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An equilibrium mixture contains 0.5, 0.1...

An equilibrium mixture contains 0.5, 0.12 and 5 moles of `SO_(2),O_(2)andSO_(3)` respectively, in a one litre vessel at a certain temperature. How many mole of `O_(2)` must be forced into the reaction mixture in order to increase the conc. Of `SO_(3)` to 5.3 mole at the same temperature? (Given `K_(c)` for the reaction, `2SO_(2)+O_(2)iff2SO_(3)` is 800)

A

`0.506`

B

`0.908`

C

`0.74`

D

`0.45`

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(,2SO_(2(g)),+,O_(2(g)),iff,2SO_(3(g))),("Initial","0.5 moles",,"0.12 moles",,"5 moles"),("At equilibrium","(0.5 - 0.3) moles",,"(0.12 + x - 0.15) moles",,"5.3 moles"):}`
Suppose .x. moles of `O_(2)` be introduced into the vessel in order to increase the conc. of `SO_(3(g))` to 5.3 moles/litre.
implies 0.3 mole `SO_(2)` and 0.15 mole `O_(2)` will be consumed according to the above equation.
Hence at equilibrium, `[SO_(2)]=0.2M`
`[O_(2)]=(x-0.03)M,[SO_(3)]=5.3M`
`implies K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)xx[O_(2)])=((5.3)^(2))/((0.2)^(2)xx(x-0.03))=800` (given)
`implies x-0.03=0.878impliesx=0.908` moles
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