The equilibrium constant K (in atm) for the reactio is 9 at 7 atm and 300 K, `A_(2(g)) hArr B_(2(g)) + C_(2(g))`. calcuate the average molar mass (in gnm/mol) of an equilibrium mixture.
Given: Molar mass of `A_(2) B_(2) and C_(2)` are 70, 49 and 21 gm/mol fespectively.

Options

A. 50
B. 45
C. 40
D. 37.5

Answer

To solve the problem, we need to calculate the average molar mass of the equilibrium mixture for the reaction:

\[ A_2(g) \rightleftharpoons B_2(g) + C_2(g) \]

### Step 1: Understanding the Equilibrium Constant
The equilibrium constant \( K_p \) is given as 9 at a pressure of 7 atm and a temperature of 300 K. The relation between \( K_p \) and \( K_c \) is given by:

\[ K_p = K_c \cdot (RT)^{\Delta n} \]

Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in moles of gas (products - reactants)

For our reaction:
- Reactants: 1 mole of \( A_2 \)
- Products: 1 mole of \( B_2 \) + 1 mole of \( C_2 \)

Thus, \( \Delta n = 2 - 1 = 1 \).

### Step 2: Calculate \( K_c \)
Using the given \( K_p \):

\[ K_p = K_c \cdot (RT) \]

Substituting the values:

\[ 9 = K_c \cdot (0.0821 \cdot 300) \]

Calculating \( RT \):

\[ RT = 0.0821 \cdot 300 = 24.63 \]

Now, substituting back to find \( K_c \):

\[ K_c = \frac{9}{24.63} \approx 0.3654 \]

### Step 3: Setting Up the Equilibrium Expression
Let \( x \) be the change in moles of \( B_2 \) and \( C_2 \) at equilibrium. Initially, we have:

- Moles of \( A_2 \) = 1
- Moles of \( B_2 \) = 0
- Moles of \( C_2 \) = 0

At equilibrium:

- Moles of \( A_2 \) = \( 1 - x \)
- Moles of \( B_2 \) = \( x \)
- Moles of \( C_2 \) = \( x \)

The equilibrium constant expression is:

\[ K_c = \frac{[B_2]^2[C_2]}{[A_2]} \]

Substituting the equilibrium concentrations:

\[ 0.3654 = \frac{x^2}{1 - x} \]

### Step 4: Solving for \( x \)
Rearranging gives:

\[ 0.3654(1 - x) = x^2 \]

This simplifies to:

\[ x^2 + 0.3654x - 0.3654 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Where \( a = 1, b = 0.3654, c = -0.3654 \):

\[ x = \frac{-0.3654 \pm \sqrt{(0.3654)^2 - 4 \cdot 1 \cdot (-0.3654)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (0.3654)^2 + 4 \cdot 0.3654 \approx 0.1335 + 1.4616 = 1.5951 \]

Calculating \( x \):

\[ x = \frac{-0.3654 \pm \sqrt{1.5951}}{2} \]

Taking the positive root:

\[ x \approx 0.6044 \]

### Step 5: Calculate Moles at Equilibrium
At equilibrium:

- Moles of \( A_2 \) = \( 1 - 0.6044 = 0.3956 \)
- Moles of \( B_2 \) = \( 0.6044 \)
- Moles of \( C_2 \) = \( 0.6044 \)

### Step 6: Calculate Mole Fractions
Total moles at equilibrium:

\[ \text{Total moles} = 0.3956 + 0.6044 + 0.6044 = 1.6044 \]

Calculating mole fractions:

\[ \text{Mole fraction of } A_2 = \frac{0.3956}{1.6044} \approx 0.246 \]
\[ \text{Mole fraction of } B_2 = \frac{0.6044}{1.6044} \approx 0.377 \]
\[ \text{Mole fraction of } C_2 = \frac{0.6044}{1.6044} \approx 0.377 \]

### Step 7: Calculate Average Molar Mass
Using the molar masses:

- Molar mass of \( A_2 = 70 \, \text{g/mol} \)
- Molar mass of \( B_2 = 49 \, \text{g/mol} \)
- Molar mass of \( C_2 = 21 \, \text{g/mol} \)

The average molar mass \( M_{avg} \) is calculated as:

\[ M_{avg} = (70 \cdot 0.246) + (49 \cdot 0.377) + (21 \cdot 0.377) \]

Calculating each term:

\[ M_{avg} = 17.22 + 18.43 + 7.91 \approx 43.56 \, \text{g/mol} \]

### Final Answer
The average molar mass of the equilibrium mixture is approximately **43.56 g/mol**.

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