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

Answer:

The equilibrium constant Kc for the reaction is :

16.07

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:

$Concentration =\frac{moles}{Volume(L)}$

$C_{H_{2}}=\frac{0.14}{2}$  

$C_{H_{2}}=0.07moles/L$

$C_{I_{2}}=\frac{0.64}{2}$  

$C_{I_{2}}=0.32moles/L$

$HI=\frac{1.2}{2}$  

$HI=0.6moles/L$

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

$Kc=\frac{[HI]^{2}}{[H_{2}][I_{2}]}$  

$Kc=\frac{0.6^{2}}{0.07\times 0.32}$

$Kc=\frac{0.36}{0.0224}$

$Kc=16.07$