When `4 mol` of `A` is mixed with `4 mol` of `B, 2 mol` of `C` and `D` are formed at equilibrium, according to the reaction
`A+B hArr C+D`
the equilibrium constant is

Options

A. (a) `sqrt(2)`
B. (b) `2`
C. (c) `1`
D. (d) `4`

Explanation

To find the equilibrium constant \( K \) for the reaction \( A + B \rightleftharpoons C + D \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ A + B \rightleftharpoons C + D \]

### Step 2: Identify the initial moles of reactants
We are given:
- Initial moles of \( A = 4 \, \text{mol} \)
- Initial moles of \( B = 4 \, \text{mol} \)

### Step 3: Identify the moles of products at equilibrium
At equilibrium, we are told that:
- Moles of \( C = 2 \, \text{mol} \)
- Moles of \( D = 2 \, \text{mol} \)

### Step 4: Determine the change in moles of reactants
Since 2 moles of \( C \) and 2 moles of \( D \) are formed, we can determine the change in moles of \( A \) and \( B \):
- Moles of \( A \) remaining = Initial moles of \( A \) - Moles of \( C \) formed
\[ \text{Moles of } A = 4 - 2 = 2 \, \text{mol} \]
- Moles of \( B \) remaining = Initial moles of \( B \) - Moles of \( D \) formed
\[ \text{Moles of } B = 4 - 2 = 2 \, \text{mol} \]

### Step 5: Write the expression for the equilibrium constant \( K \)
The equilibrium constant \( K \) is defined as:
\[ K = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
For our reaction, where \( a = b = c = d = 1 \):
\[ K = \frac{[C][D]}{[A][B]} \]

### Step 6: Substitute the equilibrium concentrations
Assuming the volume of the reaction vessel is \( V \), the concentrations at equilibrium are:
- \([C] = \frac{2}{V}\)
- \([D] = \frac{2}{V}\)
- \([A] = \frac{2}{V}\)
- \([B] = \frac{2}{V}\)

Substituting these values into the expression for \( K \):
\[ K = \frac{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)}{\left(\frac{2}{V}\right) \left(\frac{2}{V}\right)} = \frac{\frac{4}{V^2}}{\frac{4}{V^2}} = 1 \]

### Step 7: Conclusion
Thus, the equilibrium constant \( K \) for the reaction is:
\[ K = 1 \]

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