Question

For 2SO2(g)+O2(g)⇌2SO3(g),
Kp=3.0×104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.18 g of SO3and 0.108 g of O2.

How many grams of SO2 are in the vessel?

Express your answer using two significant figures.

Answer 1

consider the given reaction

2S02 + 02 ---> 2S03

now the equilibrium constant is given by

Kp = (pS03)^2 / (p02) (pS02)^2

now

we know that

PV = nRT

also

moles (n) = mass / molar mass

so for S03

P x 2 = ( 1.18 / 80) x 0.0821 x 700

P = 0.424 atm

so

pS03 = 0.424 atm

similarly for 02

P x 2 = ( 0.108 / 32) x 0.0821 x 700

P = 0.097

so

p02 = 0.097 atm

now

Kp = (pS03)^2 / (p02) (pS02)^2

so

3 x 10^4 = (0.424)^2 / ( 0.097) (pS02)^2

pS02 = 7.86 x 10-3

now

apply PV = nRT

so

7.86 x 10-3 x 2 = n x 0.0821 x 700

n = 2.735 x 10-4

now

mass = moles x molar mass

so

mass of S02 = 2.735 x 10-4 x 64

mass of S02 = 0.0175 g

so

0.0175 grams of S02 is present