The rate constant for the reaction in gaseous phase
`O+O_(3) rarr 2O_(2)` is `8.0xx10^(-15)" cm"^(3)" molecule"^(-1) s^(-1)` at 298 K. Corresponding value in `dm^(3)" mol"^(-1) s^(-1)` is :

Options

A. `4.8xx10^(6)" dm"^(3)" mol"^(-1)s^(-1)`
B. `1.33xx10^(-6)" dm"^(3)" mol"^(-1) s^(-1)`
C. `4.8xx10^(8)" dm"^(3)" mol"^(-1) s^(-1)`
D. `1.33xx10^(6)" dm"^(3)" mol"^(-1) s^(-1)`

Explanation

To convert the rate constant from the units of \( \text{cm}^3 \, \text{molecule}^{-1} \, \text{s}^{-1} \) to \( \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \), we will follow these steps:

### Step 1: Understand the conversions needed
We need to convert:
1. From \( \text{cm}^3 \) to \( \text{dm}^3 \)
2. From \( \text{molecule}^{-1} \) to \( \text{mol}^{-1} \)

### Step 2: Convert \( \text{cm}^3 \) to \( \text{dm}^3 \)
1 decimeter (dm) is equal to 10 centimeters (cm), therefore:
\[ 1 \, \text{dm} = 10 \, \text{cm} \implies 1 \, \text{dm}^3 = (10 \, \text{cm})^3 = 1000 \, \text{cm}^3 \]
Thus, to convert \( \text{cm}^3 \) to \( \text{dm}^3 \), we use:
\[ 1 \, \text{cm}^3 = \frac{1}{1000} \, \text{dm}^3 \]

### Step 3: Convert \( \text{molecule}^{-1} \) to \( \text{mol}^{-1} \)
Using Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{molecules/mol} \):
\[ 1 \, \text{molecule} = \frac{1}{N_A} \, \text{mol} \]
Thus, \( 1 \, \text{molecule}^{-1} = N_A \, \text{mol}^{-1} \).

### Step 4: Substitute the conversions into the rate constant
Given the rate constant:
\[ k = 8.0 \times 10^{-15} \, \text{cm}^3 \, \text{molecule}^{-1} \, \text{s}^{-1} \]
We can substitute the conversions:
\[ k = 8.0 \times 10^{-15} \, \text{cm}^3 \, \text{molecule}^{-1} \, \text{s}^{-1} \times \left(\frac{1}{1000} \, \text{dm}^3/\text{cm}^3\right) \times (N_A \, \text{mol}^{-1}) \]
Substituting \( N_A = 6.022 \times 10^{23} \):
\[ k = 8.0 \times 10^{-15} \times \frac{1}{1000} \times (6.022 \times 10^{23}) \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \]

### Step 5: Calculate the final value
\[ k = 8.0 \times 10^{-15} \times 6.022 \times 10^{20} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \]
Calculating this:
\[ k = 4.8176 \times 10^{6} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \approx 4.8 \times 10^{6} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \]

### Final Answer:
The corresponding value in \( \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \) is approximately \( 4.8 \times 10^{6} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \).

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