For the equation
`N_(2)O_(5)(g)=2NO_(2)(g)+(1//2)O_(2)(g)`, calculate the mole fraction of `N_(2)O_(5)(g)` decomposed at a constant volume and temperature, if the initial pressure is `600 mm Hg` and the pressure at any time is `960 mm Hg`. Assume ideal gas behaviour.

Explanation

To solve the problem, we need to calculate the mole fraction of \( N_2O_5 \) that has decomposed during the reaction given the initial and final pressures. Here’s the step-by-step solution:

### Step 1: Write the balanced chemical equation
The balanced equation for the decomposition of \( N_2O_5 \) is:
\[ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \]

### Step 2: Define initial conditions
- Initial pressure of \( N_2O_5 \) = \( 600 \, \text{mm Hg} \)
- Initial pressures of \( NO_2 \) and \( O_2 \) = \( 0 \, \text{mm Hg} \)

### Step 3: Define changes during the reaction
Let \( P \) be the pressure of \( N_2O_5 \) that has decomposed. According to the stoichiometry of the reaction:
- Pressure of \( N_2O_5 \) at time \( t \) = \( 600 - P \)
- Pressure of \( NO_2 \) at time \( t \) = \( 2P \) (since 2 moles of \( NO_2 \) are produced for every mole of \( N_2O_5 \) decomposed)
- Pressure of \( O_2 \) at time \( t \) = \( 0.5P \) (since 0.5 moles of \( O_2 \) are produced for every mole of \( N_2O_5 \) decomposed)

### Step 4: Write the expression for total pressure
The total pressure at any time \( t \) can be expressed as:
\[ P_{\text{total}} = P_{N_2O_5} + P_{NO_2} + P_{O_2} \]
Substituting the expressions we have:
\[ P_{\text{total}} = (600 - P) + 2P + 0.5P \]
This simplifies to:
\[ P_{\text{total}} = 600 + 1.5P \]

### Step 5: Set up the equation with the given total pressure
We know the total pressure at time \( t \) is \( 960 \, \text{mm Hg} \):
\[ 600 + 1.5P = 960 \]

### Step 6: Solve for \( P \)
Rearranging the equation gives:
\[ 1.5P = 960 - 600 \]
\[ 1.5P = 360 \]
\[ P = \frac{360}{1.5} = 240 \, \text{mm Hg} \]

### Step 7: Calculate the mole fraction of \( N_2O_5 \) decomposed
The mole fraction of \( N_2O_5 \) decomposed can be calculated as:
\[ \text{Mole fraction decomposed} = \frac{P}{\text{Initial pressure of } N_2O_5} = \frac{240}{600} \]
\[ \text{Mole fraction decomposed} = 0.4 \]

### Final Answer
The mole fraction of \( N_2O_5 \) decomposed is \( 0.4 \).

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