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. `250 atm`
C. `480 atm`
D. `420 atm`

Explanation

To solve the problem step by step, we will use the first-order reaction kinetics formula. The reaction given is:

\[ 2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \]

### Step 1: Identify the given data
- Initial pressure (\(P_0\)) = 500 atm
- Rate constant (\(K\)) = \(3.38 \times 10^{-5} \, \text{sec}^{-1}\)
- Time (\(t\)) = 10 minutes = \(10 \times 60 = 600 \, \text{seconds}\)

### Step 2: Use the first-order kinetics formula
For a first-order reaction, the relationship between the initial pressure, final pressure, rate constant, and time is given by:

\[ K = \frac{2.303}{t} \log\left(\frac{P_0}{P_t}\right) \]

Where:
- \(P_t\) is the final pressure after time \(t\).

### Step 3: Rearranging the formula
We can rearrange the formula to solve for \(P_t\):

\[ \log\left(\frac{P_0}{P_t}\right) = \frac{K \cdot t}{2.303} \]

Thus,

\[ \frac{P_0}{P_t} = 10^{\left(\frac{K \cdot t}{2.303}\right)} \]

And therefore,

\[ P_t = \frac{P_0}{10^{\left(\frac{K \cdot t}{2.303}\right)}} \]

### Step 4: Substitute the values into the equation
Substituting the values we have:

\[ P_t = \frac{500}{10^{\left(\frac{3.38 \times 10^{-5} \cdot 600}{2.303}\right)}} \]

### Step 5: Calculate the exponent
First, calculate the exponent:

\[ K \cdot t = 3.38 \times 10^{-5} \cdot 600 = 0.02028 \]

Now calculate:

\[ \frac{0.02028}{2.303} \approx 0.00881 \]

### Step 6: Calculate \(10^{\text{exponent}}\)
Now calculate \(10^{0.00881}\):

\[ 10^{0.00881} \approx 1.020 \]

### Step 7: Calculate \(P_t\)
Now substitute back to find \(P_t\):

\[ P_t = \frac{500}{1.020} \approx 490.2 \, \text{atm} \]

### Final Answer
The final pressure of \(N_{2}O_{5}\) after 10 minutes is approximately:

\[ P_t \approx 490 \, \text{atm} \]