Saved Bookmarks
| 1. |
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. |
|
Answer» The equilibrium constant `(K_(p))` for the reaction is `H_(2)(g) +I_(2)(g) hArr 2HI (g)` `K_(p) = (p^(2)HI(g))/(pH_(2) (g) xx pI_(2) (g)) = 0.35` Now , in all the cases , solid iodine is present in the mixture. Therefore, the partial pressure of `I_(2) i.e. pI_(2)` is taken as 1 atm in each case. Let us calculate the value of `Q_(p)` for the different reactions. `(i) " "P_(H_2) = -0.10 atm and P_(HI) =0.80 atm` `:. " "Q_(P) = (P_(HI)^(2) (g))/(P_(H_2)(g))=(0.44xx0.44)/(0.55) =0.35` As the value of `Q_(p)` (6.4) in more than that of `K_(p)` (0.35) this means that the equilibrium has not been achieved in the reaction and the reaction has proceeded more in the forward direction. `(II)" "P_(H_2) =0.55 atm and P_(HI) =0.44 atm` `:. " "Q_(p) = (P_(HI)^(2)(g))/(P_(H_2)(g)) = (0.8xx 0.8)/(0.1) = 6.4` As the value of `Q_(p)` is the same as that of `K_(p)` this means that the system is in a state of equilibrium `(iiI) " "P_(H_2) = 2.5 atm and P_(HI) = 0.15 atm` `:." "Q_(p) = (P_(HI)^(2)(g))/(P_(H_2) (g)) = (0.15xx0.15)/(2.5) = 0.009` As the value of `Q_(p)` (0.009) is less than that of `K_(p)` (0.35) , this means that the equilibrium has not been achived in the reaction and the reaction has proceeded more in the backward direction. |
|