5.0(2)

Other AP Chemistry unit study guides

AP Chemistry Ultimate Guide

Unit 1: Atomic Structure and Properties

Unit 2: Molecular and Ionic Compound Structure and Properties

Unit 3: Intermolecular Forces and Properties

Unit 4: Chemical Reactions

Unit 5: Kinetics

Unit 6: Thermodynamics

Unit 7: Equilibrium

Unit 8: Acids and Bases

Unit 9: Applications of Thermodynamics

Studying for another AP Exam?

Check out our other AP study guides

Tags

Chapter 10: Kinetics

Rate of a chemical reaction is the rate of change in concentration per unit of time, usually moles per liter that react each second.

**Reaction rates**are determined by measuring the concentration of a substance at different times during a reactionA

**kinetic curve**, or a**concentration versus time curve**, illustrates thisRate of product production (or reactant disappearance) is obtained by determine the slope of a kinetic curve

Concentration: If a substance is included in the rate law, increase the concentration (or pressure if a gas) will increase the reaction rate

Temperature: exponentially related to Kelvin temperature

Ability to meet: Gas or solutions react rapidly because of the individual molecules. Solids will react more rapidly if grinded finely

Catalyst present: Catalysts lower activation energy and provide a larger rate constant

Similar to giving a car more gas to make it go faster, if the

**concentration of reactants**is increased the rate will increase. Reactant concentration is directly related to the rate**Temperature**is carefully controlled in reactions because it changes the rate. Increasing temperature always increases reaction rate**Catalysts**increase reaction rate by lowering activation energy, or energy required for reaction to start. It does not appear in balanced equation

Rate law is the effect of concentration on a chemical reaction

Rate = k[A]^x[B]^y[C]^z

k is

**rate constant**.exponents of the concentrations are determined experiementally, not through the chemical equation itself such as the coefficients

Changing the concentration of one reactant while holding the other constant can determine if that one reactant has an effect on the rate, and therefore the exponent

For the reaction S2O8 + 3I- → 2SO4 + I3

Experiment No. | [S2O8] (mol/L) | [I-] (mol/L) | Initial Rate of Rxn |
---|---|---|---|

1 | .2 | .2 | 2.2x10^-3 |

2 | .4 | .2 | 4.4x10^-3 |

3 | .4 | .4 | 8.8x10^-3 |

This makes the rate law Rate = k[S2O8]^x[I-]^y

Experiment 1 and 2 can be used to find the exponent for S2O8 because I- is held constant

Experiment 3 and 2 can be used for [I-] exponent

To then find k, use the values from any ONE experiment and solve. any experiment values can be used.

**Order of reaction**is used to indicate the exponents in the rate lawAn exponent of one is

**first order**. The reaction previously is first order with respect to peroxydisulfate and first order with respect to iodide ionsIt is

**second order**overall because of the sum of the exponents

The

**Arrhenius equation**defines the relationship between the rate constant and temperaturek = Ae^(-Ea/RT)

k is rate constant

Ea is activation energy

R is the universal gas constant = 8.314

T is temperature in Kelvin

A is a proportionality constant

e is the base of natural logarithms

ln k = (-Ea/RT) + ln A

ln is natural log

A larger rate constant wil always be associated with the higher temperature

Activation energy always is positive

**Zero order**reactions have all exponents as zeroRate = k

This reactions does not depend on reactant concentration

Integrated rate equation is [A] = [A]0 - kt

[A]0 is the initial

Half life is t(½) = [A]0 / 2k

t(½) is half life

**First-order**reactions have an exponent sum of one.Rate = k[A]

As the concentration of A decreases, so does the rate

If the concentration is half the initial, the rate will be half the initial rate

This is the

**half life**

The integrated first order equation is

ln [A] 0 - ln [A]t = kt

t is time

k is rate constant

[A]0 is initial concentration

[A]t is concentration at some point in time

First order reactions have a constant half life

t(½) = .693/k

Most prominent first-order process is radioactive decay which follows first-order kinetics

After one half life, half of the isotope is left. After two half live, one quarter is left. After three half lives, one eighth is left

The half life of Pb 210 is 25 years. In a sample of 50 µg of Pb 210, how much is left after 100 years

