Hi can anyone show me the solution of this please

Question

Consider the following reaction which has a Kc = 25.0 at 1100 K.

H2 (g) + I2 (g) ↔ 2 HI (g)

1.50 mol H2 and 3.00 mol I2 are placed in a 2.00 L reaction vessel at 1100 K. What is the equilibrium concentration of each gas? 

Answer 1

For this type of problem, it's probably best to use an ICE table.

Since the volume is 2 L, we can calculate the concentrations as

[H₂] = 1.50 mol / 2 L = 0.75 M

[I₂] = 3.00 mol / 2 L = 1.50 M

H2 (g)  +  I2 (g)  ↔  2 HI (g)

0.75.........1.50............0...........Initial

-x............-x................+2x........Change

0.75-x....1.50-x...........2x..........Equilibrium

Kc = 25.0 = [HI]² / [H₂][I₂]

25.0 = (2x)² / (0.75-x)(1.50-x)

25.0 = 4x² / x² - 2.25x + 1.125

4x² = 25x² - 56.25x + 28.125

21x² - 56.25x + 28.125 = 0

x = 0.665 (be sure to check the math)

Concentrations at equilibrium are as follows:

[H₂] = 0.75 - 0.665 = 0.085 M

[I₂] = 1.50 - 0.665 = 0.835 M

[HI] = 2 x 0.665 = 1.33 M