The dissociation of nitrogen pentaoxide is a first order reaction. In first `24 min`, `75%` of nitrogen pentaoxide is dissociated. What amount of nitrogen pentaoxide will be left behind after one hour of the start of reaction?

Options

A. Approximately `1%`
B. Approximately `2%`
C. Approximately `3%`
D. None

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the problem
The dissociation of nitrogen pentaoxide (N2O5) is a first-order reaction. We know that in the first 24 minutes, 75% of N2O5 is dissociated. This means that 25% of N2O5 remains.

### Step 2: Calculate the half-life (t1/2)
For a first-order reaction, the time taken for 75% dissociation (T75) is equal to twice the half-life (t1/2). Given that T75 = 24 minutes, we can calculate the half-life:

\[ t_{1/2} = \frac{T_{75}}{2} = \frac{24 \text{ min}}{2} = 12 \text{ min} \]

### Step 3: Determine the number of half-lives in 1 hour
Next, we need to find out how many half-lives fit into 1 hour (60 minutes):

\[ n = \frac{\text{Total time}}{t_{1/2}} = \frac{60 \text{ min}}{12 \text{ min}} = 5 \]

### Step 4: Calculate the remaining concentration after 5 half-lives
For a first-order reaction, the remaining concentration can be calculated using the formula:

\[ \text{Remaining concentration} = \frac{[\text{Initial concentration}]}{2^n} \]

Assuming the initial concentration of N2O5 is 100%:

\[ \text{Remaining concentration} = \frac{100\%}{2^5} = \frac{100\%}{32} \approx 3.125\% \]

### Step 5: Conclusion
After 1 hour (which is 5 half-lives), approximately 3.125% of nitrogen pentaoxide will be left behind.

### Final Answer
Approximately 3% of nitrogen pentaoxide will be left after one hour.

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