1.

For the equilibrium `2SO_(2)+O_(2)hArr 2SO_(3)` we start with 2 moles of `SO_(2)` and 1 mole of `O_(2)` at 3 atm. When equilibrium is attained, pressure changes to 2.5atm. Hence, `K_(p)` is `:`A. `3 atm^(-1)`B. `2.5 atm^(-1)`C. `2 atm ^(-1)`D. `0.5 atm^(-1)`

Answer» Correct Answer - C
Initially , `P=3 atm`
Total no. of moles `=2+1=3`
`PV=nRT`
`3xx V =3RT`
`V=RT`
`{:(,2SO_(2),+,O_(2),hArr,2SO_(3)),("Initial",2,,1,,-),("At eqm.",2-2x,,1-x,,2x):}`
Total no. of moles at eqm. `=3-x`
`PV=nRT`
`2.5xxV=(3-x)RT`
`2.5 xx RT =(3-x)RT`
`3-x=2.5`
`x=0.5`
No. of moles at eqm.
`SO_(2)=2(1-x)=2(1-0.5)=1`
`O_(2)=1-x=1-0.5=0.5`
`SO_(3)=2x=2xx0.5=1`
We know `p_(A)=x_(A)P`
`p_(SO_(2))=(1)/(2.5)xx2.5=1 "atm"`
`p_(O_(2))=(0.5)/(2.5)xx2.5=0.5 "atm"`
`p_(SO_(3))=(1)/(2.5)xx2.5=1"atm"`
`K_(p)=((p_(SO_(3)))^(2))/((p_(O_(2)))(p_(SO_(2)))^(2))`
`=((1)^(2))/((0.5)(1)^(2))=2 atm^(-1)`


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