In areversible reaction, study of its mechanism says that both the forward and reverse reaction follows first-order kinetics. If the halflife of forward reaction `(t_(1//2))_(f)` is `400 s` and that of reverse reaction `(t_(1//2))_(b)` is `250 s`, the equilibrium of the reaction is

Options

A. `1.6`
B. `0.433`
C. `0.625`
D. `1.109`

Explanation

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reversible reaction given the half-lives of the forward and reverse reactions. Here’s a step-by-step solution:

### Step 1: Understand the relationship between half-life and rate constant for first-order reactions
For a first-order reaction, the relationship between the half-life (\( t_{1/2} \)) and the rate constant (\( k \)) is given by the formula:
\[ k = \frac{0.693}{t_{1/2}} \]

### Step 2: Calculate the rate constant for the forward reaction
Given the half-life of the forward reaction \( t_{1/2}^f = 400 \, s \):
\[ k_f = \frac{0.693}{400} \approx 0.0017325 \, s^{-1} \]

### Step 3: Calculate the rate constant for the reverse reaction
Given the half-life of the reverse reaction \( t_{1/2}^b = 250 \, s \):
\[ k_b = \frac{0.693}{250} \approx 0.002772 \, s^{-1} \]

### Step 4: Determine the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction can be expressed as the ratio of the rate constants:
\[ K_c = \frac{k_f}{k_b} \]
Substituting the values we calculated:
\[ K_c = \frac{0.0017325}{0.002772} \approx 0.625 \]

### Step 5: Final answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately \( 0.625 \).

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