The half life for the reaction `2N_2O_5 to 4NO_2 + O_2` in 24 hrs at `30^@C`. Starting with 10 g of `N_2O_5` how many grams of `N_2O_5` will remain after a period of 96 hours?

Options

A. 1.25 g
B. 0.63 g
C. 1.77 g
D. 0.5 g

Answer

To solve the problem of how many grams of \( N_2O_5 \) will remain after 96 hours, given that the half-life of the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \) is 24 hours, we can follow these steps:

### Step 1: Understand the Half-Life Concept
The half-life of a reaction is the time required for half of the reactant to be converted into products. In this case, the half-life of \( N_2O_5 \) is given as 24 hours.

### Step 2: Determine the Number of Half-Lives in 96 Hours
To find out how many half-lives fit into 96 hours, we divide the total time by the half-life:

\[ \text{Number of half-lives} = \frac{96 \text{ hours}}{24 \text{ hours}} = 4 \]

### Step 3: Calculate the Remaining Amount of \( N_2O_5 \)
Starting with 10 grams of \( N_2O_5 \), we can calculate the remaining amount after each half-life:

1. After 1 half-life (24 hours):
\[ \text{Remaining} = \frac{10 \text{ g}}{2} = 5 \text{ g} \]

2. After 2 half-lives (48 hours):
\[ \text{Remaining} = \frac{5 \text{ g}}{2} = 2.5 \text{ g} \]

3. After 3 half-lives (72 hours):
\[ \text{Remaining} = \frac{2.5 \text{ g}}{2} = 1.25 \text{ g} \]

4. After 4 half-lives (96 hours):
\[ \text{Remaining} = \frac{1.25 \text{ g}}{2} = 0.625 \text{ g} \]

### Step 4: Conclusion
After 96 hours, the amount of \( N_2O_5 \) remaining is **0.625 grams**.