Question

For the reaction A(g) + B(g) ⇌ C(g) + D(g); ΔH = -QKJ
The equilibrium constant cannot be disturbed by

• A. Addition of A
• B. Addition of D
• C. Increasing of pressure
• D. Increasing of temperature

Answer 1

The Correct Option is C

To solve the problem, we need to analyze how each option affects the equilibrium constant ($K$) for the given reaction:

1. Understanding the Reaction and $K$:
The reaction is given as:

$ \text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g); \quad \Delta H = -Q \, \text{kJ} $

The equilibrium constant $K$ is defined as:

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

where $[\text{A}]$, $[\text{B}]$, $[\text{C}]$, and $[\text{D}]$ are the equilibrium concentrations of the respective species.

2. Analyzing Each Option:
We will evaluate how each option affects $K$.

(A) Addition of A:
- Adding more of reactant A shifts the equilibrium to the right (Le Chatelier's principle).
- This changes the equilibrium concentrations of all species but does not change the value of $K$ because $K$ is temperature-dependent and independent of the initial concentrations.
- Effect on $K$: No change

(B) Addition of D:
- Adding more of product D shifts the equilibrium to the left (Le Chatelier's principle).
- This changes the equilibrium concentrations of all species but does not change the value of $K$ because $K$ is temperature-dependent and independent of the initial concentrations.
- Effect on $K$: No change

(C) Increasing of Pressure:
- The reaction involves gases, and increasing the pressure favors the side with fewer moles of gas.
- However, since the number of moles of gas is the same on both sides of the equation (2 moles on the reactant side and 2 moles on the product side), the equilibrium position does not shift.
- Effect on $K$: No change

(D) Increasing of Temperature:
- The reaction is exothermic ($\Delta H = -Q \, \text{kJ}$), meaning heat is released during the forward reaction.
- Increasing the temperature shifts the equilibrium to the left (endothermic direction) to counteract the added heat.
- This changes the equilibrium concentrations of all species and also changes the value of $K$ because $K$ is temperature-dependent.
- Effect on $K$: Change

Final Answer:
The correct answer is $ {\text{(C) Increasing of pressure}} $.

Answer 2

The equilibrium constant of a reaction is dependent on the temperature and the concentrations of reactants and products. Increasing pressure generally affects the position of equilibrium for reactions involving gases, but it does not affect the equilibrium constant itself. Changes in pressure typically shift the equilibrium to favor the side with fewer moles of gas, but the constant remains unchanged. Thus, increasing pressure does not disturb the equilibrium constant.

The correct answer is (C) : Increasing of pressure.