Question

For the equilibrium
SO3(g)--->SO2(g)+1/2 O2(g)
the molar mass at equilibrium was observed to be
60, then the degree of dissociation of SO3 would be:-

A) 0.33
B) 0.66
C) 0.25
D) 0.50

""Please answer subjectively"" ​

Answer 1

For the equilibrium

SO3(g)--->SO2(g)+1/2 O2(g)

the molar mass at equilibrium was observed to be  60, then the degree of dissociation of SO3 is 0.66 (option - b)

The degree of dissociation of a substance is nothing but the fraction of the molecules (of the same substance) dissociating at a particular time. It is generally denoted as 'α'.

Let the initial moles (i.e. at time = 0) of SO3 be 1 and let 'α' moles of SO3 dissociate at equilibrium. So, for the reaction,

                               SO3(g) ---> SO2(g) + 1/2 O2(g)

Initially(when t=0):     1                 0                0         [Concentrations of SO2 and O2 are NIL as no products are formed as dissociation does not take place initially.]      

At equilibrium:           1-α              α               α/2

∴Total number of moles of mixture at equilibrium = 1-α + α + α/2 = 1 + α/2.

Also, we know that Vapor Density = [(Molar Mass)/2].

For the equilibrium mixture, V.D. 'd' = 60/2 = 30.    [Molar mass of equilibrium mixture = 60 given in question.]

For the reactant, V.D. 'D' = 80/2 = 40.                      [Molar mass of SO3 is 80]

We know that, total no. of moles in a mixture = D/d.

∴ 1 + α/2 = 40/30 = 4/3.

⇒ α/2 = 4/3 - 1 = 1/3.

⇒ α = 2/3 = 0.66

∴ The degree of dissociation of SO3 (α) = 0.66