Question

The reaction, Cl_2(aq)   +   2 Cu⁺_(aq) ⇄    2 Cl^–_(aq)   +   2 Cu²⁺_(aq), has a Kc = 4.80 x 10¹⁸. You mix 200.0 mL of a 0.800 M Cl₂ with 200.0 mL of a 0.800 M Cu⁺. What would be the concentrations of each reactant at equilibrium in the 400.0 mL volume?

Please include an ICE table, as well i'm wondering why it needs to go to completion? And how?

Please and Thank you.

Answer 1

Cl_2(aq)   +   2 Cu⁺_(aq)    2 Cl^–_(aq)   +   2 Cu²⁺(aq)

kc=[Cl^-]^2 * [Cu2+]^2/[Cl2] * [Cu+]^2

given,conc, reactant [Cl2]=0.8 mol/L * 200 * 10^-3 L= 0.16 moles/400 * 10^-3 L=0.16/0.4=0.4M

                                 [Cu+] =0.8 mol/L * 200 * 10^-3 L= 0.16 moles/400 * 10^-3 L=0.4M

                     ,   [Cl2] [Cu+]      [Cl^-]                          [Cu2+]

Initial conc         0.4                              0.4                    0                              0

change               -x                                     -2x                 +2x                             +2x

At equilibrium,    0.4 M-x                       0.4 M-2x                 2x                                2x

So from kc=(2x)^2 * (2x)^2/ ( 0.4M-x) * ( 0.4 M-2x)^2

kc=(2x)^4/ ( 0.4M-x) * ( 0.4 M-2x)^2

4.8 * 10^-18=16 x^4/ ( 0.4-x) * ( 0.4 -2x)^2

4.8 * 10^18=16 x^4/ ( 0.4-x) * ( 0.4 -2x)^2

4.8 * 10^18=16 x^4/ ( 0.4-x) * ( 0.4 -2x)^2

4.8 * 10^-18=16 x^4/(0.064 + 3.2x2-0.8x-4x3)

Ignoring the higher terms in the denominator as x is negligible

4.8 * 10^-18=16 x^4/(0.064)

x=1.17 * 10^ -5M

[Cl2]=0.4 -1.17 * 10^ -5M=0.399M

[Cu+] =0.4-2*1.17 * 10^ -5M=0.399M