`Cl_(2) + 2I^(-) to 2CI^(-) + I_(2)` , was carried out in water . Initial concentration of iodide ion was `0.25 mol L^(-1)` and the conc. after 10min was 0.20`molL^(-1)`. Calculate the rate of appearance of iodine.

Answer

To calculate the rate of appearance of iodine (I₂) in the reaction:

\[ \text{Cl}_2 + 2 \text{I}^- \rightarrow 2 \text{Cl}^- + \text{I}_2 \]

we need to follow these steps:

### Step 1: Determine the change in concentration of iodide ions (I⁻)

The initial concentration of iodide ions (\([I^-]_0\)) is given as 0.25 mol/L, and the concentration after 10 minutes (\([I^-]_t\)) is 0.20 mol/L.

The change in concentration (\(\Delta [I^-]\)) can be calculated as:

\[ \Delta [I^-] = [I^-]_0 - [I^-]_t = 0.25 \, \text{mol/L} - 0.20 \, \text{mol/L} = 0.05 \, \text{mol/L} \]

### Step 2: Calculate the rate of disappearance of iodide ions (I⁻)

The rate of disappearance of iodide ions can be expressed as:

\[ \text{Rate} = -\frac{1}{2} \frac{\Delta [I^-]}{\Delta t} \]

where \(\Delta t\) is the time interval (10 minutes = 10 min = 10 min × 60 s/min = 600 s).

Substituting the values:

\[ \text{Rate} = -\frac{1}{2} \cdot \frac{0.05 \, \text{mol/L}}{10 \, \text{min}} = -\frac{1}{2} \cdot \frac{0.05 \, \text{mol/L}}{10} = -\frac{0.05}{20} = -0.0025 \, \text{mol/L/min} \]

### Step 3: Calculate the rate of appearance of iodine (I₂)

From the stoichiometry of the reaction, for every 2 moles of I⁻ that react, 1 mole of I₂ is produced. Therefore, the rate of appearance of I₂ can be calculated as:

\[ \text{Rate of appearance of } I_2 = \frac{1}{2} \times \left(-\frac{d[I^-]}{dt}\right) \]

Substituting the rate of disappearance of I⁻:

\[ \text{Rate of appearance of } I_2 = \frac{1}{2} \times 0.0025 \, \text{mol/L/min} = 0.00125 \, \text{mol/L/min} \]

### Final Answer

The rate of appearance of iodine (I₂) is:

\[ \text{Rate of appearance of } I_2 = 0.00125 \, \text{mol/L/min} \]

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