For the reaction, `Cl_(2) + 2I^(-) rarr I_(2) +2CI^(-)`, second order w.r.t `I^(-)` and zero order w.r.t `Cl_(2)` the initial concentration of `I^(-)` was `0.20" mol L"^(-1)`. Then the rate of formation of `I_(2)" in mol L"^(-1)" sec"^(-1)` would be: [Given rate constant for the reaction is `2.5xx10^(-2)" mol L"^(-1)" sec"^(-1)`

Options

A. `1xx10^(-4)`
B. `5xx10^(-4)`
C. `1xx10^(-3)`
D. `5xx10^(-3)`

Explanation

To solve the problem, we need to determine the rate of formation of \( I_2 \) for the given reaction:

\[ Cl_2 + 2I^- \rightarrow I_2 + 2Cl^- \]

### Step-by-Step Solution:

1. **Identify the Order of Reaction**:
- The reaction is second order with respect to \( I^- \) and zero order with respect to \( Cl_2 \).

2. **Write the Rate Law**:
- The rate law can be expressed as:
\[ \text{Rate} = k [Cl_2]^0 [I^-]^2 \]
- Since \( [Cl_2]^0 = 1 \), the rate simplifies to:
\[ \text{Rate} = k [I^-]^2 \]

3. **Substitute Known Values**:
- Given:
- Rate constant, \( k = 2.5 \times 10^{-2} \, \text{mol L}^{-1} \text{sec}^{-1} \)
- Initial concentration of \( I^- = 0.20 \, \text{mol L}^{-1} \)
- Substitute these values into the rate equation:
\[ \text{Rate} = (2.5 \times 10^{-2}) \times (0.20)^2 \]

4. **Calculate \( [I^-]^2 \)**:
- Calculate \( (0.20)^2 \):
\[ (0.20)^2 = 0.04 \]

5. **Calculate the Rate**:
- Now substitute \( [I^-]^2 \) back into the rate equation:
\[ \text{Rate} = (2.5 \times 10^{-2}) \times 0.04 \]
- Perform the multiplication:
\[ \text{Rate} = 2.5 \times 0.04 \times 10^{-2} = 0.1 \times 10^{-2} = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{sec}^{-1} \]

6. **Final Result**:
- The rate of formation of \( I_2 \) is:
\[ \text{Rate} = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{sec}^{-1} \]

### Conclusion:
The rate of formation of \( I_2 \) is \( 1 \times 10^{-3} \, \text{mol L}^{-1} \text{sec}^{-1} \).

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