The reaction `A+B hArr C+D` proceeds to right hand side upto 99.9% . The equilibrium constant K of the reaction will be

Options

A. `10^4`
B. `10^5`
C. `10^6`
D. `10^8`

Answer

To solve the problem regarding the equilibrium constant \( K \) for the reaction \( A + B \rightleftharpoons C + D \) that proceeds to the right-hand side up to 99.9%, we can follow these steps:

### Step 1: Understand the Reaction and Define Variables
The reaction is given as:
\[ A + B \rightleftharpoons C + D \]

Let’s denote:
- The initial concentration of \( A \) and \( B \) as 1 M each (for simplicity).
- The degree of completion (conversion) of the reaction, \( \alpha = 0.999 \) (since it proceeds to 99.9%).

### Step 2: Calculate Concentrations at Equilibrium
At equilibrium, the concentrations of the reactants and products can be expressed as:
- Concentration of \( A \) at equilibrium:
\[ [A] = 1 - \alpha = 1 - 0.999 = 0.001 \, \text{M} \]
- Concentration of \( B \) at equilibrium:
\[ [B] = 1 - \alpha = 1 - 0.999 = 0.001 \, \text{M} \]
- Concentration of \( C \) at equilibrium:
\[ [C] = \alpha = 0.999 \, \text{M} \]
- Concentration of \( D \) at equilibrium:
\[ [D] = \alpha = 0.999 \, \text{M} \]

### Step 3: Write the Expression for the Equilibrium Constant \( K \)
The equilibrium constant \( K \) for the reaction is defined as:
\[ K = \frac{[C][D]}{[A][B]} \]

### Step 4: Substitute the Equilibrium Concentrations into the Expression
Substituting the equilibrium concentrations into the expression for \( K \):
\[ K = \frac{(0.999)(0.999)}{(0.001)(0.001)} \]

### Step 5: Simplify the Expression
Calculating the numerator and denominator:
- Numerator:
\[ 0.999 \times 0.999 = 0.998001 \]
- Denominator:
\[ 0.001 \times 0.001 = 0.000001 \]

Thus, we have:
\[ K = \frac{0.998001}{0.000001} = 998001 \]

### Step 6: Approximate the Result
For practical purposes, we can express this in scientific notation:
\[ K \approx 10^6 \]

### Conclusion
The equilibrium constant \( K \) for the reaction is approximately \( 10^6 \).

### Final Answer
The equilibrium constant \( K \) is \( 10^6 \).

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