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

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction is given as \( A + B \rightleftharpoons C + D \). We know that the reaction proceeds to the right-hand side (towards products) up to 99.9%.

2. **Defining Alpha**:
The extent of the reaction can be represented by \( \alpha \), which is the fraction of reactants converted to products. Here, \( \alpha = 0.999 \) (since 99.9% is equivalent to 0.999).

3. **Initial Concentrations**:
Assume the initial concentrations of \( A \) and \( B \) are both 1 M (1 mole per liter). Therefore, at time \( t = 0 \):
- \([A]_0 = 1\) M
- \([B]_0 = 1\) M
- \([C]_0 = 0\) M
- \([D]_0 = 0\) M

4. **Change in Concentrations**:
At equilibrium, since 99.9% of \( A \) and \( B \) have reacted:
- Change in concentration of \( A \) and \( B \) = \( -\alpha \) = -0.999
- Therefore, at equilibrium:
- \([A] = 1 - 0.999 = 0.001\) M
- \([B] = 1 - 0.999 = 0.001\) M
- \([C] = 0 + 0.999 = 0.999\) M
- \([D] = 0 + 0.999 = 0.999\) M

5. **Equilibrium Concentrations**:
At equilibrium, we have:
- \([A] = 0.001\) M
- \([B] = 0.001\) M
- \([C] = 0.999\) M
- \([D] = 0.999\) M

6. **Calculating the Equilibrium Constant \( K \)**:
The equilibrium constant \( K \) is given by the formula:
\[ K = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K = \frac{(0.999)(0.999)}{(0.001)(0.001)} = \frac{0.998001}{0.000001} = 998001 \]

7. **Expressing \( K \) in Scientific Notation**:
The value \( 998001 \) can be approximated as \( 10^6 \) (since \( 998001 \) is very close to \( 10^6 \)).

8. **Final Answer**:
Thus, the equilibrium constant \( K \) for the reaction is approximately \( 10^6 \).