Question

Considering H2(g) + I2(g) ↔ 2HI and temperature = 731K, 1.20 mol of H2 and 1.20 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of I2 in the gaseous mixture? The equilibrium constant is K = 49.0

Answer 1

The equilibrium concentration of I₂ is 0.933 mol/L.

We start with the balanced chemical equation for the equilibrium.

H₂ + I₂ ⇌ 2HI

Then we write the `K_"eq"` expression.

`K_"eq" = ["HI"]^2/(["H"_2]["I"_2])`

Next, we set up an ICE table:

`"H"_2 + "I"_2 ⇌ 2"HI"`

I/mol·L⁻¹: 1.20; 1.20; 0
C/ mol·L⁻¹: -`x`; -`x`; +`2x`
E/ mol·L⁻¹: 1.20-`x`; 1.20-`x`; `2x`

Insert these values into the `K_"eq"` expression:

`K_"eq" = ["HI"]^2/(["H"_2]["I"_2]) = (2x)^2/((1.20-x)(1.20-x)) = 49.0`

`(4x^2)/(1.20 - x)^2 = 49.0`

`(2x)/(1.20 – x) = 7.00`

`2x= 8.40 – 7.00x`

`9.00x = 8.40`

`x= 0.933`

[I₂] = (1.20 – `x`) mol/L = (1.20 – 0.933) mol/L = 0.27 mol/L

Here's a video that does a similar calculation.