The equilibrium constant `K_(c)` for the reaction `P_(4)(g) hArr 2P_(2)(g)` is `1.4` at `400^(@)C`. Suppose that 3 moles of `P_(4)(g)` and 2 moles of `P_(2)(g)` are mixed in 2 litre container at `400^(@)C`. What is the value of reaction quotient `(Q)` ?

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. 1
D. None of these

Explanation

To find the value of the reaction quotient \( Q \) for the reaction \( P_4(g) \rightleftharpoons 2P_2(g) \), we will follow these steps:

### Step 1: Write the expression for the reaction quotient \( Q \)

The reaction quotient \( Q \) is defined as:
\[ Q = \frac{[P_2]^2}{[P_4]} \]
where \([P_2]\) and \([P_4]\) are the molar concentrations of \( P_2 \) and \( P_4 \) respectively.

### Step 2: Calculate the concentrations of \( P_4 \) and \( P_2 \)

We are given:
- Moles of \( P_4 = 3 \) moles
- Moles of \( P_2 = 2 \) moles
- Volume of the container = 2 liters

Now, we calculate the concentrations:
\[ [P_4] = \frac{\text{moles of } P_4}{\text{volume}} = \frac{3 \text{ moles}}{2 \text{ L}} = 1.5 \, \text{M} \]
\[ [P_2] = \frac{\text{moles of } P_2}{\text{volume}} = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \, \text{M} \]

### Step 3: Substitute the concentrations into the \( Q \) expression

Now, substituting the concentrations into the expression for \( Q \):
\[ Q = \frac{[P_2]^2}{[P_4]} = \frac{(1)^2}{1.5} = \frac{1}{1.5} = \frac{2}{3} \]

### Step 4: Final answer

Thus, the value of the reaction quotient \( Q \) is:
\[ Q = \frac{2}{3} \]