1.

A 3 L flask contains 0.5 moles of sulphur dioxide at 127^(@)C temperature. Some amount of oxygen is introduced into the flask along with a catalyst. When the pressure reached 8.2 atm, 0.3 moles of SO_(3) is formed and there is no further change in the pressure of the reaction mixture. Calculate K_(c ) value for the above reaction.

Answer»

Solution :V= 3L
P= 8.2 ATM
R=0.0821 L atm `K^(-1) " mole"^(-1)`
T=127+123=400 K
TOTAL number of moles at equilibrium is
`n=(PV)/(RT)=(8.2xx3)/(0.0821xx400)=0.75`
`2SO_(2)+O_(2) hArr 2SO_(3)`
Number of moles of `SO_(2)` at equilibrium is `0.5-0.3 n_(O_(3))0.3=0.2`
`therefore` Number of moles of `O_(2)` at equilibrium
`n_(O_(2))=n-(n_(SO_(3))+n_(SO_(2)))=0.75-0.5=0.25`
`K_(c )=((0.3)^(2))/((0.2)^(2)xx(0.25))=9`


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