Question

The equilibrium constant kc of the reaction H2+I2 produces HI is 54.3at430°C. Starting with 6.0moles,of HI moles of HI in 1.0dm^3 reaction vessel, calculate th5 concentration of H2,I2 and HI at equilibrium

Answer 1

[H2] = [I2] = 0.64M; [HI] = 4.72M

Step-by-step explanation:

Based on the reaction:

H2 + I2 ⇄ 2HI

The K is defined as:

k = 54.3 = [HI]² / [H2] [I2]

Where [] is molar concentration of each reactant at equilibrium

As the initial concentration of HI is 6mol/dm^3 = 6M the equilibrium concentration of each reactant is:

[H2] = X

[I2] = X

[HI] = 6 - 2X

Where X is reaction coordinate

Replacing:

54.3 = [6-2X]² / [X] [X]

54.3X² = 4X² - 24X + 36

0 = -50.3X² - 24X + 36

Solving for X:

X = -1.12. False solution, produce negative concentrations

X = 0.64M. Right solution

Replacing:

[H2] = 0.64M

[I2] = 0.64M

[HI] = 6 - 2*0.64M = 4.72M

Equilibrium concentrations are:

[H2] = [I2] = 0.64M; [HI] = 4.72M