.061 litre^(2) "mole"^(-2)`. At a particular time, the analysis shows that composition of the reaction mixture is `3.00 mol litre^(-1)N_(2)`, `2.00 mol litre^(-1)H_(2)`, and `0.500 mol litre^(-1)NH_(3)`. Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Explanation

To determine whether the reaction is at equilibrium and in which direction it tends to proceed, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

For the reaction:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

The equilibrium constant expression is given by:
\[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \]

### Step 2: Substitute the equilibrium constant value

Given that \( K_c = 0.061 \, \text{litre}^2 \text{mole}^{-2} \) at 500 K, we will compare this with the reaction quotient \( Q_c \).

### Step 3: Calculate the reaction quotient \( Q_c \)

Using the concentrations provided:
- \([N_2] = 3.00 \, \text{mol/litre}\)
- \([H_2] = 2.00 \, \text{mol/litre}\)
- \([NH_3] = 0.500 \, \text{mol/litre}\)

Substituting these values into the expression for \( Q_c \):
\[ Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.500)^2}{(3.00)(2.00)^3} \]

Calculating \( Q_c \):
\[ Q_c = \frac{0.25}{3.00 \times 8.00} = \frac{0.25}{24} = 0.0104 \]

### Step 4: Compare \( Q_c \) with \( K_c \)

Now, we compare \( Q_c \) and \( K_c \):
- \( Q_c = 0.0104 \)
- \( K_c = 0.061 \)

Since \( Q_c < K_c \), the reaction is not at equilibrium.

### Step 5: Determine the direction of the reaction

When \( Q_c < K_c \), the reaction will shift to the right (towards the products) to reach equilibrium.

### Conclusion

1. The reaction is **not at equilibrium**.
2. The reaction tends to proceed in the **forward direction** (towards the formation of NH3).

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