At a certain temperature the equilibrium constant, K_c, equals 0.11 for the reaction: 2ICl(g) ⇌ I₂(g) + Cl₂(g)
What is the equilibrium concentration of ICl if 0. 75 mol of I₂ and 0. 75 mol of Cl₂ are initially mixed in a 2.0 L flask?

Question

Quizplus · Accepted Answer

First, we need to set up the equilibrium expression: K_c = [I₂]^[Cl₂] / [ICl]^2 Next, we need to set up an ICE table: Initial: 0.75 mol 0.75 mol 0 mol Change: -2x -x +2x Equilibrium: 0.75-2x 0.75-x 2x Substituting the equilibrium concentrations into the equilibrium expression, we get: 0.11 = [(0.75 - x)(0.75 - x)] / (2x)^2 Solving for x, we get x = 0.225 mol/L, which corresponds to the equilibrium concentration of ICl. Therefore, the answer is C.

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

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