Law Of Chemical Equilibrium And Equilibrium Constant

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

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



The equilibrium constant is defined as:



where , , , and are the equilibrium concentrations of the respective species.

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

(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 because is temperature-dependent and independent of the initial concentrations.
- Effect on : 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 because is temperature-dependent and independent of the initial concentrations.
- Effect on : 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 : No change

(D) Increasing of Temperature:
- The reaction is exothermic ( ), 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 because is temperature-dependent.
- Effect on : Change

Final Answer:
The correct answer is .

Correct Answer: 0.25

Explanation:
The reaction is: 2HI ⇌ H₂ + I₂

Initially: HI = 1 mole, H₂ = 0, I₂ = 0

At equilibrium (since half a mole of HI is dissociated): → HI dissociates = 0.5 mol → 0.5 mol HI gives 0.25 mol H₂ and 0.25 mol I₂ Remaining HI = 1 − 0.5 = 0.5 mol

Now using equilibrium concentrations (in mol/L): Volume = 2 L [HI] = 0.5 / 2 = 0.25 mol/L [H₂] = 0.25 / 2 = 0.125 mol/L [I₂] = 0.25 / 2 = 0.125 mol/L

Equilibrium constant (K_c):

Equilibrium Constant with Catalyst

Given:

• At 500 K,
• Equilibrium is attained 10 times faster in the presence of a catalyst

Objective: Determine the equilibrium constant in the presence of a catalyst at the same temperature.

The equilibrium constant for the reaction: is given by:

The presence of a catalyst affects the rate of reaction but does not change the equilibrium constant. Therefore, the equilibrium constant remains: