Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2008
Exercise: Chemical Equilibrium	© R. Paselk 2008
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Chemical Equilibrium Problems

Answers

1. Consider the reaction

N₂(g) + 3H₂(g) 2NH₃(g) + 92.2 kJ

Note that heat appears on the product side - the system is giving up heat, therefore H is negative, H = - 92.2 kJ

Using Le Châtelier's Principle, predict what will happen to [NH₃] if:

• N₂ is added?
  • The system will respond by shifting to the right in order to use up some of the added nitrogen and reduce the stress-
  • as a result [NH₃] will increase.
• H₂ is removed?
  • The system will respond by shifting to the left in order to replace some of the lost hydrogen and reduce the stress-
  • as a result [NH₃] will decrease.
• T is decreased?
  • If temperature is decreased heat must have been removed. The system will respond by shifting to the right in order to replace some of the lost heat and reduce the stress by bringing the temperature back up-
  • as a result [NH₃] will increase.
• Ar is added?
  • Since Ar is an inert gas and will not interact with either reactants or products, there will be no change.

Predict how the equilibrium will shift (right, left, or neither) if:

• V is increased?
  • This is a bit more subtle - we have to look at how the change in volume will affect the two sides of the equation:

N₂(g) + 3H₂(g) 2NH₃(g)

The important consideration is how many moles are on each side. In this case we see that on the left (reactant) side we have 4 moles while on the right (product) side we have 2 moles. If the volume increases th econcentration of particles decreases, so the system tries to bring the concentration of particles back up.

This can be done by shifting the equilibrium to the left (more moles), so the equilibriium shifts left.

2. Consider the gas phase reaction:

2 HI H₂ + I ₂

K_eq = 2.06 x 10^-2 @ 458°C

If both hydrogen and iodine are measured to have concentrations of 0.0135 M each at 458°C, what is the concentration hydrogen iodide?

First write out the equilibrium expression:

K_eq = [H₂] [I ₂] / [HI]² = 2.06 x 10^-2

and [HI]² = [H₂] [I ₂] / (2.06 x 10^-2) = [0.0135] [0.0135] / (2.06 x 10^-2)

[HI]² = 0.008847

[HI] = 0.09406 M = 0.0941 M

 

2 HBr H₂ + Br₂

K_eq = 1.5 x 10^-5 @ 1400 K

Calculate the concentrations of all species at equilibrium if we start with 0.15 moles each of hydrogen bromide and bromine in a 0.500 L container at 1400 K.

	2 HBr		H2	+	Br2
Before reaction	0.15/0.500 = 0.30 M		0		0.15/0.500 = 0.30 M
From the equation some HBr will breakdown to give hydrogen and bromine:
@ Equilibrium	0.30 - x		x		0.30 + x

At equilibrium then, K_eq = [Br₂] [H₂] / [HBr]²

Substituting, K_eq = (0.30 + x)(x) / (0.30 - x)² = 1.5 x 10^-5

Since K_eq is small, assume x<<0.30. Can then simplify:

K_eq = (0.30)(x) / (0.30)² = 1.5 x 10^-5

0.30x = (0.090)(1.5 x 10^-5) = 1.35 x 10^-6

and x = 4.5 x 10^-6 <<0.30, Assumption OK!

At equilibrium:

HBr = 0.30 M

H₂ = 4.5 x 10^-6 M

Br₂ = 0.30 M

 

4. Consider the gas phase dissociation of carbon dioxide to carbon monoxide and oxygen @ 1000 K.

If 0.200 moles of carbon dioxide is placed in a 1.00 L container at 1000 K calculate the concentrations of all species at equilibrium. K_eq = 4.5 x 10^-23 @ 1000 K.

2 CO2 2 CO + O2 Before reaction 0.200mol/1.00L= 0.200M   0 0 Since the coefficient for oxygen is 1, let [oxygen] = x and [CO] = 2x @ Equilibrium 0.200M - 2 x   2 x   x

At equilibrium then, K_eq = [CO]² [H₂] / [CO₂]²

Substituting, K_eq = (x)(2x)² / (0.200 - 2x)² = 4.5 x 10^-23

Since K_eq is small, assume x<<0.200. Can then simplify:

K_eq = 4x³/ (0.200)² = 4.5 x 10^-23

4x³ = 0.0400(4.5 x 10^-23)

x³ = 4.5 x 10^-25

x = 7.7 x 10^-9

and x = 7.7 x 10^-9<<0.200, Assumption OK!

At equilibrium:

CO₂ = 0.200 M

CO = 1.5 x 10^-8M

O₂ = 7.7 x 10^-9M