When 50 mL of 2M solution of `N_2 O_5` was heated, 0.28 L of `O_2` at NTP was formed after 30 minutes. Calculate the concentration of unreacted `N_2 O_5` at that time and also find the average rate of reaction.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the moles of \( O_2 \) produced
We know that the volume of \( O_2 \) produced is 0.28 L at NTP (Normal Temperature and Pressure). To find the number of moles of \( O_2 \), we can use the formula:

\[ \text{Number of moles} = \frac{\text{Volume (L)}}{22.4} \]

Substituting the values:

\[ \text{Number of moles of } O_2 = \frac{0.28 \, \text{L}}{22.4 \, \text{L/mol}} = 0.0125 \, \text{mol} \]

### Step 2: Relate moles of \( O_2 \) to moles of \( N_2O_5 \)
The balanced chemical reaction for the decomposition of \( N_2O_5 \) is:

\[ 2 N_2O_5 \rightarrow 2 N_2O_4 + O_2 \]

From the balanced equation, we see that 1 mole of \( O_2 \) is produced from 2 moles of \( N_2O_5 \). Therefore, the moles of \( N_2O_5 \) that decomposed can be calculated as:

\[ \text{Moles of } N_2O_5 \text{ decomposed} = 2 \times \text{Moles of } O_2 = 2 \times 0.0125 \, \text{mol} = 0.025 \, \text{mol} \]

### Step 3: Calculate the initial moles of \( N_2O_5 \)
The concentration of the initial \( N_2O_5 \) solution is given as 2 M, and the volume is 50 mL (or 0.050 L). Therefore, the initial moles of \( N_2O_5 \) can be calculated as:

\[ \text{Initial moles of } N_2O_5 = \text{Concentration} \times \text{Volume} = 2 \, \text{mol/L} \times 0.050 \, \text{L} = 0.1 \, \text{mol} \]

### Step 4: Calculate the moles of unreacted \( N_2O_5 \)
To find the moles of unreacted \( N_2O_5 \):

\[ \text{Moles of unreacted } N_2O_5 = \text{Initial moles} - \text{Moles decomposed} = 0.1 \, \text{mol} - 0.025 \, \text{mol} = 0.075 \, \text{mol} \]

### Step 5: Calculate the concentration of unreacted \( N_2O_5 \)
The concentration of unreacted \( N_2O_5 \) can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \]

Substituting the values:

\[ \text{Concentration of unreacted } N_2O_5 = \frac{0.075 \, \text{mol}}{0.050 \, \text{L}} = 1.5 \, \text{M} \]

### Step 6: Calculate the average rate of reaction
The average rate of reaction can be calculated using the formula:

\[ \text{Rate} = -\frac{\Delta [N_2O_5]}{\Delta t} \]

Where \( \Delta [N_2O_5] \) is the change in concentration of \( N_2O_5 \) over the time period \( \Delta t \). The initial concentration was 2 M, and the final concentration is 1.5 M after 30 minutes:

\[ \Delta [N_2O_5] = 2 \, \text{M} - 1.5 \, \text{M} = 0.5 \, \text{M} \]

Now, converting 30 minutes to seconds (30 minutes = 1800 seconds):

\[ \text{Rate} = -\frac{0.5 \, \text{M}}{1800 \, \text{s}} = -\frac{5}{18000} = -\frac{1}{3600} \, \text{M/s} \]

### Final Answers
- Concentration of unreacted \( N_2O_5 \): **1.5 M**
- Average rate of reaction: **\( \frac{1}{3600} \, \text{M/s} \)**