For the reaction, `A(g)+2B(g) hArr 2C(g)`, the rate constant for forward and the reverse reactions are `1xx10^(-4)` and `2.5xx10^(-2)` respectively. The value of equilibrium constant, K for the reaction would be

Options

A. `2xx10^(-4)`
B. `3xx10^(-2)`
C. `4xx10^(-3)`
D. `3xx10^(2)`

Explanation

To find the equilibrium constant \( K_c \) for the reaction

\[ A(g) + 2B(g) \rightleftharpoons 2C(g) \]

given the rate constants for the forward and reverse reactions, we can use the relationship between the equilibrium constant and the rate constants.

### Step 1: Identify the given values
- Rate constant for the forward reaction, \( K_f = 1 \times 10^{-4} \)
- Rate constant for the reverse reaction, \( K_b = 2.5 \times 10^{-2} \)

### Step 2: Use the formula for the equilibrium constant
The equilibrium constant \( K_c \) is related to the rate constants by the formula:

\[ K_c = \frac{K_f}{K_b} \]

### Step 3: Substitute the values into the formula
Substituting the given values into the formula:

\[ K_c = \frac{1 \times 10^{-4}}{2.5 \times 10^{-2}} \]

### Step 4: Perform the calculation
To perform the division:

1. Divide the coefficients:
\[ \frac{1}{2.5} = 0.4 \]

2. Subtract the exponents (since we are dividing powers of ten):
\[ 10^{-4} / 10^{-2} = 10^{-4 + 2} = 10^{-2} \]

Combining these results gives:

\[ K_c = 0.4 \times 10^{-2} \]

### Step 5: Convert to scientific notation
To express \( 0.4 \times 10^{-2} \) in proper scientific notation:

\[ K_c = 4 \times 10^{-3} \]

### Final Answer
Thus, the value of the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 4 \times 10^{-3} \]

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