In the reaction `A + B hArr C + D,` 4 moles of A
react with 4 moles of B and form 2 moles each of C and D. The value of Ke for the reaction is

Options

A. 2
B. 3
C. 1
D. ` 1.56`

Answer

To find the equilibrium constant \( K_e \) for the reaction \( A + B \rightleftharpoons C + D \) given that 4 moles of A react with 4 moles of B to form 2 moles of C and 2 moles of D, we can follow these steps:

### Step 1: Write the balanced equation
The balanced equation based on the information provided is:
\[ 4A + 4B \rightleftharpoons 2C + 2D \]

### Step 2: Write the expression for the equilibrium constant \( K_e \)
The equilibrium constant \( K_e \) is defined in terms of the concentrations of the products and reactants. For the reaction:
\[ K_e = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]
where \( a, b, c, \) and \( d \) are the coefficients from the balanced equation.

### Step 3: Substitute the coefficients into the \( K_e \) expression
From the balanced equation \( 4A + 4B \rightleftharpoons 2C + 2D \), we can identify:
- Coefficient of \( A \) (reactant) = 4
- Coefficient of \( B \) (reactant) = 4
- Coefficient of \( C \) (product) = 2
- Coefficient of \( D \) (product) = 2

Thus, we can write:
\[ K_e = \frac{[C]^2 [D]^2}{[A]^4 [B]^4} \]

### Step 4: Determine the concentrations at equilibrium
Since we started with 4 moles of A and 4 moles of B and formed 2 moles each of C and D, we can assume the total volume of the reaction mixture is 1 L for simplicity (this can be adjusted based on actual volume if provided). Therefore, the concentrations at equilibrium will be:
- \([A] = \frac{4 - 4}{1} = 0 \, \text{M}\)
- \([B] = \frac{4 - 4}{1} = 0 \, \text{M}\)
- \([C] = \frac{2}{1} = 2 \, \text{M}\)
- \([D] = \frac{2}{1} = 2 \, \text{M}\)

### Step 5: Substitute the equilibrium concentrations into the \( K_e \) expression
Now we can substitute these concentrations into the \( K_e \) expression:
\[ K_e = \frac{(2)^2 (2)^2}{(0)^4 (0)^4} \]

### Step 6: Evaluate \( K_e \)
Since the concentrations of A and B at equilibrium are zero, this indicates that the reaction has gone to completion, and thus, \( K_e \) is considered to be very large (approaching infinity).

### Final Answer:
Thus, the value of \( K_e \) for the reaction is very large, indicating that the products are favored at equilibrium.

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