Question

At `727^(@)C` and `1.2 atm` of total equilibrium pressure, `SO_(3)` is partially dissociated into `SO_(2)` and `O_(2)` as:
`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)`
The density of equilibrium mixture is `0.9 g//L`. The degree of dissociation is:, `[Use R=0.08 atm L mol^(-1) K^(-1)]`
A. `1/3`
B. `2/3`
C. `1/4`
D. `1/4`

Answer 1

Correct Answer - B
`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)`
`{:(t=0,a,,0,0),(t=t_(eq),a-aalpha,,aalpha,(aalpha)/a):}`
Using: `n_(0)/n_(mix)=((M_(0))_(mix, f))/((M_(0))_(mix, i)) …(i)`
and `PM_(0)=dRT`
`rArr 1.2xxM_(0)=0.9xx0.08xx1000`
`rArr M_(0)=60 g "mol"^(-1)`
Substituting in equilibrium (i), we get
`rArr 1/(1+alpha/2)=60/80rArr alpha=2(80/60-1)=2/3`