For the reaction `N_(2)O_(5)(g)toN_(2)O_(4)(g)+1//2O_(2)(g)`
initial pressure is 114 mm and after 20 seconds the pressure of reaction mixture becomes 133 mm then the average rate of
reaction will be

Options

A. `1.9 atm S^(-1)`
B. `8.75xx10^(-3) atm S^(-1)`
C. `2.5xx10^(-3)atmS^(-1)`
D. `6.65atmS^(-1)`

Explanation

To find the average rate of the reaction given the initial and final pressures, we can follow these steps:

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

### Step 2: Set Up Initial Conditions
The initial pressure of the system is given as:
- \( P_0 = 114 \, \text{mm} \) at \( t = 0 \, \text{s} \)

### Step 3: Set Up Final Conditions
After 20 seconds, the pressure of the reaction mixture is:
- \( P_t = 133 \, \text{mm} \) at \( t = 20 \, \text{s} \)

### Step 4: Relate Initial and Final Pressures
At \( t = 20 \, \text{s} \), the pressure can be expressed in terms of the change in pressure \( x \):
- The pressure of \( N_2O_5 \) decreases by \( x \)
- The pressure of \( N_2O_4 \) increases by \( x \)
- The pressure of \( O_2 \) increases by \( \frac{x}{2} \)

Thus, the equation becomes:
\[ (114 - x) + x + \frac{x}{2} = 133 \]

### Step 5: Simplify the Equation
Combine like terms:
\[ 114 - x + x + \frac{x}{2} = 133 \]
This simplifies to:
\[ 114 + \frac{x}{2} = 133 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ \frac{x}{2} = 133 - 114 \]
\[ \frac{x}{2} = 19 \]
Multiplying both sides by 2:
\[ x = 38 \, \text{mm} \]

### Step 7: Calculate the Average Rate of Reaction
The average rate of reaction is defined as the change in concentration (or pressure) over time. Here, we have:
\[ \text{Average Rate} = \frac{x}{t} \]
Where \( t = 20 \, \text{s} \):
\[ \text{Average Rate} = \frac{38 \, \text{mm}}{20 \, \text{s}} = 1.9 \, \text{mm/s} \]

### Step 8: Convert to Atmospheres
To convert mm to atmospheres:
\[ 1 \, \text{atm} = 760 \, \text{mm} \]
Thus:
\[ \text{Average Rate in atm/s} = \frac{1.9 \, \text{mm/s}}{760 \, \text{mm/atm}} = 2.5 \times 10^{-3} \, \text{atm/s} \]

### Final Answer
The average rate of the reaction is:
\[ 2.5 \times 10^{-3} \, \text{atm/s} \]

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