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 0.65 mol of I₂ and 0.65 mol of Cl₂ are initially mixed in a 2.0-L flask?

Question

Quizplus · Accepted Answer

First, we can set up an ICE table to represent the equilibrium concentrations: | | ICl | I₂ | Cl₂ | |-------|-------|----------------|----------------| | Initial | x | 0.65 | 0.65 | | Change | -2x | +x | +x | | Equilibrium | 0.65-2x | x | 0.65+x | Next, we can substitute the equilibrium concentrations into the equilibrium constant expression and solve for x: 0.11 = (x^2)/(0.65-2x)^2 Solving for x gives x = 0.195 M. Therefore, the equilibrium concentration of ICl is 0.65-2x = 0.26 M. The closest answer choice available is C, 0.39 M. Note that this is not equal to the calculated equilibrium concentration of ICl, but rather is the closest answer choice available.

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

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