Question

Consider reaction

I_{2(g)+Cl2(g)<======>2ICL(g) Kp=81.9 at 25 C}

_{A mixture at 25 C initially contains} Pcl12=0.100 atm P_I2=0.100 P_ICl=0.100

Find the equilibrium partial pressures of I₂,Cl_2,and ICL at this temp.

I want to know how to slove this problem with ICE table and show tips and ways around thanks.

Answer 1

First, write

Kp expression

Kp = Pproducts^p / )(P-reactants^r)

Kp = (ICl)^2/ (I2)(Cl2)

so...

ICE chart is:

(I) initially:

I2 = 0.1

Cl2 = 0.1

ICl = 0.1

(C) the change (stoichiometric)

I2 = -x

Cl2 = -x

ICl = +x

(E) in equilibrium

I2 = 0.1 - x

Cl2 = 0.1 - x

ICl = 0.1 + x

substitute in KP

knowing that Kp = 81.9

81.9 = (0.1+x)^2 /(0.1 - x)(0.1 - x)

solve for x

sqrt(81.9) = sqrt( (0.1+x)^2 /(0.1 - x)(0.1 - x))

9.0498= (0.1+x) /(0.1 - x)

9.0498*0.1 - 9.0498x = 0.1 + x

9.0498*0.1-0.1 = (1+9.0498)x

x = (9.0498*0.1-0.1) / (1+9.0498)

x = 0.08

I2 = 0.1 - x = 0.1- 0.08 = 0.02

Cl2 = 0.1 - x = 0.1- 0.08 = 0.02

ICl = 0.1 + x = 0.1+ 0.08 = 0.18

Proof:

Q = (0.18^2) / (0.02*0.02) = 81, which is pretty near to 81.9