1.

SO_(3) (g) is produced as : SO_(2) (g) + (1)/(2) O_(2) (g) hArr SO_(3) (g) At 900 K, 0.2 mol of SO_(2) and 0.4 mol of O_(2) are taken in 2L vesset. When equilibrium is reaches by concentration of SO_(3) (g) is 0.08 M. Then K_(C)^(0) for reaction is :

Answer»

`(10)/(sqrt2)`
`10^(-1)`
`10`
`100`

Solution :`{:(,SO_(2) (g),+,(1)/(2)O_(2) (g),hArr,SO_(3) (g),T = 900 K,),("Concentration",(0.2)/(2),,(0.4)/(2),,,,),("initially",,,,,,,),("concentration",0.1 - x,,0.2 - (1)/(2) x,,x,,),(,x = 0.08 M,,,,,,),(,,,,,,,):}`
`K_(C) = ([SO_(3)])/([SO_(2)] [O_(2)]^(1//2)) = (0.08)/(sqrt(0.16) XX 0.02) = 10`


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