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)`.

Explanation

`2HI(g)hArr H_(2)(g)+I_(2)(g)`
The initial pressure of HI `(p_(HI))=0.2` atm and the partial pressure of HI at equilibrium=0.04 atm. So, decrease in pressure of HI=0.2-0.04=0.16 atm
According to the above reaction, decomposition of 2 mol of HI gives 1 mol each of `H_(2) and I_(2)`.
Therefore, the increase in pressure for each of `H_(2) and I_(2)` will be half of the decrease in pressure of HI.
Hence, at equilibrium
`p_(HI)=0.04atm," "p_(H_(2))=p_(I_(2))=(1)/(2)xx0.16=0.08atm`.
`therefore K_(p)=(p_(H_(2))xxp_(I_(2)))/(p_(HI)^(2))=(0.08xx0.08)/((0.04)^(2))=4`.