When `N_(2)O_(5)` is heated at certain temperature, it dissociates as `N_(2)O_(5)(g)hArrN_(2)O_(3)(g)+O_(2)(g),K_(c)=2.5` At the same time `N_(2)O_(3)` also decomposes as :
`N_(2)O_(3)(g)hArrN_(2)O(g)+O_(2)(g).` "If initially" `4.0` moles of `N_(2)O_(5)` "are taken in" `1.0` litre flask and alowed to dissociate. Concentration of `O_(2)` at equilibrium is `2.5` M. "Equilibrium concentratio of " `N_(2)O_(5)` is :

Options

A. `1.0` M
B. `1.5`M
C. `2.166 M`
D. `1.846` M

Answer

To solve the problem, we need to analyze the dissociation of \( N_2O_5 \) and the subsequent decomposition of \( N_2O_3 \) at equilibrium. Let's break it down step by step.

### Step 1: Write the reactions and initial conditions
We have two reactions:
1. \( N_2O_5(g) \rightleftharpoons N_2O_3(g) + O_2(g) \) with \( K_c = 2.5 \)
2. \( N_2O_3(g) \rightleftharpoons N_2O(g) + O_2(g) \)

Initially, we have:
- \( N_2O_5 \): 4.0 moles in a 1.0 L flask, so the initial concentration is \( 4.0 \, M \).
- \( N_2O_3 \) and \( O_2 \): 0 moles initially.

### Step 2: Set up the changes in concentration
Let \( x \) be the amount of \( N_2O_5 \) that dissociates at equilibrium. Therefore, at equilibrium:
- Concentration of \( N_2O_5 = 4.0 - x \)
- Concentration of \( N_2O_3 = x \)
- Concentration of \( O_2 = x \)

### Step 3: Use the equilibrium expression for \( N_2O_5 \)
The equilibrium constant expression for the first reaction is:
\[ K_c = \frac{[N_2O_3][O_2]}{[N_2O_5]} = 2.5 \]
Substituting the equilibrium concentrations:
\[ 2.5 = \frac{x \cdot x}{4.0 - x} = \frac{x^2}{4.0 - x} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:
\[ 2.5(4.0 - x) = x^2 \]
\[ 10 - 2.5x = x^2 \]
Rearranging gives:
\[ x^2 + 2.5x - 10 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2.5, c = -10 \):
\[ x = \frac{-2.5 \pm \sqrt{(2.5)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \]
\[ x = \frac{-2.5 \pm \sqrt{6.25 + 40}}{2} \]
\[ x = \frac{-2.5 \pm \sqrt{46.25}}{2} \]
Calculating \( \sqrt{46.25} \approx 6.8 \):
\[ x = \frac{-2.5 \pm 6.8}{2} \]
Calculating the positive root:
\[ x = \frac{4.3}{2} = 2.15 \]

### Step 6: Calculate the equilibrium concentration of \( N_2O_5 \)
Now, substituting \( x \) back to find the equilibrium concentration of \( N_2O_5 \):
\[ [N_2O_5] = 4.0 - x = 4.0 - 2.15 = 1.85 \, M \]

### Final Answer
The equilibrium concentration of \( N_2O_5 \) is \( 1.85 \, M \).