The Rates of Chemical Reactions That Consume Reactants and Form Products Are Affected By

Name______Date______Period______

Kinetics - Reaction Rate Standards: 8. Chemical reaction rates depend on factors that influence the frequency of collision of reactant molecules[1]

The rates of chemical reactions that consume reactants and form products are affected by several factors: temperature, pressure, and concentration. Chemical reactions can be explained at the molecular level where kinetic energy at the molecular level is measured by temperature. An important way of describing the potential energy is to plot potential energy versus course of reaction for endothermic and exothermic reactions. The rates of change can be calculated from slopes of lines on the potential energy graph. Steep lines have reactions that are fast.

8. a. The rate of reaction is determined by measuring the decrease in concentration of reactants or the increase in concentration of products with time. Qualitative knowledge of the reaction rate measures how fast reactions proceed and general scales of rate: explosive and biological/cellular reactions are very fast, while rusting (iron oxidation) is very slow.

Reaction rate is defined as the rate of decrease in concentration of reactants or as the rate of increase in concentration of products, and these reciprocal changes form a balanced equation (mole ratios) that reflects the conservation of matter.

Conservation of matter means that the total mass does not change.

8.b. Reaction rates depend on such factors as concentration, temperature, and pressure.

Concentration, temperature, and pressure are major factors affecting the collision of reactant molecules and, thus, affecting reaction rates.

• Increasing the concentration of reactants increases the number of collisions per unit time.
• Increasing temperature (which increases the average kinetic energy of molecules) also increases the number of collisions per unit time. Though the collision rate modestly increases, the greater kinetic energy dramatically increases the chances of each collision leading to a reaction.
• Increasing pressure increases the reaction rate only when one or more of the reactants or products are gases. With gaseous reactants, increasing pressure is the same as increasing concentration and results in an elevated reaction rate.

Additionally, the surface area of the reactants affect the rate with pulverized powders being more reactive than lumps or chunks.

8.c. Catalyst increasing the reaction rate. A catalyst increases the rate of a chemical reaction without taking part in the net reaction. A catalyst lowers the energy barrier (activation energy) between reactants and products by promoting a more favorable pathway for the reaction. Surfaces often play important roles as catalysts for many reactions. One reactant might be temporarily held on the surface of a catalyst. There the bonds of the reactant may be weakened, allowing another substance to react with it more quickly. Living systems speed up life-dependent reactions with biological catalysts called enzymes. Catalysts are not consumed in the reaction. Catalysts are used in automobile exhaust systems to reduce the emission of smog-producing unburned hydrocarbons.

8.d.* Activation energy in a chemical reaction. Even in a spontaneous reaction, reactants are usually required to pass through a transition state that has a higher energy than either the reactants or the products. The additional energy, called the activation energy, or the activation barrier, is related to such factors as strength of bonding within the reactants. The more energy required to go from reactants to activated transition complex, the higher the activation barrier, and the slower a reaction will be. Catalysts speed up rates by lowering the activation barrier along the reaction pathway between products and reactants.

KINETICS

A Chemical Reaction

Three things are required for a reaction to occur:

1. Molecules must collide.
2. They must collide with enough energy to break old bonds so new ones can form.
3. They must collide in the correct orientation.

BE A SUCCESSFUL COLLISON

Kinetics

1. Definition of RATE: change in concentration over change in time

1. Factors that Affect RATE:

a) temp= increasing temp increases frequency and effectiveness of collisions, so rate increases

b) concentration: increasing [ ] of reactants means more collisions per unit time per unit space. The rate of a reaction is proportional to concentration.As the compounds react, the concentration decreases and the rate slows.

c) pressure (gases only): greater the pressure the more “concentrated” the gas, so the faster the rate

d) surface area: the greater the surface area the faster the rate. Ex powdered NaHCO3 with vinegar produced CO2 faster than chunky NaHCO3

e) catalyst: increases both forward and reverse reaction rates because it lowers forward and reverse activation energy

f) nature of reactants: phase=aqueous faster than gas faster than liquid faster than solid because water has already “broken bonds” dissociated ions so they are free to recombine. Strength and # of bonds: the more bonds to be broken the slower the rate. The stronger the bonds the slower the rate

g) Reaction mechanism: the more steps, the slower rate

3. Reaction Mechanisms: “behind the scene” steps to a reaction.

a) slow step= rate-determining step. THIS can be ANY numbered step.

b) You are only as strong as your weakest link so changing the [ ] of a substance NOT in the rate-determining step will NOT affect rate

c) To get the NET equation, cross out “identicals” (otherwise known as intermediates), add like substances on the same side, cross out substances on different sides leaving “leftovers” on the greater side

Ex. Step #1: O3 O2 + O fast Step #2: O + O3 2O2 slow

Net= 2 O3 3 O2 Intermediate = O Rate-determining step = #2

Identify the type of reactions that each graph depicts

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1. The rate equation for a chemical reaction is determined by

(A) theoretical calculations.

(B) measuring reaction rate as a function of concentration of reacting species.

(C) determining the equilibrium constant for the reaction.

