Consider the reaction
`A(g)+B(g) hArr C(g)+D(g)`
Which occurs in one step. The specific rate constant are `0.25` and `5000` for the forward and reverse reaction, respectively. The equilibrium constant is

Options

A. `2.0xx10^(-4)`
B. `4.0xx10^(2)`
C. `5.0xx10^(-5)`
D. `2.5xx10^(-6)`

Explanation

To find the equilibrium constant (K) for the reaction \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \), we can follow these steps:

### Step 1: Understand the relationship between the equilibrium constant and rate constants
The equilibrium constant \( K \) is defined as the ratio of the rate constant of the forward reaction (\( K_f \)) to the rate constant of the reverse reaction (\( K_r \)):
\[ K = \frac{K_f}{K_r} \]

### Step 2: Identify the given values
From the problem, we have:
- The specific rate constant for the forward reaction \( K_f = 0.25 \)
- The specific rate constant for the reverse reaction \( K_r = 5000 \)

### Step 3: Substitute the values into the equilibrium constant formula
Now, we can substitute the values of \( K_f \) and \( K_r \) into the formula:
\[ K = \frac{0.25}{5000} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ K = \frac{0.25}{5000} = \frac{0.25}{5 \times 10^3} = \frac{25}{50000} = \frac{1}{2000} = 5 \times 10^{-5} \]

### Step 5: Conclusion
Thus, the equilibrium constant \( K \) for the reaction is:
\[ K = 5 \times 10^{-5} \]

### Final Answer
The equilibrium constant is \( 5 \times 10^{-5} \).

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