Answer :
Answer:
The equilibrium constant Kc for the reaction is :
16.07
Explanation:
The balanced equation is :
[tex]H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)[/tex]
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,
[tex]H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)[/tex]
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:
[tex]Concentration =\frac{moles}{Volume(L)}[/tex]
[tex]C_{H_{2}}=\frac{0.14}{2}[/tex]
[tex]C_{H_{2}}=0.07moles/L[/tex]
[tex]C_{I_{2}}=\frac{0.64}{2}[/tex]
[tex]C_{I_{2}}=0.32moles/L[/tex]
[tex]HI=\frac{1.2}{2}[/tex]
[tex]HI=0.6moles/L[/tex]
The equilibrium constant for this reaction "Kc" can be written as:
[tex]Kc=\frac{[HI]^{2}}{[H_{2}][I_{2}]}[/tex]
[tex]Kc=\frac{0.6^{2}}{0.07\times 0.32}[/tex]
[tex]Kc=\frac{0.36}{0.0224}[/tex]
[tex]Kc=16.07[/tex]