Question

1 mole of HI is heated in a closed container of capacity of 2 L. At equilibrium half a mole of HI is dissociated. The equilibrium constant of the reaction is 

• A. 0.25
• B. 1
• C. 0.35
• D. 0.5

Answer 1

The Correct Option is A

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): \[ K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.125 \times 0.125}{(0.25)^2} = \frac{0.015625}{0.0625} = 0.25 \]

Answer 2

The dissociation of HI is given by the balanced equation: \[ 2HI \rightleftharpoons H_2 + I_2 \] The equilibrium constant expression for this reaction is: \[ K_c = \frac{[H_2] \cdot [I_2]}{[HI]^2} \] We are given that at equilibrium, half a mole of HI dissociates in a 2 L container. So, the equilibrium concentrations are as follows: \[ [HI] = \frac{0.5 \, \text{mole}}{2 \, \text{L}} = 0.25 \, \text{M} \] Since half of the HI dissociates, the concentrations of \(H_2\) and \(I_2\) will each be 0.125 M: \[ [H_2] = 0.5 \times 0.25 = 0.125 \, \text{M} \] \[ [I_2] = 0.5 \times 0.25 = 0.125 \, \text{M} \] Substituting these values into the equilibrium constant expression: \[ K_c = \frac{(0.125) \cdot (0.125)}{(0.25)^2} = \frac{0.015625}{0.0625} = 0.25 \] Thus, the equilibrium constant of the reaction is \(K_c = 0.25\). Therefore, the correct_