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
                              Â
2SO3 ( g)Â Â Â Â Â Â Â Â
⇔          2SO2
(g )Â Â Â Â Â Â
+Â Â Â Â Â Â Â Â O2
Initial
pressure              1
atm                       Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
Change in pressure      Â
-2x                        Â
2x                           Â
x
Equilibrium                    Â
1-2x                       Â
2x                          Â
x
K =4.077= [PSO2] [PO2]1/2 /
[PSO3]
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
