Answer:
The equilibrium concentration are :
[tex]{H_{2}}[/tex] = 0.04 M
[tex]{I_{2}}[/tex] = 0.04 M
[tex]{HI}[/tex] = 0.16 M
Explanation:
[tex]{H_{2}} + {I_{2}} \rightleftharpoons 2 HI[/tex]
it means 1 mole of iodine and hydrogen produce 2 mole of HI
Concentration(C) : Moles per unit volume.It is expressed in Molarity
(M=mol/L )
[tex]concentration = \frac{moles}{volume}[/tex]
Initial moles :
[tex]{H_{2}}[/tex] = 1.00
So, [tex]C = \frac{1}{V}[/tex]
[tex]{I_{2}}[/tex] = 1.00
[tex]C = \frac{1}{V}[/tex]
[tex]{HI}[/tex] = 0
[tex]C = \frac{0}{V}[/tex]
let during the reaction x moles of both [tex]{H_{2}}[/tex] and [tex]{I_{2}}[/tex] get dissociated , then
At equilibrium ,
[tex]{H_{2}}[/tex] = 1.00 - x
[tex]C = \frac{1-x}{V}[/tex]
For iodine
[tex]{I_{2}}[/tex] = 1.00 - x
[tex]C = \frac{1-x}{V}[/tex]
1.00 - x mole of hydrogen will produce 2x of HI
[tex]{HI}[/tex] = 2x
[tex]C = \frac{2x}{V}[/tex]
[tex]K_{eq} = \frac{[products]^{coefficient}}{[reactants]^{coefficient}}[/tex]
On solving for x , (look at the image)
[tex]{H_{2}}[/tex] = 0.04 M
[tex]{I_{2}}[/tex] = 0.04 M
[tex]{HI}[/tex] = 0.16 M
