# 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

Oct 8, 2014

The equilibrium concentration of I₂ is 0.933 mol/L.

H₂ + I₂ ⇌ 2HI

Then we write the ${K}_{\text{eq}}$ expression.

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

Next, we set up an ICE table:

$\text{H"_2 + "I"_2 ⇌ 2"HI}$

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

Insert these values into the ${K}_{\text{eq}}$ expression:

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

$\frac{4 {x}^{2}}{1.20 - x} ^ 2 = 49.0$

(2x)/(1.20 – x) = 7.00

2x= 8.40 – 7.00x

$9.00 x = 8.40$

$x = 0.933$

[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.