Which of the following relation is correct for `k_(f)` and `k_(b)` in an equibilirium process that contains equal moles of rectants and products.

Options

A. `k_(f)=k_(b)`
B. `k_(f)gtk_(b)`
C. `k_(f)ltk_(b)`
D. we cannot predict

Answer

To solve the problem regarding the relationship between the forward rate constant \( k_f \) and the backward rate constant \( k_b \) in an equilibrium process with equal moles of reactants and products, we can follow these steps:

### Step 1: Understand the Equilibrium Process
In a chemical reaction, we can represent the forward reaction as:
\[ A \rightleftharpoons B \]
where \( A \) is the reactant and \( B \) is the product.

### Step 2: Define the Rate Constants
The forward reaction rate constant is denoted as \( k_f \), and the backward reaction rate constant is denoted as \( k_b \).

### Step 3: Write the Rate of Reaction
The rate of the forward reaction can be expressed as:
\[ \text{Rate}_{\text{forward}} = k_f [A] \]
And the rate of the backward reaction can be expressed as:
\[ \text{Rate}_{\text{backward}} = k_b [B] \]

### Step 4: Establish Equilibrium Condition
At equilibrium, the rate of the forward reaction equals the rate of the backward reaction:
\[ k_f [A] = k_b [B] \]

### Step 5: Consider Equal Moles of Reactants and Products
Since the problem states that there are equal moles of reactants and products, we can assume:
\[ [A] = [B] \]

### Step 6: Substitute into the Equilibrium Expression
Substituting \( [A] \) for \( [B] \) in the equilibrium expression gives:
\[ k_f [A] = k_b [A] \]

### Step 7: Simplify the Equation
Since \( [A] \) is not zero, we can divide both sides of the equation by \( [A] \):
\[ k_f = k_b \]

### Conclusion
The correct relationship between the forward and backward rate constants \( k_f \) and \( k_b \) in this equilibrium process is:
\[ k_f = k_b \]

### Final Answer
Thus, the correct option is that \( k_f \) is equal to \( k_b \).

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