4 mole of `N_(2)O_(4)` is taken in container of unit volume at any temperature. After some time, equlibrium is attained and vapour density of mixture is `34.5.` The value of `Delta G` will be

Options

A. Zero
B. 9.2 Kcal
C. 50 Kcal
D. Data is insufficient to colculate

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understanding the Reaction
We start with 4 moles of \( N_2O_4 \) in a container. The equilibrium reaction is:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

### Step 2: Setting Up the Initial Conditions
At the beginning (time \( t = 0 \)):
- Moles of \( N_2O_4 = 4 \)
- Moles of \( NO_2 = 0 \)

### Step 3: Establishing the Change at Equilibrium
Let \( x \) be the number of moles of \( N_2O_4 \) that dissociate at equilibrium. Therefore:
- Moles of \( N_2O_4 \) at equilibrium = \( 4 - x \)
- Moles of \( NO_2 \) at equilibrium = \( 2x \)

### Step 4: Total Moles at Equilibrium
The total number of moles at equilibrium will be:
\[ \text{Total moles} = (4 - x) + 2x = 4 + x \]

### Step 5: Using Vapor Density to Find Molecular Weight
The vapor density of the mixture is given as 34.5. The molecular weight (M) of the gas mixture can be calculated using the formula:
\[ M = 2 \times \text{Vapor Density} = 2 \times 34.5 = 69 \text{ g/mol} \]

### Step 6: Relating Molecular Weight to Moles
The molecular weight of the mixture can also be expressed in terms of the moles at equilibrium:
\[ M = \frac{\text{Total mass}}{\text{Total moles}} = \frac{(4 - x) \times M_{N_2O_4} + (2x) \times M_{NO_2}}{4 + x} \]
Where \( M_{N_2O_4} = 92 \text{ g/mol} \) and \( M_{NO_2} = 46 \text{ g/mol} \).

### Step 7: Setting Up the Equation
Substituting the known values:
\[ 69 = \frac{(4 - x) \times 92 + (2x) \times 46}{4 + x} \]

### Step 8: Solving for \( x \)
Now, we can solve this equation for \( x \):
1. Multiply both sides by \( (4 + x) \):
\[ 69(4 + x) = (4 - x) \times 92 + (2x) \times 46 \]
2. Expand and simplify:
\[ 276 + 69x = 368 - 92x + 92x \]
3. Combine like terms:
\[ 276 + 69x = 368 \]
4. Isolate \( x \):
\[ 69x = 368 - 276 \]
\[ 69x = 92 \]
\[ x = \frac{92}{69} \approx 1.33 \]

### Step 9: Finding the Reaction Quotient (Q)
Now, we can find the reaction quotient \( Q \):
\[ Q = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2x)^2}{(4 - x)} = \frac{(2 \times 1.33)^2}{(4 - 1.33)} = \frac{(2.66)^2}{2.67} \approx \frac{7.0756}{2.67} \approx 2.65 \]

### Step 10: Calculating \( \Delta G \)
At equilibrium, \( \Delta G \) is given by:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
However, at equilibrium, \( \Delta G = 0 \):
\[ 0 = \Delta G^\circ + RT \ln Q \]
Thus, we can conclude that:
\[ \Delta G^\circ = -RT \ln Q \]

### Final Answer
Since we are asked for the value of \( \Delta G \) at equilibrium, the answer is:
\[ \Delta G = 0 \]