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Starting with 1 mol of O(2) : 2 mol of S...

Starting with 1 mol of `O_(2) : 2` mol of `SO_(2)`, the equilibrium for the formation of `SO_(3)(g)` was established a certain temperature. If V is the volume of the vessel and 2x is the number of moles of `SO_(3)` present, equilibrium constant for the reaction `2SO_(2(g)) + O_(2(g)) hArr 2SO_(3(g))` would be :

A

`(x^(2)V)/((1-x)^(3))`

B

`(4x^(2))/((2-x)(1-x))`

C

`((1-x)^(3))/(2V)`

D

`(x^(2))/((2-x)(1-x))`

Text Solution

Verified by Experts

The correct Answer is:
A
`{:(,2SO_(2(g)),+,O_(2(g)),hArr,2SO_(3(g))),(t = 0,2,,1,,0),(t_("eqm"),[(2-2x)/(V)],,[(1-x)/(V)],,[(2x)/(V)]):}`
`K_(c) = ([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)]) = ([2x//V^(2)])/([(2-2x)/(V)]^(2)[(1-x)/(V)]) = (x^(2)V)/((1-x)^(3))`
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Knowledge Check

  • For the equilibrium reaction 2 NO2(g)↔ N2 O4(g)+60.0 kJ The increase in temperature (1)Favour the formation of N O (2)Favours the decomposition of N2 O4 (3)does not effect the equilibrium (4)stops the reaction

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  • The volume of the reaction vessel containing an equilibrium mixture in the reaction, SO_(2)Cl_(2(g)) hArr SO_(2(g)) + Cl_(2(g)) is increased. When equilibrium is re-established :

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    the amount of `SO_(2(g))` will decrease
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    D
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