Question

For the system SO3 ( g) ⇔ SO2 (g ) + 1 2 O2 ( g) at 1000 K, K = 0.45. Sulfur trioxide, originally at 1.00 atm pressure, partially dissociates to SO2 and O2 at 1000 K. what is its partial pressure at equilibrium?

Answer 1

we know that

Kp=Kc*(RT)^del n

Kp=0.45X(0.0821X1000)^(3/2-1)

Kp=0.45 X 9.06 = 4.077

Kp=x*x/(1-x)

36.945=x^2/(1-x)

x^2+36.945x-36.945=0

x=-36.945+sqrt(36.945^2+4*36.945) /2

x=0.9743

so partial pressue of SO3 at equilibrium=1-x=1-0.9743=0.0257atm

for the given system

                                2SO3 ( g)         ⇔           2SO2 (g )       +         O2

Initial pressure              1 atm                         0                               0

Change in pressure        -2x                          2x                             x

Equilibrium                      1-2x                         2x                            x

K =4.077= [P_SO2] [P_O2]^1/2 / [P_SO3]

0.45 = 2x (x)² / 1-2x

0.45 - 0.9x = 4x³

On solvng for x

x= 0.334

so partial pressure at equilibrium of SO3 = 1-2X0.334 = 0332atm