Law Of Chemical Equilibrium And Equilibrium Constant

Step 1: Le Chatelier's Principle.
According to Le Chatelier's principle, if the concentration of is increased, the equilibrium will shift to the right to produce more , releasing heat since the reaction is exothermic.
Thus, the correct answer is (3).

Step 1: Analyze the equilibrium expression.
The equilibrium constant for the dissociation of iodine is given by: Since is very small, this means that most of the iodine remains in the molecular form, and the concentration of will be greater than that of .
Step 2: Conclusion.
At equilibrium, the concentration of will be less than that of . Final Answer:

Step 1: Use the relation between the equilibrium constant and temperature.
For the given reaction, the equilibrium constant has a temperature dependence that involves , where is the gas constant and is the temperature in Kelvin.
Step 2: Conclusion.
The value of for this reaction is . Final Answer:

Step 1: According to Le-Chatelier’s principle, the system opposes the applied change.
Step 2: Increase in temperature favours the endothermic direction.
Step 3: Increase in pressure favours the side with fewer gaseous moles, hence (C) is incorrect.

Step 1: One mole forms two moles on dissociation.
Step 2: For such equilibria: α ∝ \frac1√(P)

Step 1: Let equilibrium concentration of N₂O₅ = x.
Step 2: Using equilibrium relations and given [O₂]=2.5M, solve for x.
Step 3: x = 1.846M

According to Le-Chatelier’s principle, increase in temperature shifts equilibrium in the direction of the endothermic reaction. Options (B) and (C) are incorrect, hence (D) is also wrong.

Step 1: Let initial moles of PCl₅=1, degree of dissociation =α.
Step 2: Equilibrium moles: PCl₅=1-α, PCl₃=α, Cl₂=α
Step 3: Equilibrium constant: Kp=(α²P)/((1-α)P) ≈ α²P
Step 4: α ∝ \frac1√(P)

We are given the equation for the temperature dependence of the equilibrium constant : This equation is a form of the van't Hoff equation, which relates the change in the equilibrium constant to the change in temperature. The general form of the van't Hoff equation is: However, we can also use the relation: By differentiating with respect to , we obtain: Equating this to the expression from the van't Hoff equation: From this, we can solve for : Now, to find , we use the relation: At equilibrium, , so: Thus, the value of is .

Entropy and Gibbs Free Energy Calculation
The entropy change ( ) is given by: Substituting values: The enthalpy change ( ): The Gibbs free energy equation: At equilibrium, , so:

Let the concentration of formed at equilibrium be M (since 2 moles of are formed for every moles reacted).
Initial concentrations: Change in concentrations at equilibrium: Equilibrium concentrations: The expression for equilibrium constant is: Substitute concentrations: Simplify denominator: So: Multiply both sides by denominator: Take square root of both sides: Calculate , so: Rearranged: Solve this equation approximately: Try : Try : Try : Try : Between and , zero crossing, so approximately . So approximate to 0.5 range, more precise calculation gives . Thus, concentration of at equilibrium is: