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5 moles of `SO_(2) and 5 " moles of " O_(2) ` are allowed to react . At equilibrium , it was found that 60% `SO_(2)` is used up . If the pressure of the mixture is one atmosphere , the partial pressure of `O_(2)` isA. `0*52` atmB. `0*21` atmC. ` 0*41`atmD. `0*82` atm |
Answer» Correct Answer - C ` {:(,2 SO_(2)(g),+,O_(2)(g),hArr,2 SO_(3)(g)),("Intial",5,,5,,0):}` As 60 % `SO_(2)` is used up, no of moles of `SO_(2) "used up" = 60/100 xx 5 =3` ` :. " No. of moles of " SO_(2) " at equilibrium " = 5-3=2` As 2 moles of `SO_(2) " react with 1 mole of " O_(2)` `=1/2 xx 3 = 1.5 " moles" ` i.e. No. of moles of `O_(2)` at equilibrium `=5-1*5 = 3.5 ` As 2 moles of `SO_(2) " produce 2 moles of " SO_(3) ` `:. " No . of moles of "SO_(3) " equilibrium = 3 moles"` `= 2 + 3*5 + 3 = 8*5 ` As 2 moles of `SO_(2)"produce 2 moles of"SO_(3)` ` :. " No. of moles of " SO_(3) " at equilibrium" = 3 moles ` `:. "Total no. of moles at equilibrium "` |
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