for the reaction `P+Q rArr R + 2S`, initially the concentration of P is equal to that of Q (1 molar) but at equilibrium the concentration of R will be twice of that of P, then the equilibrium constant of the reaction is

Options

A. `4/3`
B. `32/3`
C. `3/10`
D. `1/10`

Answer

To solve the problem, we need to analyze the reaction and the information given about the concentrations at equilibrium. The reaction is:

\[ P + Q \rightleftharpoons R + 2S \]

### Step 1: Define Initial Concentrations
Initially, the concentrations of P and Q are both given as 1 M. Therefore:

- \([P]_0 = 1 \, \text{M}\)
- \([Q]_0 = 1 \, \text{M}\)
- \([R]_0 = 0 \, \text{M}\)
- \([S]_0 = 0 \, \text{M}\)

### Step 2: Define Change in Concentration
Let \(x\) be the amount of P and Q that reacts at equilibrium. Then, at equilibrium, the concentrations will be:

- \([P] = 1 - x\)
- \([Q] = 1 - x\)
- \([R] = x\)
- \([S] = 2x\) (since 2 moles of S are produced)

### Step 3: Use Given Condition
According to the problem, at equilibrium, the concentration of R is twice that of P:

\[ x = 2(1 - x) \]

### Step 4: Solve for \(x\)
Now, we can solve the equation:

\[ x = 2 - 2x \]
\[ x + 2x = 2 \]
\[ 3x = 2 \]
\[ x = \frac{2}{3} \]

### Step 5: Calculate Equilibrium Concentrations
Now we can substitute \(x\) back into the expressions for the equilibrium concentrations:

- \([P] = 1 - x = 1 - \frac{2}{3} = \frac{1}{3}\)
- \([Q] = 1 - x = 1 - \frac{2}{3} = \frac{1}{3}\)
- \([R] = x = \frac{2}{3}\)
- \([S] = 2x = 2 \times \frac{2}{3} = \frac{4}{3}\)

### Step 6: Write the Expression for the Equilibrium Constant \(K_c\)
The equilibrium constant \(K_c\) for the reaction is given by:

\[ K_c = \frac{[R][S]^2}{[P][Q]} \]

Substituting the equilibrium concentrations we found:

\[ K_c = \frac{\left(\frac{2}{3}\right) \left(\frac{4}{3}\right)^2}{\left(\frac{1}{3}\right) \left(\frac{1}{3}\right)} \]

### Step 7: Simplify the Expression
Calculating the numerator:

\[ \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
So,
\[ K_c = \frac{\left(\frac{2}{3}\right) \left(\frac{16}{9}\right)}{\left(\frac{1}{3}\right) \left(\frac{1}{3}\right)} = \frac{\frac{32}{27}}{\frac{1}{9}} = \frac{32}{27} \times 9 = \frac{32 \times 9}{27} = \frac{288}{27} = \frac{32}{3} \]

### Final Answer
Thus, the equilibrium constant \(K_c\) is:

\[ K_c = \frac{32}{3} \]