Question

For a chemical reaction: Cl2 (g) + O2 (g) ⇄2 ClO (g) has a KP = 0.155, and the initial concentrations are: 0.55 atm Cl2 0.55 atm O2 0.45 atm ClO Solve for the final concentration of Cl2 in atm:

A. 0.098 B. 0.12 C. 0.42 D. 0.65 E. 0.67

Answer 1

Kp = 0.155

Kp = ClO^2 / (Cl2*O2)

intially

CL2 = 0.55

O2 = 0.55

ClO = 0.45

in equilbriuim

CL2 = 0.55 -x

O2 = 0.55- x

ClO = 0.45 +2x

then

substitute in

Kp = ClO^2 / (Cl2*O2)

0.155 = ( 0.45 +2x)^2 / (0.55 -x)^2

solve for x

sqrt(0.155 ) = (0.45 +2x) / (0.55 -x)

0.3937*(0.55-x) = 0.45 + 2x

0.3937*0.55 - 0.3937x = 0.45 + 2x

-2.3937x =0.233465

x = 0.233465/(-2.3937) = -0.0975

then

CL2 = 0.55 - -0.0975 = 0.6475

O2 = 0.55- -0.0975 = 0.6475

ClO = 0.45 +2* -0.0975 = 0.255

Prove

Kp = ClO^2 / (Cl2*O2)

Kp = (0.255^2)/(0.6475^2) = 0.155 which is pretty accurate

then

Cl = 0.6475