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 reaction for the given chemical reaction \(2HI \rightarrow H_2 + I_2\), we need to calculate the average rate of change in concentration of HI over the specified time interval. Here’s how to do it step by step:

### Step 1: Identify the initial and final concentrations of HI
- **Initial concentration of HI** (\([HI]_0\)): 0.5 mol/L
- **Final concentration of HI** (\([HI]_t\)): 0.4 mol/L

### Step 2: Calculate the change in concentration of HI
The change in concentration (\(\Delta [HI]\)) can be calculated as:
\[ \Delta [HI] = [HI]_t - [HI]_0 = 0.4 \, \text{mol/L} - 0.5 \, \text{mol/L} = -0.1 \, \text{mol/L} \]
(Note: The negative sign indicates a decrease in concentration.)

### Step 3: Determine the time interval
The time interval (\(\Delta t\)) is given as:
\[ \Delta t = 10 \, \text{minutes} \]

### Step 4: Calculate the average rate of reaction
The average rate of reaction for a reactant is given by the formula:
\[ \text{Average Rate} = -\frac{1}{\Delta t} \Delta [HI] \]
Substituting the values we have:
\[ \text{Average Rate} = -\frac{1}{10 \, \text{minutes}} \times (-0.1 \, \text{mol/L}) = \frac{0.1 \, \text{mol/L}}{10 \, \text{minutes}} = 0.01 \, \text{mol/L/min} \]

### Step 5: Adjust for stoichiometry
Since the reaction involves 2 moles of HI, we need to divide the rate by 2 to find the rate of reaction:
\[ \text{Rate of Reaction} = \frac{0.01 \, \text{mol/L/min}}{2} = 0.005 \, \text{mol/L/min} = 5 \times 10^{-3} \, \text{mol/L/min} \]

### Final Answer
The rate of reaction during this interval is:
\[ 5 \times 10^{-3} \, \text{mol/L/min} \]