At a certain temperature the equilibrium constant,K_c,equals 0.11 for the reaction:
2 ICl(g) ⇌ I₂(g) + Cl₂(g)
What is the equilibrium concentration of ICl if 0.45 mol of I₂ and 0.45 mol of Cl₂ are initially mixed in a 2.0 L flask?

Question

Quizplus · Accepted Answer

Let x be the equilibrium concentration of ICl. Then, the equilibrium concentrations of ISB12 and ClSB12 are both 0.45 - x. Using the equilibrium constant expression: K_c=([I₂][Cl₂])/([ICl]2) 0.11 = [(0.45 - x)2] / [x2] 0.11x^2 = 0.45 - x 0.11x^2 + x - 0.45 = 0 Solving for x using the quadratic formula, we get: x ≈ 0.14 or x ≈ -3.94 Since the concentration of ICl cannot be negative, the answer is: ICl = 0.14 mol/L

At a Certain Temperature the Equilibrium Constant,Kc,equals 0

At a Certain Temperature the Equilibrium Constant,K_c,equals 0