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_(c))` ?

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. 1
D. `(1)/(2) `

Answer

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

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

The reaction quotient \( Q_c \) is given by the formula:

\[ Q_c = \frac{[P_2]^2}{[P_4]} \]

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 L

To find the concentrations, we use the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Calculating for \( P_4 \):

\[ [P_4] = \frac{3 \text{ moles}}{2 \text{ L}} = 1.5 \, \text{M} \]

Calculating for \( P_2 \):

\[ [P_2] = \frac{2 \text{ moles}}{2 \text{ L}} = 1 \, \text{M} \]

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

Now we substitute the concentrations into the \( Q_c \) expression:

\[ Q_c = \frac{[P_2]^2}{[P_4]} = \frac{(1)^2}{1.5} = \frac{1}{1.5} \]

### Step 4: Simplify the expression

Now we simplify \( \frac{1}{1.5} \):

\[ Q_c = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 \]

### Final Answer

Thus, the value of the reaction quotient \( Q_c \) is:

\[ Q_c = \frac{2}{3} \]