Question

The equilibrium constant, Kc, for the following

Answer 1

Answer:

2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)

0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x

i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L

Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)

10.5 = x*x/(0.205-2x)^2

=> 10.5(4x^2-0.82x+0.042) = x^2

=>42x^2-8.61x+0.441=x^2

=>41x^2-8.61x+0.441 = 0

This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121

The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.

Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L