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, we will follow these steps:

### Step 1: Determine the rate constant (k) for the reaction
The half-life (t₁/₂) for a first-order reaction is given as 24 hours. The rate constant (k) can be calculated using the formula:

\[ k = \frac{0.693}{t_{1/2}} \]

Substituting the value of the half-life:

\[ k = \frac{0.693}{24 \text{ hours}} \approx 0.028875 \text{ hr}^{-1} \]

### Step 2: Calculate 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: Determine the remaining mass of \(N_2O_5\) after 4 half-lives
The amount of substance remaining after n half-lives can be calculated using the formula:

\[ \text{Remaining mass} = \text{Initial mass} \times \left( \frac{1}{2} \right)^n \]

Substituting the initial mass (10 g) and the number of half-lives (4):

\[ \text{Remaining mass} = 10 \text{ g} \times \left( \frac{1}{2} \right)^4 \]

Calculating this gives:

\[ \text{Remaining mass} = 10 \text{ g} \times \frac{1}{16} = 0.625 \text{ g} \]

### Conclusion
After 96 hours, the mass of \(N_2O_5\) remaining will be **0.625 grams**.

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