1 mole of `A_((g))` is heated to `200^@ C` in a one litre closed flask, till the following equilibrium is reached. `A_((g)) Leftrightarrow B_(g))`. The rate of forward reaction at equilibrium is 0.02 mol lit`""^(-1) "min"^(-1)`. What is the rate `("mol. Lit"^(-1)" min"^(-1))` of the backward reaction at equilibrium?

Options

A. 0.04
B. 0.01
C. 0.02
D. 1

Answer

To solve the problem, we need to understand the concept of chemical equilibrium and the relationship between the rates of the forward and backward reactions.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ A_{(g)} \leftrightarrow B_{(g)} \]
This indicates that gas A is converting to gas B and vice versa.

2. **Understand the Given Information**:
- We have 1 mole of A in a 1-liter closed flask.
- The temperature is maintained at \(200^\circ C\).
- The rate of the forward reaction (\(R_F\)) at equilibrium is given as \(0.02 \, \text{mol L}^{-1} \text{min}^{-1}\).

3. **Apply the Equilibrium Condition**:
At equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction. This is a fundamental property of chemical equilibrium:
\[ R_F = R_B \]
where \(R_B\) is the rate of the backward reaction.

4. **Substitute the Given Rate**:
Since we know the rate of the forward reaction:
\[ R_F = 0.02 \, \text{mol L}^{-1} \text{min}^{-1} \]
Therefore, we can conclude:
\[ R_B = 0.02 \, \text{mol L}^{-1} \text{min}^{-1} \]

5. **Final Answer**:
The rate of the backward reaction at equilibrium is:
\[ R_B = 0.02 \, \text{mol L}^{-1} \text{min}^{-1} \]

### Summary:
The rate of the backward reaction at equilibrium is \(0.02 \, \text{mol L}^{-1} \text{min}^{-1}\).

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