In an experiment the equilibrium constant for the reaction `A+BhArr C+D` is `K_(C)` when the initial concentration of A and B each is 0.1 mole. Under similar conditions in an another experiment if the initial concentration of A and B are taken to be 2 and 3 moles respectively then the value of equilibrium constant will be

Options

A. `(K_(C))/(2)`
B. `K_(C)`
C. `K_(C)^(2)`
D. `(1)/(K_(C))`

Answer

To solve the problem, we need to understand the concept of the equilibrium constant (Kc) and how it relates to the concentrations of reactants and products at equilibrium.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ A + B \rightleftharpoons C + D \]

2. **Initial Conditions for First Experiment**:
In the first experiment, the initial concentrations of A and B are both 0.1 moles. Therefore:
- \([A]_0 = 0.1\) M
- \([B]_0 = 0.1\) M
- \([C]_0 = 0\) M
- \([D]_0 = 0\) M

3. **Change in Concentrations**:
Let \(x\) be the amount of A and B that reacts at equilibrium. Then at equilibrium:
- \([A] = 0.1 - x\)
- \([B] = 0.1 - x\)
- \([C] = x\)
- \([D] = x\)

4. **Expression for Kc**:
The equilibrium constant \(K_c\) is given by the formula:
\[ K_c = \frac{[C][D]}{[A][B]} = \frac{x^2}{(0.1 - x)(0.1 - x)} \]

5. **Initial Conditions for Second Experiment**:
In the second experiment, the initial concentrations of A and B are:
- \([A]_0 = 2\) M
- \([B]_0 = 3\) M
- \([C]_0 = 0\) M
- \([D]_0 = 0\) M

6. **Change in Concentrations for Second Experiment**:
Let \(x'\) be the amount of A and B that reacts at equilibrium in the second experiment. Then at equilibrium:
- \([A] = 2 - x'\)
- \([B] = 3 - x'\)
- \([C] = x'\)
- \([D] = x'\)

7. **Expression for Kc in Second Experiment**:
The equilibrium constant \(K_{c}'\) for the second experiment is given by:
\[ K_{c}' = \frac{[C][D]}{[A][B]} = \frac{(x')^2}{(2 - x')(3 - x')} \]

8. **Conclusion About Kc**:
The key point is that the equilibrium constant \(K_c\) depends only on the temperature and not on the initial concentrations of the reactants. Since the problem states that both experiments are conducted under similar conditions (implying the same temperature), we conclude that:
\[ K_{c}' = K_c \]

### Final Answer:
The value of the equilibrium constant in the second experiment will be the same as in the first experiment:
\[ K_{c}' = K_c \]