In the equilibrium `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` , the partial pressure of `SO_(2),O_(2)` and `SO_(3)` are 0.662,0.10 and 0.331 atm respectively . What should be the partial pressure of Oxygen so that the equilibrium concentrations of `SO_(3)` are equal ?

Options

A. 0.36
B. 0.036
C. 3.6
D. 36

Answer

b

Explanation

(b) We have `P_(O_(2))=X_(O_(2))xxP_("total")`
`:." " 0.60=(n_(O_(2)))/(n_(O_(2))+n_(SO_(2)))xxP_("total")`
Let no. of moles of `O_(2)` are `x,`
`0.60=(x//32)/(x//32+3x//64)xxP_("total")`
`:. 0.60=0.4xxP_("total")`
`:. " "P_("total")=(0.60)/(0.4)=1.5"atm"`
`:'P_(O_(2))+P_(SO_(2))=1.5"atm"`
`implies P_(SO_(2)) =1.5-P_(O_(2))=1.5-0.60=0.9"atm"`
`:. "Concentration of "SO_(2)=(0.9)/(0.0821xx300)`
`=0.036`