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When sulphur is heated at 900 K, S(8) is...

When sulphur is heated at 900 K, `S_(8)` is converted to `S_(2)`. What will be the equilibrim constant for the reaction if initial pressure of 1 atm falls by `25%` at equilibrium ?

A

`0.75 atm^(3)`

B

`2.55atm^(3)`

C

`25.0 atm^(3)`

D

`1.33 atm^(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(,S_(8(g)),hArr,4S_(2(g)),),("Intial pressure",1-(25)/(100),,4xx(25)/(100),),(,=0.75,,=1,):}`
`K_(p)=((P_(S_(2)))^(4))/(P_(S_(8)))=((1)^(4))/(0.75)=1.33" atm"^(3)`
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