(D) measuring reaction rates as a function of temperature

2. Which line in the diagram represents the activation energy for a forward reaction?

(A) A(B) B (C) C (D) D

3. The addition of a catalyst in a chemical reaction

(A) increases the concentration of products at equilibrium.

(B) increases the fraction of reactant molecules with a given kinetic energy.

(C) provides an alternate path with a different activation energy.

(D) lowers the energy change in the overall reaction.

6. Given that

2

2SO2(g) +O2(g) 2SO3(g)

1

The forward reaction (1) is proceeding at a certain rate at some temperature and pressure; when the pressure increased, we may expect for the forward reaction (1)

(A) a greater rate of reaction and a greater yield of SO3 at equilibrium.

(B) a greater rate of reaction and the same yield of SO3 at equilibrium.

(C) a lesser rate of reaction and a lesser yield of SO3 at equilibrium.

(D) a lesser rate of reaction and a greater yield of SO3 at equilibrium.

7.Consider the reaction: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)

At certain conditions 0.26 moles of O2 is consumed in 3.0 minutes.
What is the rate of production of CO2 in g/s ?

8. Consider the reaction: C2H5OH (l) + 3O2(g)  2CO2(g) + 3H2O(g)

At certain conditions 13.44 L of CO2 is produced in 180.0 sec at STP.What is the rate of consumption of C2H5OH in g/min ?

9. A chemist wishes to determine the rate of reaction of zinc with hydrochloric acid. The equation for the reaction is:

Zn(s) + 2HCl(aq)  H2(g) + ZnCl2(aq)

A piece of zinc is dropped into 1.00 L of 0.100 M HCl and the following data were obtained:

Time Mass of Zinc

a)Calculate the Rate of Reaction in grams of Zn consumed per second.

b) Calculate the Rate of Reaction in moles of Zn consumed per second.

c)Write out the complete ionic equation for the reaction showing the acid dissociation.
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d)What will happen to the [H+] as the reaction proceeds? ______

e)What will happen to the [Cl-] as the reaction proceeds? ______

Equilibrium Notes and Standards

9. Chemical equilibrium is a dynamic process at the molecular level. As a basis for understanding this concept:

a. Students know how to use Le Chatelier’s principle to predict the effect of changes in concentration, temperature, and pressure.

Le Chatelier’s principle can be introduced by emphasizing the balanced nature of an equilibrium system. If an equilibrium system is stressed or disturbed, the system will respond (change or shift) to partially relieve or undo the stress. A new equilibrium will eventually be established with a new set of conditions. When the stress is applied, the reaction is no longer at equilibrium and will shift to regain equilibrium. For instance, if the concentration of a reactant in a system in dynamic equilibrium is decreased, products will be consumed to produce more of that reactant. Students need to remember that heat is a reactant in endothermic reactions and a product in exothermic reactions. Therefore, increasing temperature will shift an endothermic reaction, for example, to the right to regain equilibrium. Students should note that any endothermic chemical reaction is exothermic in the reverse direction.

Pressure is proportional to concentration for gases; therefore, for chemical reactions that have a gaseous product or reactant, pressure affects the system as a whole. Increased pressure shifts the equilibrium toward the smaller number of moles of gas, alleviating the pressure stress. If both sides of the equilibrium have an equal number of moles of gas, increasing pressure does not affect the equilibrium. Adding an inert gas, such as argon, to a reaction will not change the partial pressures of the reactant or product gases and therefore will have no effect on the equilibrium.

9. b. Students know equilibrium is established when forward and reverse reaction rates are equal.

Forward and reverse reactions at equilibrium are going on at the same time and at the same rate, causing overall concentrations of each reactant and product to remain constant over time.

9. c.* Students know how to write and calculate an equilibrium constant expression for a reaction.

Because the concentrations of substances in a system at chemical equilibrium are constant over time, chemical expressions related to each concentration will also be constant. Here is a general equation for a reaction at equilibrium:

aA + bB  c C + d D

The general expression for the equilibrium constant of a chemical reaction is Keq, defined at a particular temperature, often 25°C. Its formula is

When Keq is being calculated, only gaseous substances and aqueous solutions are considered. Equilibrium concentrations of products, in moles per liter, are in the numerator, and equilibrium concentrations of reactants are in the denominator. The exponents are the corresponding coefficients from the balanced chemical equation. A large Keq means the forward reaction goes almost to completion; that is, little reverse reaction occurs. A very small Keq means the reverse reaction goes almost to completion, or little forward reaction occurs. The solubility product constant Ksp is the equilibrium constant for salts in solution.

Ammonia Gas Production Equilibrium

N2(g)+ 3H2(g) 2NH3(g)

Not equal concentration totals, but the number of NH3 decomposition is equal to the number of NH3creation.

Opposing processes are in equilibrium:

Forward rate (right) equals backwards rate (left)

Factor which affects only heterogeneous reactions (more than one phase)

Surface area -when 2 different phases react, reaction can only take place on surface.

