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H(2)S((g)) intially at a pressure of 10...

`H_(2)S_((g))` intially at a pressure of 10 atm anda temperature of 800K, dissociates as `2H_(2)S_((g)) hArr 2H_(2) +S_(2(g))` At equilibrium, the partial pressure of S, vapour is 0.02 atm. Thus, `K_(p)` is

A

`3.23 xx 10^(-7)`

B

`6.45 xx 10^(-7)`

C

`1.55 xx 10^(6)`

D

`6.2 xx 10^(-7)`

Text Solution

Verified by Experts

The correct Answer is:
A

x=0.020 atm
`PH_(2)S=10-2x=10-0.02 xx 2=9.96` atm
`PH_(2)=x=0.02`
`K_(P)=(PS_(2) xx PH_(2)^(2))/(P^(2)H_(2)S)=(0.02 xx (0.04)^(2))/((9.96)^(2))`
`= 3.23 xx 10^(-7)` atm
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