PLEASE SOMEBODY GIVE ME THE RIGHT TECHNIQUE<> these problems have been plaguing

Question

PLEASE SOMEBODY GIVE ME THE RIGHT TECHNIQUE<> these problems have been plaguing me for days now.... Please somebody help. This is my THIRD time posting.. I need a step by stepp. for the first problem, answers should be [c610I2]= .0775M [C6H10]= .022 M [I2]= .172 M. For the second problem asnwers should be [Cl2]= .200 M [F2] =.563M [ClF]=2.72M questions each require a lot of work.....so I will give YOU ALL POINTS IF U GET IT RIGHT AND SHOW ME HOW TO DO BOTH OF THEM.

For the reaction C6H10 + I2--> C6H10 I2   Kc=20.0 If [C6H10] initial= .100M and [I2] initial= .250 M, what are the equilibrium concentrations of all species?

Kc= 19.9 at 2500 K for Cl2 + F2--> 2 ClF. What direction will the reaction go if [Cl2]= .200 M [F2]= .100 M and [ClF]= 3.65 M? What will be the equilibrium concentrations of all species?

Explanation / Answer

C6H10 + I2 --> C6H10 I2

0.1 0.25 0 initially

0.1-x 0.25-x x at equilibrium

Kc = [C6H10I2]/[I2][C6H10] = 20

20 = x/(0.1-x)(0.25-x) = x/(x^2 + 0.025 - 0.35x)

x = 0.0775

so

[C6H10I2] = x = 0.0775 M

[C6H10] = 0.1-x = 0.0225M

[I2] = 0.25-x = 0.1725M

Cl2 + F2 <-------------> 2ClF

0.2 0.1 3.65 at given condition

0.2-x 0.1-x 3.65+2x at equilibrium

Q = [ClF]^2/[Cl2][F2] at given condition

Q = 3.65^2/0.1*0.2 = 666.125

as Q >Kc the reaction will move in backward direction

Kc = [ClF]^2/[Cl2][F2] at equilibrium

Kc = 19.9 = (3.65+2x)^2/(0.1-x)(0.2-x)

19.9 = (13.3225 + 4x^2 + 14.6x)/(0.02 + x^2 - 0.3x)

0.398 + 19.9x^2 - 5.97x = 13.3225 +4x^2 + 14.6x

15.9x^2 - 20.57x - 12.9245 = 0

x = -0.463

[ClF] = 3.65-2*0.463 = 2.724 M

[F2] = 0.1+0.463 = 0.563 M

[I2] = 0.2+0.463 = 0.663 M

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