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 over time, we can follow these steps:

### Step 1: Understand the Concept of Equilibrium
At equilibrium, the rate of the forward reaction (formation of \( B \) from \( A \)) is equal to the rate of the reverse reaction (formation of \( A \) from \( B \)). This means that the concentrations of \( A \) and \( B \) remain constant over time.

**Hint:** Remember that at equilibrium, the rates of the forward and reverse reactions are equal.

### Step 2: Analyze the Rate Expressions
Let’s denote the rate of the forward reaction as \( r_f \) and the rate of the reverse reaction as \( r_b \). At equilibrium, we can express this as:
\[ r_f = k_f [A] \]
\[ r_b = k_b [B] \]
Where \( k_f \) and \( k_b \) are the rate constants for the forward and reverse reactions, respectively.

**Hint:** Write down the rate expressions for both the forward and reverse reactions.

### Step 3: Set Up the Equilibrium Condition
At equilibrium, we have:
\[ k_f [A] = k_b [B] \]
This equation indicates that the concentrations of \( A \) and \( B \) are related through their respective rate constants.

**Hint:** Use the equality of the rates to set up an equation involving concentrations.

### Step 4: Determine the Changes Over Time
Initially, as the reaction proceeds, the concentration of \( A \) will decrease while the concentration of \( B \) will increase. Over time, the rates of the reactions will change until they reach equilibrium.

**Hint:** Consider how the concentrations of \( A \) and \( B \) change over time until they stabilize.

### Step 5: Conclusion
At equilibrium, since the rates are equal, we can conclude that:
\[ \text{At equilibrium, } r_f = r_b \]
This means that the system has reached a state where the concentrations of \( A \) and \( B \) do not change, and the forward and reverse reactions continue to occur at the same rate.

**Hint:** Remember that equilibrium is a dynamic state where reactions continue to occur, but there is no net change in concentrations.

### Final Answer
The correct option regarding the equilibrium of the reaction \( A \rightleftharpoons B \) is that the rate of the forward reaction equals the rate of the reverse reaction at equilibrium.

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