- increase surface area by cutting solid into smaller pieces (liquids in smaller droplets)

- In general - reactants with solids are slow (except powdered)

- gaseous reactants are faster (but watch for diatomic bonds!)

- reactants in ionic solution. are fastest if no bonds to break

Example: Precipitation reaction Ag+(aq) + Cl-(aq)  AgCl (s)

(aqueous ions are mobile (unlike in a solid ) and more concentrated than molecules in a gas)

Equilibrium

1. Definition: Rate of forward reaction = Rate of reverse reaction
2. Writing Keq’s:

a) products in NUMERATOR,

reactants in DENOMINATOR

b) use [ ] around substances and include their phases

c) (+) signs separating substances in equation BECOME (x) signs

d) coefficients BECOME exponents

e) NO liquids or solids !!!!!! Only aqueous and gas

1. Solving for Keq’s: put theENTIRE numerator in ( ) and the ENTIRE denominator in ( )

1. Interpreting the equilibrium constant

If the Keq= >1 then PRODUCTS are favoredMore products than reactants at equilibrium.

If Keq near one= significant amounts of both reactants and products

If Keq= <1 the reactants favoredMore reactants than products at equilibrium.

LeChatelier: If stressed, shift to relieve stress

1)[ ] as a stress; If increase [ ] of reactants then shift RIGHT to use them up

If decrease [ ] reactants shift LEFT to put them Back

Same idea for products

2)If temp increases shift AWAY from heat

If temp decreases shift TOWARD heat

3)Pressure: Only directly affects gases (indirectly it can affect [ ] of other substances)

If pressure increase shift to side with LESS moles of gases

If pressure decreases, shift to side with GREATER moles of gases

Equilibrium Practice Problems

For the reaction: SiH4(g) + O2(g) SiO2(g) + H2O(g)

a. Write the equilibrium equation in the forward reaction:

b. Write the equilibrium equation in the reverse reaction:

c. What is the equilibrium constant

if [SiH4] = 0.45M; [O2] = 0.25M; [SiO2] = 0.15M; and [H2O] = 0.10M at equilibrium?

d. What is the equilibrium constant in the reverse reaction?

5. If [SiH4] = 0.34M; [O2] = 0.22M; [SiO2] = 0.35M; and [H2O] = 0.20M, what would be the reaction quotient?

1. Which direction would the reaction go? (Towards products or reactants?)

2A(g) + B(s) ↔ C(g)

1. For the reaction above, predict which way the reaction will shift: left, right, or none:

a. the concentration of B is increased ______

b. the amount of C is decreased ______

c. the amount of A is decreased ______

2. For the reaction below, what would speed up the reaction?

NH4+ + H2O ↔ NH3 + H3O+

a. the forward reaction

b. the reverse reaction

3. Define reversible reaction equilibrium:

4. An equilibrium shift to the right means:

a. more products are produced

b. more reactants are produced

c. nothing happens

5. Use this reaction for each of the following:

2NO2 (g) + O2 (g) ↔ 2NO3 (g)

a. Write the expression for the equilibriumconstant.

b. If gas concentrations are as follows, 2.10 M NO2, 1.75 M O2, and 1.00 M NO3, calculateKeq

c. Using Keq from part c, are the reactants or products favored?

d. Using Keq from part c, calculate [NO3] if [NO2] = [O2] = 4.3 x 10-6

POTENTIAL ENERGY diagrams

Endo: 1. graph ends high

2. A + B + heat  AB

3. H = (+)

Exo 1. graph ends low

2. A + B  AB + heat

3. H = (-)

Equilibrium Worksheet

For the reaction: SiH4(g) + O2(g) SiO2(g) + H2O(g)

1. Write the equilibrium equation in the forward reaction:

2. Write the equilibrium equation in the reverse reaction:

3. What is the equilibrium constant if [SiH4] = 0.45M; [O2] = 0.25M; [SiO2] = 0.15M; and [H2O] = 0.10M at equilibrium?

4. What is the equilibrium constant in the reverse reaction?

5. If [SiH4] = 0.34M; [O2] = 0.22M; [SiO2] = 0.35M; and [H2O] = 0.20M, what would be the reaction quotient?

1. Which direction would the reaction go? (Towards products or reactants?)

1. H2(g) + CO2(g)  H2O(g) + CO(g)

a)It is found at 986oC that there are 11.2 atm each of CO and water vapor and 8.8atm each of H2 and CO2 at equilibrium. Calculate the equilibrium constant.

b)If there were 8.8 moles of H2 and CO2 in a 500.0mL container at equilibrium, how many moles of CO(g) and H2O(g) would be present?

1. The equilibrium constant for the decomposition of COBr2 : COBr2 CO + Br2 (all gases)

Is 0.190 at 73oC. If the concentrations of both CO and Br2 are 0.402M, and the concentration of COBr2 is 0.950M, is the system at equilibrium? If not, which way does it proceed?

9. You place some COBr2 in a 5.0 L flask and heat it to form CO and Br2. If you want a Br2 concentration of 0.0500 M at equilibrium, how many grams of COBr2 will you use in the beginning?

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