The following rate data were obtained for the thermal decomposition of `N_(2)O_(5)(g)`
`2N_(2)O_(5)(g)to2N_(2)O_(4)(g)+O_(2)(g)`
`{:("Time (sec)",,,0,,,50),("Total pressure (atm)",,,0.2,,,0.25):}`
Calculate the reaction rate when the total pressure is 0.28 atm

Explanation

To solve the problem of calculating the reaction rate for the thermal decomposition of \( N_2O_5(g) \) when the total pressure is 0.28 atm, we can follow these steps:

### Step 1: Understand the Reaction
The reaction is given as:
\[ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) \]

### Step 2: Analyze Initial and Final Conditions
From the data provided:
- At \( t = 0 \) seconds, the total pressure \( P_{total} = 0.20 \) atm.
- At \( t = 50 \) seconds, the total pressure \( P_{total} = 0.25 \) atm.

### Step 3: Calculate the Change in Pressure
The change in pressure over the 50 seconds can be calculated as:
- Initial pressure of \( N_2O_5 \) at \( t = 0 \): \( P_{N_2O_5} = 0.20 \) atm.
- At \( t = 50 \) seconds, the total pressure is \( 0.25 \) atm.

The increase in pressure is due to the formation of products. The relationship can be expressed as:
\[ P_{total} = P_{N_2O_5} + P_{N_2O_4} + P_{O_2} \]

### Step 4: Determine the Pressure at \( t = 50 \) seconds
From the stoichiometry of the reaction:
- For every 2 moles of \( N_2O_5 \) that decompose, 2 moles of \( N_2O_4 \) and 1 mole of \( O_2 \) are produced.
- Let \( P \) be the pressure of \( N_2O_5 \) that decomposed.

At \( t = 50 \) seconds:
\[ P_{total} = 0.20 + \frac{P}{2} \]
Setting this equal to \( 0.25 \) atm gives:
\[ 0.20 + \frac{P}{2} = 0.25 \]
\[ \frac{P}{2} = 0.05 \]
\[ P = 0.10 \text{ atm} \]

### Step 5: Calculate the Pressure of \( N_2O_5 \) at \( t = 50 \) seconds
The remaining pressure of \( N_2O_5 \) is:
\[ P_{N_2O_5} = 0.20 - P = 0.20 - 0.10 = 0.10 \text{ atm} \]

### Step 6: Calculate the Rate Constant \( K \)
Using the first-order rate equation:
\[ K = \frac{2.303}{t} \log \left( \frac{A}{A - x} \right) \]
Where:
- \( A = 0.20 \) atm (initial pressure of \( N_2O_5 \))
- \( x = P = 0.10 \) atm (pressure of \( N_2O_5 \) decomposed)

Substituting into the equation:
\[ K = \frac{2.303}{50} \log \left( \frac{0.20}{0.10} \right) \]
\[ K = \frac{2.303}{50} \log(2) \]
Calculating gives:
\[ K \approx 0.01386 \text{ s}^{-1} \]

### Step 7: Calculate the Rate at \( P_{total} = 0.28 \) atm
Now, when \( P_{total} = 0.28 \) atm:
\[ 0.28 = 0.20 + \frac{P}{2} \]
Solving for \( P \):
\[ \frac{P}{2} = 0.08 \]
\[ P = 0.16 \text{ atm} \]

### Step 8: Calculate the Remaining Pressure of \( N_2O_5 \)
The pressure of \( N_2O_5 \) at this point:
\[ P_{N_2O_5} = 0.20 - P = 0.20 - 0.16 = 0.04 \text{ atm} \]

### Step 9: Calculate the Reaction Rate
The rate of the reaction is given by:
\[ \text{Rate} = K \times P_{N_2O_5} \]
Substituting the values:
\[ \text{Rate} = 0.01386 \times 0.04 \]
Calculating gives:
\[ \text{Rate} \approx 5.54 \times 10^{-4} \text{ atm/s} \]

### Final Answer
The reaction rate when the total pressure is 0.28 atm is approximately \( 5.54 \times 10^{-4} \text{ atm/s} \).

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