On doubling P and V at constant temperature, the equilibrium constant will

Options

A. Remain constant
B. Become double
C. Become one-fourth
D. None of these

Explanation

To solve the question regarding the effect of doubling pressure (P) and volume (V) on the equilibrium constant (K) at constant temperature, we can follow these steps:

### Step 1: Understand the Equilibrium Constant
The equilibrium constant (K) for a reaction at equilibrium is defined in terms of the concentrations (or partial pressures) of the reactants and products. For a general reaction:
\[ aA + bB \rightleftharpoons cC + dD \]
The equilibrium constant expression is given by:
\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
where [A], [B], [C], and [D] are the concentrations of the reactants and products.

### Step 2: Analyze the Effect of Pressure
According to Le Chatelier's principle, changes in pressure do not affect the equilibrium constant (K) as long as the temperature remains constant. This is because K is dependent on the ratio of concentrations of products to reactants, which remains unchanged when pressure is altered without changing temperature.

### Step 3: Analyze the Effect of Volume
When the volume of a system is doubled at constant temperature, the concentrations of all gaseous species are halved. For example, if the initial concentration of a species is [X], after doubling the volume, the new concentration becomes:
\[ [X]_{new} = \frac{[X]}{2} \]
This means that both the concentrations of reactants and products are halved.

### Step 4: Substitute into the Equilibrium Constant Expression
If we substitute the new concentrations into the equilibrium constant expression, we get:
\[ K_{new} = \frac{([C]/2)^c([D]/2)^d}{([A]/2)^a([B]/2)^b} \]
This simplifies to:
\[ K_{new} = \frac{1}{2^{c+d}} \cdot \frac{[C]^c[D]^d}{[A]^a[B]^b} \cdot \frac{1}{2^{a+b}} \]
However, since both the numerator and denominator are halved, the factors of 1/2 cancel out, leading to:
\[ K_{new} = K \]
Thus, the equilibrium constant remains unchanged.

### Step 5: Conclusion
Therefore, when both pressure and volume are doubled at constant temperature, the equilibrium constant (K) remains constant.

### Final Answer
The equilibrium constant will remain constant.

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