Question

A sealed container was filled with 0.300mol H2(g), 0.400mol I2(g), and 0.200mol HI(g) at 870K and total pressure 1.00bar. Calculate the amounts of the components in the mixture at equilibrium given that K.= 70 for the reaction H2(g)+I2(g) --> 2HI(g).

Answer 1

Answer:

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol

Explanation:

Based on the equilibrium:

H2(g)+I2(g) --> 2HI(g)

The equilibrium constant, K, is defined as:

K = 70 = [HI]² / [H2] [I2]

Where [] could be taken as the moles in equilibrium of each reactant

To know the direction of the equilibrium we need to find Q with the initial moles of each species:

Q = [0.200mol]² / [0.300mol] [0.400mol]

Q = 0.333

As Q < K, the reaction will shift to the right producing more HI. The equilibrium moles are:

[HI] = 0.200mol + 2X

[H2] = 0.300mol - X

[I2] = 0.400mol - X

Replacing in K:

70 = [0.200 + 2X ]² / [0.300 - X] [0.400 - X]

70 = 0.04 + 0.8 X + 4 X² / 0.12 - 0.7 X + X²

8.4 - 49 X + 70 X² = 0.04 + 0.8 X + 4 X²

8.36 - 49.8X + 66X² = 0

Solving for X:

X = 0.252 moles. Right solution

X = 0.502 moles. False solution. Produce negative moles.

Replacing:

[HI] = 0.200mol + 2*0.252 mol

[H2] = 0.300mol - 0.252 mol

[I2] = 0.400mol - 0.252 mol

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol