Question

If 3.4 mol of HI was placed in a 2.0 L container and allowed to reach equilibrium, what would the equilibrium concentrations be for H2(g), I2(g) and HI(g) if the Ke = 49?	(5 marks)

H2(g) + I2(g) ⇆ 2HI(g)

Answer 1

C_M (HI) = n/V = 3.4mol / 2L = 1.7 mol/l

Ke= 49

2HI(g) → H₂(g) + I₂(g)

1.7-2x — x — x

$Ke= \cfrac {[H_2][I_2]}{[HI]²}\\~\\ 49 = \cfrac {x²}{(1.7-2x)²}\\~\\ 7 = \cfrac{x} {(1.7-2x)}\\~\\ 11.9-14x=x\\ 15x=11.9\\x=0.793$

[HI] = 1.7 – 2*0.793 = 0.114 M

[H₂] = 0.793 M

[I₂] = 0.793 M