3.1 µg

How many years of 50 µg of Pb 210 takes to decrease to 5.0µg ( t(½) =25yrs)

83 years

Rate = k[A]^2

Rate = k [A][B]

Integrated rate law

Half-life equation

**Collision Theory:**Rate of chemical equation is equal to the collision rate decreased by multiplying by an orientation factor and minimum energy factor.**Transition State Theory**: During the collision process, kinetic energy is converted to potential. If the potential energy exceeds or meets activation energy, the reaction occurs. The geometry of reactants changes as they become products, which is called the transition state as they convert from to the other and occurs when the max kinetic energy has been converted to potential energy

According to the transition state theory, as molecules approach each other, they slow down and potential energy increases. After they get as close as they can, the potential energy decreases and kinetic increases.

The minimum amount of kinetic energy that must be converted is the activation energy

Reaction profiles help determine if a reaction is endothermic or exothermic.

Heat of reaction (ΔH) can be determine from potential energy of product and reactants

ΔH = PE product - PE reactants

If ΔH is positive, heat is absorbed and is

**endothermic**potential energy of products is greater than reactants

If ΔH is negative, heat is release and is

**exothermic**reactants have more potential energy

Catalysts allows a reaction to speed up by providing an alternate reaction pathway with a lower activation energy. However, this does not change the amount of product formed or the equilibrium concentrations.

**Reaction mechanisms**are the series of steps that occur in a reaction from start to finish. the steps in between are call elementary reactions which add up to the overall balanced equations

**Elementary reactions**are usually one or two molecules breaking apart of combining, rarely three. The**coefficients are the exponents**from the rate lawThe slowest elementary step is the rate-determining step, or the rate-limiting step.

Example

For the reaction H2 +2ICl → I2 + HCl with the rate law Rate = k[H2][ICl] it has the two-step mechanism

H2 + ICl → HI +HCl

HI + ICl → I2 + HCl

The HI does not appear in the balance equation and is called and

**intermediate**.

Mechanisms with an intermediate in the slowest step must have the rate law adjusted to not include the intermediate in the rate law

Example

For the reaction 2NO + O2 → 2NO2

NO + O2 → NO3

NO3 + NO → NO2

The rate for the first step is Rate = k [NO][O2]

the rate for the second step includes and intermediate Rate = k [NO3][NO]

The rate of disappearance and formation are opposite and equal, which gives the equation

kf [NO][O2] = kr [NO3]

kf is formation and kr is reverse

Solve for NO3

[NO3] = (kf/kr)[NO][O2]

Substituting this into the elementary step rate law

Rate = k (kf/kr)[NO][O2][NO]

Combining the rates and rate constants

Rate = k[NO]^2[O2]

Rate of a chemical reaction is the rate of change in concentration per unit of time, usually moles per liter that react each second.

**Reaction rates**are determined by measuring the concentration of a substance at different times during a reactionA

**kinetic curve**, or a**concentration versus time curve**, illustrates thisRate of product production (or reactant disappearance) is obtained by determine the slope of a kinetic curve

Concentration: If a substance is included in the rate law, increase the concentration (or pressure if a gas) will increase the reaction rate

Temperature: exponentially related to Kelvin temperature

Ability to meet: Gas or solutions react rapidly because of the individual molecules. Solids will react more rapidly if grinded finely

Catalyst present: Catalysts lower activation energy and provide a larger rate constant

Similar to giving a car more gas to make it go faster, if the

**concentration of reactants**is increased the rate will increase. Reactant concentration is directly related to the rate**Temperature**is carefully controlled in reactions because it changes the rate. Increasing temperature always increases reaction rate**Catalysts**increase reaction rate by lowering activation energy, or energy required for reaction to start. It does not appear in balanced equation

Rate law is the effect of concentration on a chemical reaction

Rate = k[A]^x[B]^y[C]^z

k is

**rate constant**.exponents of the concentrations are determined experiementally, not through the chemical equation itself such as the coefficients

Changing the concentration of one reactant while holding the other constant can determine if that one reactant has an effect on the rate, and therefore the exponent

For the reaction S2O8 + 3I- → 2SO4 + I3

Experiment No. | [S2O8] (mol/L) | [I-] (mol/L) | Initial Rate of Rxn |
---|---|---|---|

1 | .2 | .2 | 2.2x10^-3 |

2 | .4 | .2 | 4.4x10^-3 |

3 | .4 | .4 | 8.8x10^-3 |

This makes the rate law Rate = k[S2O8]^x[I-]^y

Experiment 1 and 2 can be used to find the exponent for S2O8 because I- is held constant

Experiment 3 and 2 can be used for [I-] exponent

To then find k, use the values from any ONE experiment and solve. any experiment values can be used.

