The composition of the equilibrium mixture for the equilibrium `Cl_(2)hArr2Cl` at `1470^(@)K`, may be determined by the rate of diffusion of mixture through a pin hole. It is found that at `1470^(@)K`, the mixture diffuses 1.16 times as fast as krypton (83.8) diffuses under the same conditions. Calculate the % degree of dissociation of Cl2 at equilibrium.

Options

A. 0.14
B. 0.41
C. 0.91
D. 0.24

Answer

To solve the problem, we need to calculate the percentage degree of dissociation of Cl2 at equilibrium based on the diffusion rate of the equilibrium mixture compared to krypton. Here’s a step-by-step solution:

### Step 1: Write the equilibrium reaction
The equilibrium reaction is given as:
\[ \text{Cl}_2 \rightleftharpoons 2\text{Cl} \]

### Step 2: Define initial and equilibrium conditions
Assume we start with 1 mole of Cl2 and no Cl at time \( t = 0 \):
- Initial moles of Cl2 = 1
- Initial moles of Cl = 0

At equilibrium, let \( \alpha \) be the degree of dissociation of Cl2. Therefore, at equilibrium:
- Moles of Cl2 = \( 1 - \alpha \)
- Moles of Cl = \( 2\alpha \)

### Step 3: Calculate total moles at equilibrium
The total moles at equilibrium can be expressed as:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 4: Calculate the mass of the mixture
The mass of the mixture can be calculated using the molecular mass of Cl2 (approximately 70.9 g/mol) and the total moles at equilibrium:
\[ \text{Mass of mixture} = \frac{1 \times 70.9}{1 + \alpha} \]

### Step 5: Use the rate of diffusion
According to Graham's law of effusion, the rate of diffusion is inversely proportional to the square root of the molar mass. Given that the mixture diffuses 1.16 times as fast as krypton (molar mass = 83.8 g/mol), we can set up the following equation:
\[ \frac{\text{Rate of diffusion of mixture}}{\text{Rate of diffusion of Kr}} = \frac{M_{\text{Kr}}}{M_{\text{mixture}}} \]
Substituting the known values:
\[ 1.16 = \frac{83.8}{\frac{70.9}{1 + \alpha}} \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ 1.16 \cdot \frac{70.9}{1 + \alpha} = 83.8 \]
\[ 70.9 \cdot 1.16 = 83.8 \cdot (1 + \alpha) \]
\[ 82.984 = 83.8 + 83.8\alpha \]

### Step 7: Solve for \( \alpha \)
Now, isolate \( \alpha \):
\[ 83.8\alpha = 82.984 - 83.8 \]
\[ 83.8\alpha = -0.816 \]
\[ \alpha = \frac{-0.816}{83.8} \]
Calculating this gives:
\[ \alpha \approx 0.1374 \]

### Step 8: Calculate the percentage degree of dissociation
The percentage degree of dissociation is given by:
\[ \text{Percentage degree of dissociation} = \alpha \times 100 \]
\[ = 0.1374 \times 100 \approx 13.74\% \]

### Final Answer
The percentage degree of dissociation of Cl2 at equilibrium is approximately **13.74%**.

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