For a reversible reaction, if the concentration of the reactants are doubled, then the equilibrium constant will

Options

A. also be doubled
B. be halved
C. become one fourth
D. remain the same.

Answer

To solve the question regarding the effect of doubling the concentration of reactants on the equilibrium constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
Let's consider a reversible reaction:
\[ A \rightleftharpoons B \]
where \( A \) is the reactant and \( B \) is the product.

2. **Write the Expression for the Equilibrium Constant**:
The equilibrium constant \( K_c \) for this reaction is defined as:
\[ K_c = \frac{[B]}{[A]} \]
where \([B]\) is the concentration of the product and \([A]\) is the concentration of the reactant at equilibrium.

3. **Initial Concentrations**:
Assume the initial concentrations of \( A \) and \( B \) at equilibrium are \([A] = a\) and \([B] = b\). Thus, the equilibrium constant can be expressed as:
\[ K_c = \frac{b}{a} \]

4. **Doubling the Concentration of Reactants**:
If the concentration of the reactants is doubled, the new concentration of \( A \) becomes \( 2a \). The concentration of \( B \) remains unchanged at \( b \) (since we are only changing the reactants).

5. **New Expression for the Equilibrium Constant**:
The new equilibrium constant \( K_c' \) when the concentration of \( A \) is doubled is given by:
\[ K_c' = \frac{[B]}{[A]_{new}} = \frac{b}{2a} \]

6. **Relate the New Constant to the Original**:
We can express \( K_c' \) in terms of \( K_c \):
\[ K_c' = \frac{b}{2a} = \frac{1}{2} \cdot \frac{b}{a} = \frac{1}{2} K_c \]

7. **Conclusion**:
Therefore, when the concentration of the reactants is doubled, the equilibrium constant \( K_c \) becomes half of its original value:
\[ K_c' = \frac{1}{2} K_c \]

### Final Answer:
The equilibrium constant will be halved.

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