Question

A reaction mixture that consisted of 0.35 molH2and 1.6 mol I2 was introduced into a 2 L flask and heated. At the equilibrium, 60% of the hydrogen gas had reacted. What is the equilibrium constant Kc for the reactionH2(g) + I2(g)⇀↽2 HI(g)at this temperature?

Answer 1

The equilibrium constant Kc for the reaction is :

16.07

Step-by-step explanation:

The balanced equation is :

`H_(2)(g)+I_(2)(g)\rightleftharpoons 2HI(g)`

First ,

"At the equilibrium, 60% of the hydrogen gas had reacted"

This means the degree of dissociation = 60% = 0.6

Here we are denoting degree of dissociation by ="x" = 0.6

Now , consider the equation again,

`H_(2)(g)+I_(2)(g)\rightleftharpoons 2HI(g)`

H2 I2 2HI

0.35 1.6 0 (Initial Concentration)

0.35(1 - x) 1.6(1 - x) 2x (At equilibrium)

0.35(1 - 0.6) 1.6(1 - 0.6) 2(0.6)

0.14 0.64 1.2

calculate the concentration of each:

`Concentration =(moles)/(Volume(L))`

`C_{H_(2)}=(0.14)/(2)`

`C_{H_(2)}=0.07moles/L`

`C_{I_(2)}=(0.64)/(2)`

`C_{I_(2)}=0.32moles/L`

`HI=(1.2)/(2)`

`HI=0.6moles/L`

The equilibrium constant for this reaction "Kc" can be written as:

`Kc=([HI]^(2))/([H_(2)][I_(2)])`

`Kc=(0.6^(2))/(0.07* 0.32)`

`Kc=(0.36)/(0.0224)`

`Kc=16.07`