In a general reaction `A+B hArr AB`, which value of equilibrium constant most favours the production of AB?

Options

A. `9.0xx10^(-3)`
B. `3.0xx10^(-3)`
C. `9.0xx10^(-7)`
D. `9.0xx10^(-12)`

Explanation

To determine which value of the equilibrium constant most favors the production of AB in the reaction \( A + B \rightleftharpoons AB \), we can follow these steps:

### Step 1: Understand the Equilibrium Constant
The equilibrium constant \( K \) for the reaction \( A + B \rightleftharpoons AB \) is defined as:
\[ K = \frac{[AB]}{[A][B]} \]
This equation indicates that \( K \) is a ratio of the concentration of products to the concentration of reactants at equilibrium.

### Step 2: Analyze the Values of \( K \)
We are given four options for the equilibrium constant:
1. \( 9 \times 10^{-3} \)
2. \( 3 \times 10^{-3} \)
3. \( 9 \times 10^{-7} \)
4. \( 9 \times 10^{-12} \)

### Step 3: Determine Favorability for Product Formation
A larger value of \( K \) indicates that at equilibrium, the concentration of products (in this case, \( AB \)) is greater relative to the concentration of reactants (\( A \) and \( B \)). Therefore, to favor the production of \( AB \), we need to choose the largest value of \( K \) from the options provided.

### Step 4: Compare the Values
- \( 9 \times 10^{-3} \) is greater than \( 3 \times 10^{-3} \), \( 9 \times 10^{-7} \), and \( 9 \times 10^{-12} \).
- Thus, \( 9 \times 10^{-3} \) is the largest value among the options.

### Step 5: Conclusion
The value of the equilibrium constant that most favors the production of \( AB \) is:
\[ \text{Option 1: } 9 \times 10^{-3} \]