The equilibrium constant for the reaction
`H_(2)(g) + I_(2) (g) hArr 2HI(g)`
is 0.35 at 298 K. In the following mixture at 298 K, has equilibrium been reached ? If not state on which side of the equilibrium the system is :
`(i) P_(H_2) =0.10` atm and `P_(HI) = 0.80` atm and there is solid `I_(2)` in the container.
`(ii) P_(H_2)= 0.55 `atm and `P_(HI) = 0.44` atm and there is solid `i_(2)` in the container.
`(III) P_(H_2) =2.5` atm and `P_(Hi) =0.15` atm and there is solid `I_(2)` in the container.

Explanation

`H_2(g) + I_2(g) hArr 2HI(g)`
Since there is solid iodine `I_2(s)` in the container in all the three cases the partial pressure of `I_2`, i.e., `pI_2` is assumed to be 1 atm in each case and is not considered in the problem so that
`Q_p=([pHI(g)]^2)/(pH_2(g))`
(i)`pH_2`=0.1 atm , pHI=0.80 atm
`Q_p=(0.8xx0.8)/0.1`=6.4
At the value of concentration quotient, `(Q_p)` is more than the equilibrium constant `(K_p),Q_p` will tend to decrease so as to become equal to `K_p`. As a result , the reaction will proceed in the backward direction .
(ii)`pH_2`=0.55 atm , pHI=0.44 atm
`Q_p=(0.44xx0.44)/0.55`=0.352
As the vaue of concentration quotient `(Q_p)` is the same as that of the equilibrium constant `(K_p)`, this means that the equilibrium has been reached in the reaction.
(iii)`pH_2`=2.5 atm, pHI=0.15atm
`Q_p=(0.15xx0.15)/2.5`=0.009
As the value of the concentration quotient `(Q_p)` is less than `K_p` , this means that equilibrium has not been reached in the reaction and `Q_p` will tend to increase. As a result , the reaction will proceed in the forward direction.