In a reaction `2X to Y`, the concentration of X decreases from 3.0 mole/litre to 2.0 moles/litre in 5 minutes. The rate of reaction is

Options

A. 0.1 mol `L^(-1) "min"^(-1)`
B. 5 mol `L^(-1) "min"^(-1)`
C. 1 mol `L^(-1) "min"^(-1)`
D. 0.5 mol `L^(-1) "min"^(-1)`

Explanation

To solve the problem, we need to calculate the rate of the reaction based on the change in concentration of reactant X over a given time period. Here’s a step-by-step solution:

### Step 1: Identify the change in concentration of X
The concentration of X decreases from 3.0 moles/litre to 2.0 moles/litre.

\[ \Delta [X] = [X]_{\text{final}} - [X]_{\text{initial}} = 2.0 \, \text{moles/litre} - 3.0 \, \text{moles/litre} = -1.0 \, \text{moles/litre} \]

### Step 2: Identify the change in time
The time period over which this change occurs is given as 5 minutes.

\[ \Delta t = 5 \, \text{minutes} \]

### Step 3: Write the rate of reaction expression
For the reaction \(2X \rightarrow Y\), the rate of reaction can be expressed as:

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

### Step 4: Substitute the values into the rate expression
Substituting the values we found:

\[ \text{Rate} = -\frac{1}{2} \left( \frac{-1.0 \, \text{moles/litre}}{5 \, \text{minutes}} \right) \]

### Step 5: Calculate the rate
Now we calculate the rate:

\[ \text{Rate} = -\frac{1}{2} \left( \frac{-1.0}{5} \right) = -\frac{1}{2} \times -0.2 = 0.1 \, \text{moles/litre/minute} \]

### Final Answer
Thus, the rate of reaction is:

\[ \text{Rate} = 0.1 \, \text{moles/litre/minute} \]

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