For the reaction `p+qhArrr+s` , initially concentrations of p and q are equal and at equilibrium the concentration of s will be twice of that of p. What be the equilibrium constant for the reacation ?

Options

A. 2
B. 4
C. `1//4`
D. 8

Answer

To find the equilibrium constant \( K_c \) for the reaction \( P + Q \rightleftharpoons R + S \), we will follow these steps:

### Step 1: Set Up Initial Concentrations
Let the initial concentrations of \( P \) and \( Q \) be \( 1 \) M each (since they are equal). Thus, we have:
- \([P]_0 = 1\)
- \([Q]_0 = 1\)
- \([R]_0 = 0\)
- \([S]_0 = 0\)

### Step 2: Define Changes at Equilibrium
Let \( \alpha \) be the degree of dissociation of \( P \) and \( Q \). At equilibrium, the concentrations will change as follows:
- \([P] = 1 - \alpha\)
- \([Q] = 1 - \alpha\)
- \([R] = \alpha\)
- \([S] = \alpha\)

### Step 3: Use Given Condition
We are given that at equilibrium, the concentration of \( S \) is twice that of \( P \):
\[ [S] = 2[P] \]
Substituting the equilibrium concentrations:
\[ \alpha = 2(1 - \alpha) \]

### Step 4: Solve for \( \alpha \)
Expanding the equation:
\[ \alpha = 2 - 2\alpha \]
Rearranging gives:
\[ \alpha + 2\alpha = 2 \]
\[ 3\alpha = 2 \]
\[ \alpha = \frac{2}{3} \]

### Step 5: Calculate Equilibrium Concentrations
Now substituting \( \alpha \) back into the equilibrium concentrations:
- \([P] = 1 - \frac{2}{3} = \frac{1}{3}\)
- \([Q] = 1 - \frac{2}{3} = \frac{1}{3}\)
- \([R] = \frac{2}{3}\)
- \([S] = \frac{2}{3}\)

### Step 6: Write the Expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[R][S]}{[P][Q]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} \]

### Step 7: Simplify the Expression
Calculating the numerator:
\[ \text{Numerator} = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \]
Calculating the denominator:
\[ \text{Denominator} = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \]
Thus,
\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = \frac{4}{9} \times \frac{9}{1} = 4 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is \( 4 \).

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