General Chemistry 2

Question

The equilibrium constant, K_c, for the following reaction is 55.6 at 698 K.

H₂ (g) + I₂ (g)  2 HI (g)

Calculate the equilibrium concentrations of reactants and product when 0.366 moles of H₂ and 0.366 moles of I₂ are introduced into a 1.00 L vessel at 698 K.

[ H2 ]	=	M
[ I2 ]	=	M
[ HI ]	=	M

Answer 1

H₂(g) + I₂(g) <=====> 2HI(g)

I 0.366M 0.366M 0

Δ -x -x +2x

Eq 0.366-x 0.366-x 2x

Kc = [HI]²/[H₂][I₂] = 55.6

(2x)²/(0.366-x)(0.366-x) = 55.6

(2x)²/(0.366-x)² = 55.6

[2x/(0.366-x)]² = 55.6

2x/(0.366-x) = √55.6 = 7.46

2x = 7.46(0.366-x)

9.46x = 2.729 => x = 0.2886

[H₂] = 0.366 - x = 0.366 - 0.289 = 0.077 M

[I₂] = 0.366 - x = 0.366 - 0.289 = 0.077 M

[HI] = 2x = (2)(0.289) = 0.578 M

Answer 2

H₂(g) + I₂(g) <==> 2HI(g) ... Kc = 55.6

Set up an ICE table:

H₂(g) + I₂(g) <==> 2HI(g)

0.366.....0.366...........0.........Initial

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

0.366-x...0.366-x.....2x........Equilibrium

Kc = [HI]² / [H₂][I₂]

55.6 = (2x)² / (0.366-x)(0.366-x)

51.6x² - 40.7x + 7.45 = 0

Solve for x

x = 0.289 (be sure to check the math)

Concentrations at equilibrium are as follows:

[H₂] = 0.366 - 0.289 = 0.077 M

[I₂] = 0.366 - 0.289 = 0.077 M

[HI] = 2x = (2)(0.289) = 0.578 M