For the reaction , `Cl_(2)+2I^(-) rarr I_(2)+2Cl^(-)`, the initial concentration of `I^(-)` was `0.20 "mol lit"^(-1)` and the concentration after 20 minutes was `0.18 "mol lit"^(-1)`. Then the rate of formation of `I_(2)` in `"mol lit"^-1min^-1` would be

Options

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

Answer

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

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

### Step 1: Understand the reaction and the concentrations
We are given:
- Initial concentration of \( I^- \) = 0.20 mol/L
- Final concentration of \( I^- \) after 20 minutes = 0.18 mol/L

### Step 2: Calculate the change in concentration of \( I^- \)
The change in concentration of \( I^- \) can be calculated as follows:

\[ \Delta [I^-] = [I^-]_{initial} - [I^-]_{final} = 0.20 \, \text{mol/L} - 0.18 \, \text{mol/L} = 0.02 \, \text{mol/L} \]

### Step 3: Determine the rate of change of \( I^- \)
The rate of change of \( I^- \) over the time period of 20 minutes is given by:

\[ \text{Rate} = -\frac{\Delta [I^-]}{\Delta t} \]

Where \( \Delta t = 20 \, \text{minutes} = 20 \times 60 \, \text{seconds} = 1200 \, \text{seconds} \).

Substituting the values:

\[ \text{Rate} = -\frac{0.02 \, \text{mol/L}}{20 \, \text{minutes}} = -\frac{0.02 \, \text{mol/L}}{20} = -0.001 \, \text{mol/L/min} \]

### Step 4: Relate the rate of \( I^- \) to the rate of formation of \( I_2 \)
From the stoichiometry of the reaction, we know that:

\[ -\frac{1}{2} \frac{d[I^-]}{dt} = \frac{d[I_2]}{dt} \]

Thus, we can express the rate of formation of \( I_2 \):

\[ \frac{d[I_2]}{dt} = -\frac{1}{2} \left(-0.001 \, \text{mol/L/min}\right) = \frac{0.001}{2} \, \text{mol/L/min} = 0.0005 \, \text{mol/L/min} \]

### Step 5: Final answer
The rate of formation of \( I_2 \) is:

\[ \frac{d[I_2]}{dt} = 5 \times 10^{-4} \, \text{mol/L/min} \]