For the system SO3 ( g) SO2 (g ) 1 2 O2 ( g) at 1000 K, K 0 45 Su...

For the system SO3 ( g) â‡” SO2 (g ) + 1 2 O2 (...

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Chemistry

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

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3.8 Ratings (554 Votes)

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 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 Â 2SO3 ( g)Â Â Â Â Â Â Â Â â‡”Â Â 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 pressureÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 1 atmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 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 pressureÂ Â Â Â Â Â Â -2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2xÂ 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 x

EquilibriumÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1-2xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2xÂ 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 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

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