For the equilibrium `AhArrB` , the variation of the rate of the forward (a) and reverse (b) reaction with time is given by :

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the equilibrium of the reaction \( A \rightleftharpoons B \) and the variation of the rates of the forward and reverse reactions with time, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction Dynamics**:
- The reaction \( A \rightleftharpoons B \) involves the conversion of reactant \( A \) into product \( B \) and vice versa.
- Initially, the concentration of reactants \( [A] \) is higher than that of products \( [B] \).

**Hint**: Consider how the concentrations of reactants and products change over time.

2. **Rate of Forward and Reverse Reactions**:
- The rate of the forward reaction is proportional to the concentration of reactants:
\[ \text{Rate}_{\text{forward}} = k_f [A] \]
- The rate of the reverse reaction is proportional to the concentration of products:
\[ \text{Rate}_{\text{reverse}} = k_r [B] \]

**Hint**: Remember that the rates depend on the concentrations of the respective species involved in the reaction.

3. **Initial Conditions**:
- At the beginning of the reaction, since \( [A] \) is greater than \( [B] \), the rate of the forward reaction will be greater than that of the reverse reaction.
- This means that the forward reaction will dominate initially, leading to a decrease in \( [A] \) and an increase in \( [B] \).

**Hint**: Think about how the reaction progresses and how the concentrations change over time.

4. **Approaching Equilibrium**:
- As the reaction proceeds, the concentration of \( A \) decreases while the concentration of \( B \) increases.
- Eventually, the concentrations will reach a point where the rate of the forward reaction equals the rate of the reverse reaction.

**Hint**: Consider what happens to the rates as the concentrations approach a balance.

5. **Equilibrium Condition**:
- At equilibrium, the rates of the forward and reverse reactions are equal:
\[ k_f [A] = k_r [B] \]
- This indicates that the system has reached a state where the concentrations of \( A \) and \( B \) remain constant over time.

**Hint**: Recall the definition of chemical equilibrium and how it is characterized by constant concentrations.

6. **Conclusion**:
- The correct interpretation of the variation of rates with time indicates that the forward reaction rate decreases while the reverse reaction rate increases until they become equal at equilibrium.

### Final Answer:
The best option that describes the variation of the rates of the forward and reverse reactions with time is that initially, the forward reaction rate is greater than the reverse reaction rate, and at equilibrium, both rates are equal.