In the reaction `2N_(2)O_(5) rarr 4NO_(2) + O_(2)`, initial pressure is `500 atm` and rate constant `K` is `3.38 xx 10^(-5) sec^(-1)`. After `10` minutes the final pressure of `N_(2)O_(5)` is

Options

A. 490 atm
B. 350 atm
C. 480 atm
D. 420 atm

Explanation

To solve the problem, we need to find the final pressure of \( N_2O_5 \) after 10 minutes, given the initial pressure and the rate constant for the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
The reaction is:
\[ 2N_2O_5 \rightarrow 4NO_2 + O_2 \]
- Initial pressure \( P_0 = 500 \, \text{atm} \)
- Rate constant \( K = 3.38 \times 10^{-5} \, \text{s}^{-1} \)
- Time \( t = 10 \, \text{minutes} = 10 \times 60 = 600 \, \text{seconds} \)

2. **Determine the Order of the Reaction:**
The reaction is first order with respect to \( N_2O_5 \).

3. **Use the First Order Rate Equation:**
The rate equation for a first-order reaction is given by:
\[ K = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right) \]
where \( P_t \) is the pressure of \( N_2O_5 \) at time \( t \).

4. **Substitute the Known Values:**
Rearranging the equation gives:
\[ \log \left( \frac{P_0}{P_t} \right) = K \cdot t \cdot \frac{2.303}{1} \]
Substitute \( K = 3.38 \times 10^{-5} \, \text{s}^{-1} \) and \( t = 600 \, \text{s} \):
\[ \log \left( \frac{500}{P_t} \right) = 3.38 \times 10^{-5} \times 600 \times 2.303 \]

5. **Calculate the Right Side:**
First, calculate \( 3.38 \times 10^{-5} \times 600 \):
\[ 3.38 \times 10^{-5} \times 600 = 2.028 \times 10^{-2} \]
Now multiply by \( 2.303 \):
\[ 2.028 \times 10^{-2} \times 2.303 \approx 0.0467 \]

6. **Set Up the Logarithmic Equation:**
Now we have:
\[ \log \left( \frac{500}{P_t} \right) = 0.0467 \]

7. **Convert from Logarithmic to Exponential Form:**
To eliminate the logarithm, we convert to exponential form:
\[ \frac{500}{P_t} = 10^{0.0467} \]
Calculate \( 10^{0.0467} \):
\[ 10^{0.0467} \approx 1.110 \]

8. **Solve for \( P_t \):**
Rearranging gives:
\[ P_t = \frac{500}{1.110} \approx 450.45 \, \text{atm} \]

9. **Final Answer:**
The final pressure of \( N_2O_5 \) after 10 minutes is approximately:
\[ P_t \approx 450.45 \, \text{atm} \]