COMEDK2024Evening ShiftChemistryChemical EquilibriumActual
At 700 K , the Equilibrium constant value for the formation of HI from H ₂ and I ₂ is 49.0 . 0.7 mole of HI(g) is present at equilibrium. What will be the concentrations of H ₂ and I ₂ gases if we initially started with HI(g) and allowed the reaction to reach equilibrium at the same temperature?
Options
- A0.3442
- B0.4692
- C0.521
- D0.1
Correct answer
D. 0.1
Step-by-step solution
The reaction is: H₂(g) + I₂(g) 2HI(g) K_c = [HI]^2 [H₂][I₂] = 49.0 Since the reaction starts with only HI(g) , it decomposes to give H₂ and I₂ in 1:1 ratio. Let [H₂] = [I₂] = x 49.0 = (0.7)^2 x^2 x^2 = 0.49 49.0 = 0.01 x = 0.1 mol/L