Answer:
[tex] C_{H_{2}} = 0.07 M [/tex]
[tex] C_{I_{2}} = 1.07 M [/tex]
[tex] C_{HI} = 1.86 M [/tex]
Explanation:
The reaction is:
H₂(g) + I₂(g) ⇄ 2HI(g)
Initially, we have the following concentrations of H₂ and I₂:
[tex] C_{H_{2}} = \frac{n}{V} = \frac{1 mol}{1.00 L} = 1 mol/L [/tex]
[tex] C_{I_{2}} = \frac{n}{V} = \frac{2 mol}{1.00 L} = 2 mol/L [/tex]
Then, in the equilibrium we have:
H₂(g) + I₂(g) ⇄ 2HI(g)
1-x 2-x 2x
[tex] Kc = \frac{[HI]^{2}}{[H_{2}][I_{2}]} = \frac{(2x)^{2}}{(1-x)(2-x)} [/tex]
[tex] 50.5*(1-x)(2-x) - (2x)^{2} = 0 [/tex]
By solving the above equation for x we have:
x₁ = 2.32 and x₂= 0.93
Hence, the concentrations of H₂, I₂ and HI are:
[tex] C_{H_{2}} = 1-x = 1 - 0.93 M = 0.07 M [/tex]
[tex] C_{I_{2}} = 2-x = 2 - 0.93 M = 1.07 M [/tex]
[tex] C_{HI} = 2*x = 2*0.93 M = 1.86 M [/tex]
I hope it helps you!
