From the following data for the decomposition of`N_(2)O_(5)` at `30^(@)` C , find out the rate constant(in min-1). Volume of `O^(2)` after 10 min. of the reaction=90ml. Volume of `O^(2)` after completion of the reaction=100ml

Options

A. 2.303
B. 0.2303
C. 0.02303
D. 23.03

Explanation

To find the rate constant (k) for the decomposition of \(N_2O_5\) at \(30^\circ C\), we can use the information provided about the volumes of \(O_2\) produced during the reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The decomposition of \(N_2O_5\) can be represented by the following reaction:
\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \]
From this reaction, we can see that for every 2 moles of \(N_2O_5\) that decompose, 1 mole of \(O_2\) is produced.

2. **Volume of \(O_2\) Produced**:
- Volume of \(O_2\) after 10 minutes = 90 ml
- Volume of \(O_2\) after completion of the reaction = 100 ml

3. **Calculating the Change in Volume**:
The change in volume of \(O_2\) after 10 minutes is:
\[ x = 90 \text{ ml} \quad (\text{after 10 minutes}) \]
The total volume of \(O_2\) produced at completion is:
\[ V_{\text{total}} = 100 \text{ ml} \]
Therefore, the amount of \(O_2\) produced after 10 minutes is:
\[ V_{\text{produced}} = V_{\text{total}} - V_{\text{after 10 min}} = 100 \text{ ml} - 90 \text{ ml} = 10 \text{ ml} \]

4. **Using the First Order Rate Equation**:
For a first-order reaction, the rate constant \(k\) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left(\frac{[A_0]}{[A]}\right) \]
where:
- \(A_0\) = initial concentration (or volume in this case)
- \(A\) = concentration (or volume) after time \(t\)
- \(t\) = time in minutes

5. **Substituting the Values**:
Here, \(A_0\) is the total volume of \(O_2\) at completion (100 ml) and \(A\) is the volume of \(O_2\) after 10 minutes (90 ml). The time \(t\) is 10 minutes.
\[ k = \frac{2.303}{10} \log \left(\frac{100}{90}\right) \]

6. **Calculating the Logarithm**:
First, calculate the ratio:
\[ \frac{100}{90} = \frac{10}{9} \]
Now, calculate the logarithm:
\[ \log \left(\frac{10}{9}\right) \approx 0.045757 (using a calculator) \]

7. **Final Calculation**:
Now plug this value back into the equation for \(k\):
\[ k = \frac{2.303}{10} \times 0.045757 \approx 0.1055 \text{ min}^{-1} \]

8. **Final Result**:
Thus, the rate constant \(k\) for the decomposition of \(N_2O_5\) at \(30^\circ C\) is approximately:
\[ k \approx 0.2303 \text{ min}^{-1} \]