Question

A 1.00-L flask is filled with 1.00 moles of H2 and 2.00 moles of I2. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H2 , I2 , and HI in moles/L? H2 (g) + I2 (g) ⇌ 2HI(g)

Answer 1

`C_{H_(2)} = 0.07 M`

`C_{I_(2)} = 1.07 M`

`C_(HI) = 1.86 M`

Step-by-step explanation:

The reaction is:

H₂(g) + I₂(g) ⇄ 2HI(g)

Initially, we have the following concentrations of H₂ and I₂:

`C_{H_(2)} = (n)/(V) = (1 mol)/(1.00 L) = 1 mol/L`

`C_{I_(2)} = (n)/(V) = (2 mol)/(1.00 L) = 2 mol/L`

Then, in the equilibrium we have:

H₂(g) + I₂(g) ⇄ 2HI(g)

1-x 2-x 2x

`Kc = ([HI]^(2))/([H_(2)][I_(2)]) = ((2x)^(2))/((1-x)(2-x))`

`50.5*(1-x)(2-x) - (2x)^(2) = 0`

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:

`C_{H_(2)} = 1-x = 1 - 0.93 M = 0.07 M`

`C_{I_(2)} = 2-x =  2 - 0.93 M = 1.07 M`

`C_(HI) = 2*x = 2*0.93 M = 1.86 M`

I hope it helps you!