The specific rate of reaction is `1.5 xx 10^(-4)` lit `"mole"^(-1)."sec"^(-1)`. If the reaction is connected with 0.2 mole/lit of of the reactant, the initial rate of the reaction in mole `"lit"^(-1) sec^(-1)` is

Options

A. `1.5 xx 10^(-4)`
B. `3 xx 10^(-5)`
C. `6 xx 10^(-6)`
D. `6 xx 10^(-5)`

Explanation

To solve the problem, we need to calculate the initial rate of the reaction given the specific rate constant and the concentration of the reactant. Here's the step-by-step solution:

### Step 1: Identify the given values
- Specific rate constant (k) = \(1.5 \times 10^{-4} \, \text{lit mole}^{-1} \text{sec}^{-1}\)
- Concentration of the reactant ([A]) = \(0.2 \, \text{mole/lit}\)

### Step 2: Determine the order of the reaction
From the video transcript, we can infer that the order of the reaction (n) is 2, as derived from the units of the rate constant. The relationship between the rate constant and the order of the reaction is given by:
\[ \text{Rate} = k \times [A]^n \]
Where:
- k is the specific rate constant
- [A] is the concentration of the reactant
- n is the order of the reaction

### Step 3: Write the rate equation
Since the order of the reaction is 2, we can write the rate equation as:
\[ \text{Rate} = k \times [A]^2 \]

### Step 4: Substitute the values into the rate equation
Now we can substitute the values of k and [A] into the rate equation:
\[ \text{Rate} = (1.5 \times 10^{-4}) \times (0.2)^2 \]

### Step 5: Calculate [A]^2
Calculate \(0.2^2\):
\[ 0.2^2 = 0.04 \]

### Step 6: Substitute back into the rate equation
Now substitute this value back into the rate equation:
\[ \text{Rate} = (1.5 \times 10^{-4}) \times 0.04 \]

### Step 7: Perform the multiplication
Now calculate the rate:
\[ \text{Rate} = 1.5 \times 0.04 \times 10^{-4} = 0.06 \times 10^{-4} \]

### Step 8: Express the final answer in scientific notation
Convert \(0.06 \times 10^{-4}\) to scientific notation:
\[ \text{Rate} = 6 \times 10^{-6} \, \text{mole lit}^{-1} \text{sec}^{-1} \]

### Final Answer
The initial rate of the reaction is:
\[ \text{Rate} = 6 \times 10^{-6} \, \text{mole lit}^{-1} \text{sec}^{-1} \]

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