In a reaction `2HI to H_(2) + I_(2)` the concentration of HI decreases from 0.5 mol `L^(-1)` to 0.4 mol `L^(-1)` in 10 minutes. What is the rate of reaction during this interval?

Options

A. `5 xx 10^(-3) M "min"^(-1)`
B. `2.5 xx 1-^(-3) M "min"^(-1)`
C. `5 xx 10^(-1) M "min"^(-1)`
D. `2.5 xx 10^(-2) M "min"^(-1)`

Explanation

To find the rate of the reaction for the given chemical equation \( 2HI \rightarrow H_2 + I_2 \), we can follow these steps:

### Step 1: Identify the change in concentration of the reactant
The concentration of HI decreases from 0.5 mol/L to 0.4 mol/L. We can calculate the change in concentration (\(\Delta [HI]\)) as follows:

\[ \Delta [HI] = [HI]_{\text{final}} - [HI]_{\text{initial}} = 0.4 \, \text{mol/L} - 0.5 \, \text{mol/L} = -0.1 \, \text{mol/L} \]

### Step 2: Determine the change in time
The time interval (\(\Delta t\)) during which this change occurs is given as 10 minutes.

### Step 3: Write the rate expression
The rate of reaction can be expressed using the formula:

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

Here, the factor of \(-\frac{1}{2}\) accounts for the stoichiometry of the reaction, as 2 moles of HI produce 1 mole of products.

### Step 4: Substitute the values into the rate expression
Now we can substitute the values we have into the rate expression:

\[ \text{Rate} = -\frac{1}{2} \frac{-0.1 \, \text{mol/L}}{10 \, \text{minutes}} \]

### Step 5: Calculate the rate
Calculating this gives:

\[ \text{Rate} = -\frac{1}{2} \cdot \frac{-0.1}{10} = \frac{0.1}{20} = 0.005 \, \text{mol/L/min} \]

This can also be expressed in scientific notation:

\[ \text{Rate} = 5 \times 10^{-3} \, \text{mol/L/min} \]

### Final Answer
Thus, the rate of the reaction during this interval is:

\[ \text{Rate} = 5 \times 10^{-3} \, \text{mol/L/min} \]

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