Reaction Progress Diagrams

Answers

1. Consider the reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction and label it completely.

First we need to recall the axis of the plot, Energy (y-axis) and Reaction Progress (x-axis).

Then we need to recall the curve shape - a hill or peak between two flat reagions.

Next we need to look at the reaction. In addition to the reactants and products we see energy appears on the right side, so energy is given off in the reaction.

• In other words the products have less energy.
• Therefore the products are more stable.
• The products will be lower on the diagram to show this lower energy.

Now we can draw the plot and label it, including: E_a, G and the transition state:

2. Consider the reaction

Reactants + Energy Products

Draw a reaction coordinate diagram for this reaction and label it completely.

Again recall the axis of the plot, Energy (y-axis) and Reaction Progress (x-axis).

Then recall the curve shape - a hill or peak between two flat reagions.

Next we need to look at the reaction. In addition to the reactants and products we see energy appears on the left side, so energy is added in the reaction.

• In other words the products have more energy.
• Therefore the products are less stable.
• The products will be higher on the diagram to show this higher energy.

Now we can draw the plot and label it, including: E_a, G and the transition state:

3 a. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction showing the uncatalyzed and catalyzed reaction curves. Label the axis, and the two curves:

b. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction as above but add the activation energy, E_a, on the appropriate curve in this diagram and label it.

c. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction as above but add the activation energy, E_a, for the catalyzed reaction on the appropriate curve in this diagram and label it.

This is a bit more subtle since there are two peaks in the catalyzed reaction curve. So, are there two activation energies? Or just one? Which one?

It turns out only the higher peak reflects the activation energy and the slow step in the reaction. Think about it - if the molecules can "jump" over the high peak, they can easily "jump" over the lower peak, so only the high peak is a real barrier.

Thus the resulting diagram looks like this:

 

d. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction as above but this time show the free energy difference. Is G positive or negative?

 

e. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction as above but this time label the reactants, the products and the transition state(s). How many are there on this diagram? How many for each curve?

This time we have to think about whst each peak represents. It turns out each is a transition state - each can go forward or back, each represents a transitional form between two more stable states:

 

f. Consider the catalyzed reaction

Reactants Products + Energy

Draw a reaction coordinate diagram for this reaction as above but this time label completely.

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 at equilibrium :

PCl₅ PCl₃ + Cl₂

What will happen to this reaction if the volume is increased?

If the volume is increased the concentration (number of particles/volume) decreases. Le Chatelier's Principle tells us the system should try to compensate. Since the equilibria has more molecules on the right, if the reaction shifts to the right the concentration will go back up, partially compensating for the initial concentration decrease, as Le Chatelier's Principle predicts.

What will happen to this reaction if chlorine (Cl₂) is added?

If chlorine is added the reaction should shift to the left to reduce the amount of Cl₂ added as Le Chatelier's Principle predicts.

3. 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

 

5. 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