Question

At a certain temperature the equilibrium constant,kc, equals 0.11 for the reaction: 2 icl(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?

Answer 1

For equilibrium reactions, it would be best to use the ICE method (Initial-Change-Equilibrium). Then, let x be the number of moles that reactedto form ICl.


2ICl ------> I2 + Cl2

I 0 0.75 0.75
C +2x -x -x
--------------------------------------------
E 2x 0.75 -x 0.75 -x

Kc = [I2][Cl2]/[ICl2]^2 = 0.11, the [I2] represents concentration of I2 in mol/L at equilibrium


`0.11= ( ((0.75-x)/(2) )( (0.75-x)/(2) ))/( ( (2x)/(2) )^(2) )`

x = 0.451 moles

Thus [ICl] = (2x/2) = x = 0.451 mol/L