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. none of these

Answer

To solve the problem, we need to calculate the reaction quotient \( Q_c \) for the reaction:

\[ P_4(g) \rightleftharpoons 2P_2(g) \]

**Step 1: Identify the given information.**
- The equilibrium constant \( K_c = 1.4 \) at \( 400^\circ C \).
- Moles of \( P_4 = 3 \) moles.
- Moles of \( P_2 = 2 \) moles.
- Volume of the container = 2 liters.

**Step 2: Calculate the concentrations of the reactants and products.**
- The concentration of a substance is calculated using the formula:

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

- For \( P_4 \):

\[ \text{Concentration of } P_4 = \frac{3 \text{ moles}}{2 \text{ L}} = 1.5 \, \text{mol/L} \]

- For \( P_2 \):

\[ \text{Concentration of } P_2 = \frac{2 \text{ moles}}{2 \text{ L}} = 1.0 \, \text{mol/L} \]

**Step 3: Write the expression for the reaction quotient \( Q_c \).**
- The expression for \( Q_c \) for the reaction is given by:

\[ Q_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}} \]

- For our reaction:

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

**Step 4: Substitute the concentrations into the \( Q_c \) expression.**
- Substitute the concentrations we calculated:

\[ Q_c = \frac{(1.0)^2}{(1.5)} = \frac{1}{1.5} \]

**Step 5: Simplify the expression.**
- Simplifying \( \frac{1}{1.5} \):

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

**Final Answer:**
The value of the reaction quotient \( Q_c \) is \( \frac{2}{3} \).

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