For the elem entary reaction `2A to C` the concentration of A after 30 minutes was found to be 0.01 mole/lit. If the rate constant of the reaction is `2.5 xx 10^(-2)` lit `"mole"^(-1)sec^(-1)` the rate of the reaction at 30 minutes is

Options

A. `2.5 xx 10^(-4)"mole"^(-1)"lit"^(-1)sec^(-1)`
B. `2.5 xx 10^(-6)"mole lit"^(-1)"sec"^(-1)`
C. `2.5 xx 10^(-2) "mole lit"^(-1)"sec"^(-1)`
D. `2.5 xx 10^(-8) "mole"^(-1)"lit"^(1)"sec"^(-1)`

Explanation

To solve the problem step by step, we need to determine the rate of the reaction at 30 minutes given the concentration of A and the rate constant.

### Step 1: Understand the Reaction
The reaction given is:
\[ 2A \rightarrow C \]
This indicates that two moles of A are consumed to produce one mole of C.

### Step 2: Identify Given Data
From the problem, we have:
- Concentration of A after 30 minutes, \([A] = 0.01 \, \text{mol/L}\)
- Rate constant, \(k = 2.5 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1}\)

### Step 3: Write the Rate Law
For an elementary reaction of the form \(2A \rightarrow C\), the rate law can be expressed as:
\[ \text{Rate} = k [A]^n \]
Since the reaction is elementary and the stoichiometry indicates that two moles of A are involved, we have:
\[ n = 2 \]
Thus, the rate law becomes:
\[ \text{Rate} = k [A]^2 \]

### Step 4: Substitute the Values
Now, we can substitute the values of \(k\) and \([A]\) into the rate equation:
\[ \text{Rate} = (2.5 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1}) \times (0.01 \, \text{mol/L})^2 \]

### Step 5: Calculate the Rate
Calculating \((0.01 \, \text{mol/L})^2\):
\[ (0.01)^2 = 0.0001 \, \text{mol}^2/\text{L}^2 \]

Now substituting back into the rate equation:
\[ \text{Rate} = (2.5 \times 10^{-2}) \times (0.0001) \]
\[ \text{Rate} = 2.5 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 6: Final Answer
Thus, the rate of the reaction at 30 minutes is:
\[ \text{Rate} = 2.5 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \]