Reaction Rates Chapter 18 CP Chemistry

Reactions can be… FAST! Liquid hydrogen and oxygen reacting to launch a shuttle

..or S L O W Concrete hardening

..or S L O W Watching paint dry

Expressing rxn rates in quantitative terms

Red  Blue Reaction Rates

Collision Theory Atoms, ions and molecules must collide in order to react. Reacting substances must collide with the correct orientation. Reacting substances must collide with sufficient energy to form the activated complex.

Orientation and the activated complex Analogy: if you start with two separate paperclips (reactants) and you wish to link them together (products), not only must the paperclips come into contact, but they also must collide with a specific orientation.

Activation energy and reaction Only collisions with enough energy to react form products

Factors affecting reaction rates 1) Temperature 2) Concentration 3) Particle Size 4) Catalysts

TEMPERATURE: Generally, ↑ temp = ↑ rate Why? Higher temp = faster molecular motion = more collisions and more energy per collision = faster rxn Analogy: imagine that you are baby-sitting a bunch of 6 year olds. You put them in a yard and you let them run around. Every now and then a couple of kids will run into each other. Now imagine that you decide to feed them some sugar. What happens? They run around faster and of course there are many more collisions. Not only that, the collisions are likely to be a lot harder/more intense.

CONCENTRATION As concentration ↑, frequency of collisions ↑, and therefore rxn rate ↑

SURFACE AREA ← slow fast  As surface area ↑, rxn rate ↑

CATALYST pg. 547 Provides an easier way to react Lowers the activation energy Catalyst: a substance that speeds up the rate of a reaction without being consumed in the reaction. (remains unchanged)

CATALYST Adding a catalyst speeds up the rxn by lowering the activation energy

Chemical Equilibrium

Consider a glass of water… Evaporation

Consider a glass of water… Now, put a lid on it….

Consider a glass of water… Evaporation continues, but condensation also occurs...

Consider a glass of water… The rates equalize, and the system reaches equilibrium.

Chemical Equilibrium Consider the following reaction(s): H 2 O (liquid)  H 2 0 (gas) H 2 O (gas)  H 2 O (liquid) H 2 O (liquid)  H 2 O (gas) Equilibrium Symbol

Chemical equilibrium occurs when opposing reactions are proceeding at equal rates. The rate at which the products are formed from the reactants equals the rate at which the reactants are formed from the products. For equilibrium to occur, neither reactant nor products can escape from the system.

N 2 + 3H 2  2NH KCal Forward Reaction

N 2 + 3H 2  2NH KCal Reverse Reaction

Reversible Reactions REVERSIBLE REACTIONS do not go to completion and can occur in either direction: aA + bB  cC + dD

CHEMICAL EQUILIBRIUM exists when the forward & reverse reactions occur at exactly the same rate R P Time (reaction progress ) EQUILIBRIUM concentration Fig pg. 550 There is no net change in the amounts of reactants and products at equilibrium PAGE 550 (key point)

Checkpoint ? Pg. 550 Why is chemical equilibrium called a dynamic state?

Answer Because both the forward and reverse reactions continue

At equilibrium: If there are more products than reactants, the products are said to be favored. If there are more reactants than products, the reactants are said to be favored.

Factors Affecting Equilibrium

When a system is at equilibrium, it will stay that way until something changes this condition.

Le Chatelier’s Principal when a change (“stress”) is applied to a system at equilibrium, the system will shift its equilibrium position to counter act the effect of the disturbance.

Factors affecting equilibrium include changes in: Concentrations of reactants or products Temperature Pressure (gases)

Changes in Concentration: Consider this reaction at equilibrium: H 2 (g) + I 2 (g)  2HI(g) What will happen to the equilibrium if we: add some H 2 ? Reaction shifts to the right (forms more product)

Changes in Concentration: Consider this reaction at equilibrium: H 2 (g) + I 2 (g)  2HI(g) What will happen to the equilibrium if we: remove some H 2 ? Reaction shifts to the left (forms more reactants)

Changes in Concentration: When a substance is added, the stress is relieved by shifting equilibrium in the direction that consumes some of the added substance. When a substance is removed, the reaction that produces that substance occurs to a greater extent.

Changes in Temperature: Consider this reaction at equilibrium: 2SO 2 (g) + O 2 (g)  2SO 3 (g) kJ What will happen to the equilibrium if we: increase the temperature? Reaction shifts to the left (forms more reactants)

Changes in Temperature: Consider this reaction at equilibrium: 2SO 2 (g) + O 2 (g)  2SO 3 (g) kJ What will happen to the equilibrium if we: decrease the temperature? Reaction shifts to the right (forms more products)

Changes in Temperature: Increasing the temperature always favors the reaction that consumes heat, and vice versa.

Changes in Pressure: Consider this reaction at equilibrium: 2NO 2 (g)  N 2 O 4 (g) What will happen to the equilibrium if we: increase the pressure? Reaction shifts to the right (forms more product)

Changes in Pressure: Consider this reaction at equilibrium: 2NO 2 (g)  N 2 O 4 (g) What will happen to the equilibrium if we: decrease the pressure? Reaction shifts to the left (forms more reactant)

Changes in Pressure: Increasing the pressure favors the reaction that produces the fewer moles of gas, and vice-versa.

Catalysts A catalyst increases the rate at which equilibrium is reached, but it does not change the composition of the equilibrium mixture.

Action of a Catalyst Activation Energy Without a catalyst

Action of a Catalyst Lower Activation Energy With a catalyst: A catalyst lowers the activation energy.

Example: consider the rxn at equilibrium: N 2 (g) + 3H 2 (g)  2NH 3 (g) + 94 kJ How would the equilibrium be influenced by: Increasing the temp: Decreasing the temp: Increasing the pressure: Adding more H 2 : Removing some NH 3 : Decreasing the pressure: Adding a catalyst: rxn shifts to the left rxn shifts to the right rxn shifts to the left rxn shifts to the right no change in equilibrium position

Example: How will an increase in pressure affect the equilibrium in the following reactions: 4NH 3 (g) + 5O 2 (g)  4NO (g) + 6H 2 O (g) RXN SHIFTS LEFT 2H 2 (g) + O 2 (g)  2H 2 O (g) RXN SHIFTS RIGHT

Example: How will an increase in temperature affect the equilibrium in the following reactions: 2NO 2 (g)  N 2 O 4 (g) + heat RXN SHIFTS LEFT H 2 (g) + Cl 2 (g)  2HCl (g) + 92 KJ RXN SHIFTS LEFT H 2 (g) + I 2 (g) + 25 kJ  2HI (g) RXN SHIFTS RIGHT

To close, let’s talk about Fritz Haber and the “process” that made him famous… Let’s say you want to manufacture ammonia gas (NH 3 ). What are the optimum conditions? N 2 + 3H 2  2 NH 3 + Heat We wish to favor the forward reaction, thus producing more NH 3 gas.

For example N 2 + 3H 2  2 NH 3 + Heat By cooling the reaction, the reactions counters by producing heat. It does this by shifting to the right, producing heat, and more NH 3 gas.

For example N 2 + 3H 2  2 NH 3 + Heat By increasing the pressure, the reaction tries to reduce the pressure. It does this by shifting to the side with the fewest moles of gas. This is the product side with 2 moles of gas. Thus the reaction shifts to the right reducing pressure, and producing more NH 3 gas. 4 moles of gas2 moles of gas

For example N 2 + 3H 2  2 NH 3 + Heat By adding additional N 2 and H 2, the reactions tries to use them up. In doing so, the reaction shifts to the right. By removing NH 3 as soon as its formed, the reactions tries to produce more. Shifting the reaction to the right.

For example N 2 + 3H 2  2 NH 3 + Heat So if you want to produce maximum ammonia gas you should Cool the reaction Conduct the reaction under high pressure Add N 2 and H 2 Remove NH 3

The Equilibrium Constant, K eq page 556 For the reaction: aA + bB  cC + dD at equilibrium, the constant, k c or k eq : K eq is a measure of the extent to which a reaction occurs; it varies with temperature.

Example (a): Write the equilibrium expression for… PCl 5  PCl 3 + Cl 2

Example (b): Write the equilibrium expression for… 4NH 3 + 5O 2  4NO + 6H 2 O

Ex: One liter of the equilibrium mixture from example (a) was found to contain mol PCl 3, mol Cl 2 and mol PCl 5. Calculate K eq. PCl 5  PCl 3 + Cl 2

What does k eq =0.53 mean to me??? When K eq > 1, most reactants will be converted to products. When K eq < 1, most reactants will remain unreacted.

The equilibrium constant allows us to …. Predict the direction in which a reaction mixture will proceed to achieve equilibrium. Calculate the concentrations of reactants and products once equilibrium has been reached.

Ex: PCl 3 (g) + Cl 2 (g)  PCl 5 (g) K eq =1.9 In a system at equilibrium in a 1.00 L container, we find 0.25 mol PCl 5, and 0.16 mol PCl 3. What equilibrium concentration of Cl 2 must be present?

Reaction Quotient (Q) Reaction Quotient (Q) is calculated the same as K eq, but the concentrations are not necessarily equilibrium concentrations. Comparing Q with K eq enables us to predict the direction in which a rxn will proceed if it is NOT at equilibrium.

Comparing Q to K eq When Q < K eq : When Q = K eq : When Q > K eq : Forward rxn predominates – “reaction proceeds to the right”(until equil. is reached) System is at equilibrium Reverse reaction predominates – “reaction proceeds to the left” (until equilibrium is reached)

Ex: H 2 (g) + I 2 (g)  2HI(g) K eq for this reaction at 450  C is 49. If 0.22 mol I 2, 0.22 mol H 2, and 0.66 mol HI are put into a 1.00-L container, would the system be at equilibrium? If not, what must occur to establish equilibrium. Q < K eq ; therefore forward reaction predominates until equilibrium is reached.