4 moles of A are mixed with 4 moles of B. At equilibrium for the raction `A+BhArrC+D`, 2 moles of C and D are formed. The equilibrium constant for the reaction will be

Options

A. `1//4`
B. `1//2`
C. 1
D. 4

Answer

To solve the problem step by step, we need to determine the equilibrium constant (K) for the reaction:

\[ A + B \rightleftharpoons C + D \]

**Step 1: Identify the initial moles of reactants and products.**
- We start with 4 moles of A and 4 moles of B.
- Initially, the moles of C and D are 0.

**Step 2: Determine the change in moles at equilibrium.**
- According to the problem, at equilibrium, 2 moles of C and 2 moles of D are formed.
- Therefore, the change in moles for A and B will be -2 moles each (since they are consumed to form C and D).

**Step 3: Write the equilibrium expression.**
- The equilibrium expression for the reaction is given by:

\[ K = \frac{[C][D]}{[A][B]} \]

**Step 4: Calculate the equilibrium moles of each species.**
- At equilibrium:
- Moles of A = Initial moles - Change = 4 - 2 = 2 moles
- Moles of B = Initial moles - Change = 4 - 2 = 2 moles
- Moles of C = 2 moles (given)
- Moles of D = 2 moles (given)

**Step 5: Calculate the concentrations.**
- Assuming the volume of the reaction mixture is V (in liters), the concentrations are:
- \([A] = \frac{2}{V}\)
- \([B] = \frac{2}{V}\)
- \([C] = \frac{2}{V}\)
- \([D] = \frac{2}{V}\)

**Step 6: Substitute the concentrations into the equilibrium expression.**
- Substituting the values into the equilibrium expression:

\[ K = \frac{\left(\frac{2}{V}\right)\left(\frac{2}{V}\right)}{\left(\frac{2}{V}\right)\left(\frac{2}{V}\right)} = \frac{\frac{4}{V^2}}{\frac{4}{V^2}} = 1 \]

**Step 7: Conclusion.**
- The equilibrium constant \( K \) for the reaction is 1.