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A sample of HI(g) is placed in flask at ...

A sample of `HI(g)` is placed in flask at a pressure of `0.2 atm`. At equilibrium. The partial pressure of `HI(g)` is `0.04 atm`. What is `K_(p)` for the given equilibrium?
`2HI(g) hArr H_(2)(g)+I_(2)(g)`

Text Solution

Verified by Experts

The correct Answer is:
4
Initial pressure of HI=0.2 atm
Partial pressure of HI at equilibrium = 0.4 atm
Decrease in pressure after dissociation = 0.2-0.4
=0.16 atm
Partial pressure of `H_2` after dissociation =`0.16/2` atm
=0.08 atm
`therefore` Partial pressure of `I_2` after dissociation = 0.08atm
`{:("Then",2HI(g)hArr , H_2(g)+ , I_2(g)),("Initial pressure","0.2 atm",0,0),("Pressure at equi.",0.04,0.08,0.08):}`
`K_p=(p_(H_2) xx p_(I_2))/p_(HI)^2=(0.08xx0.08)/(0.04)^2` = 4.0
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