InterviewSolution
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InterviewSolution
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A mixture of `SO_(3), SO_(2)` and `O_(2)` gases is maintained in a `10 L` flask at a temperature at which the equilibrium constant for the reaction is `100`: `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` a. If the number of moles of `SO_(2)` and `SO_(3)` in the flask are equal. How many moles of `O_(2)` are present? b. If the number of moles of `SO_(3)` in flask is twice the number of moles of `SO_(2)`, how many moles of oxygen are present? |
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Answer» Correct Answer - (a) `0.1` (b) `0.4` (a) `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g) K_(C)=100` molelit Initial mole `{:(a,b,0),(a-2x,b-x,2x),((a)/(2),b-(a)/(4),(a)/(2)):}` But according to question. No. of mole of `SO_(2)="No. of mole of" SO_(3)` `=a-2x=2x` `a=4x`. `x=(a)/(4)`. Now, `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])` But No. of mole of `SO_(3) and SO_(2)` are equal at eq. so. `K_(C)=(1)/([O_(2)])` `[O_(2)]=(1)/(K_(C)) , [O_(2)]= (1)/(100)` But `[O_(2)]=("mole of"O_(2) "at" eq.)/(10)=(1)/(100)` So No. of mole of `O_(2)=(1)/(10)=0.1` (b) `K_(C)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=``(((2n_(So_(2)))/(V))^(2))/((n_(SO_(2))/(v))^(2)xx(n_(O_(2))/(V))` `K_(C)=(4)/((n_(O_2)/(V))) : n_(O_2)=(4xxV)/(K_(c))=(4xx10)/(100)=0.4` |
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