Read the following paragraph and answer the questions given below,
for a general reaction, `aA+bB hArr cC+dD`, equilibrium constant `K_c` is given by `K_c=([C]^c[D]^d)/([A]^a[B]^b)` However, when all reactants and products are gases the equilibrium constant is generally expressed in terms of partial pressure.
`K_p=(P_C^cxx P _D^d)/(P_A^axxP_B^b)`
For a reversible reaction , if concentration of the reactants are doubled, the equilibrium constant will be

Options

A. `1/2` times
B. `1/4` times
C. 2 times
D. remain same

Answer

To solve the question regarding the effect of doubling the concentration of reactants on the equilibrium constant \( K_c \) for a reversible reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of Equilibrium Constant**:
The equilibrium constant \( K_c \) for a general reaction \( aA + bB \rightleftharpoons cC + dD \) is defined as:
\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
This expression shows the ratio of the concentrations of products to the concentrations of reactants at equilibrium.

2. **Recognize the Independence of \( K_c \)**:
The equilibrium constant \( K_c \) is a value that is dependent on the temperature of the system and is independent of the concentrations of the reactants and products. This means that changing the concentrations does not change the value of \( K_c \).

3. **Analyze the Effect of Doubling Concentrations**:
If the concentrations of the reactants \( [A] \) and \( [B] \) are doubled, the new concentrations can be represented as:
\[ [A]' = 2[A], \quad [B]' = 2[B] \]
Substituting these new concentrations into the equilibrium expression gives:
\[ K_c' = \frac{[C]^c[D]^d}{(2[A])^a(2[B])^b} = \frac{[C]^c[D]^d}{2^a[A]^a \cdot 2^b[B]^b} = \frac{[C]^c[D]^d}{2^{a+b}[A]^a[B]^b} \]
This shows that while the concentrations of the reactants have changed, the equilibrium constant \( K_c' \) remains the same as \( K_c \) because the changes in the numerator (products) will also adjust to maintain the equilibrium.

4. **Conclusion**:
Therefore, regardless of the doubling of the reactants' concentrations, the equilibrium constant \( K_c \) remains unchanged. The answer to the question is that the equilibrium constant will remain the same.

### Final Answer:
The equilibrium constant \( K_c \) will remain the same.