**Order of reaction**is used to indicate the exponents in the rate lawAn exponent of one is

**first order**. The reaction previously is first order with respect to peroxydisulfate and first order with respect to iodide ionsIt is

**second order**overall because of the sum of the exponents

The

**Arrhenius equation**defines the relationship between the rate constant and temperaturek = Ae^(-Ea/RT)

k is rate constant

Ea is activation energy

R is the universal gas constant = 8.314

T is temperature in Kelvin

A is a proportionality constant

e is the base of natural logarithms

ln k = (-Ea/RT) + ln A

ln is natural log

A larger rate constant wil always be associated with the higher temperature

Activation energy always is positive

**Zero order**reactions have all exponents as zeroRate = k

This reactions does not depend on reactant concentration

Integrated rate equation is [A] = [A]0 - kt

[A]0 is the initial

Half life is t(½) = [A]0 / 2k

t(½) is half life

**First-order**reactions have an exponent sum of one.Rate = k[A]

As the concentration of A decreases, so does the rate

If the concentration is half the initial, the rate will be half the initial rate

This is the

**half life**

The integrated first order equation is

ln [A] 0 - ln [A]t = kt

t is time

k is rate constant

[A]0 is initial concentration

[A]t is concentration at some point in time

First order reactions have a constant half life

t(½) = .693/k

Most prominent first-order process is radioactive decay which follows first-order kinetics

After one half life, half of the isotope is left. After two half live, one quarter is left. After three half lives, one eighth is left

The half life of Pb 210 is 25 years. In a sample of 50 µg of Pb 210, how much is left after 100 years

3.1 µg

How many years of 50 µg of Pb 210 takes to decrease to 5.0µg ( t(½) =25yrs)

83 years

Rate = k[A]^2

Rate = k [A][B]

Integrated rate law

Half-life equation

**Collision Theory:**Rate of chemical equation is equal to the collision rate decreased by multiplying by an orientation factor and minimum energy factor.**Transition State Theory**: During the collision process, kinetic energy is converted to potential. If the potential energy exceeds or meets activation energy, the reaction occurs. The geometry of reactants changes as they become products, which is called the transition state as they convert from to the other and occurs when the max kinetic energy has been converted to potential energy

According to the transition state theory, as molecules approach each other, they slow down and potential energy increases. After they get as close as they can, the potential energy decreases and kinetic increases.

The minimum amount of kinetic energy that must be converted is the activation energy

Reaction profiles help determine if a reaction is endothermic or exothermic.

Heat of reaction (ΔH) can be determine from potential energy of product and reactants

ΔH = PE product - PE reactants

If ΔH is positive, heat is absorbed and is

**endothermic**potential energy of products is greater than reactants

If ΔH is negative, heat is release and is

**exothermic**reactants have more potential energy

Catalysts allows a reaction to speed up by providing an alternate reaction pathway with a lower activation energy. However, this does not change the amount of product formed or the equilibrium concentrations.

**Reaction mechanisms**are the series of steps that occur in a reaction from start to finish. the steps in between are call elementary reactions which add up to the overall balanced equations

**Elementary reactions**are usually one or two molecules breaking apart of combining, rarely three. The**coefficients are the exponents**from the rate lawThe slowest elementary step is the rate-determining step, or the rate-limiting step.

Example

For the reaction H2 +2ICl → I2 + HCl with the rate law Rate = k[H2][ICl] it has the two-step mechanism

H2 + ICl → HI +HCl

HI + ICl → I2 + HCl

The HI does not appear in the balance equation and is called and

**intermediate**.

Mechanisms with an intermediate in the slowest step must have the rate law adjusted to not include the intermediate in the rate law

Example

For the reaction 2NO + O2 → 2NO2

NO + O2 → NO3

NO3 + NO → NO2

The rate for the first step is Rate = k [NO][O2]

the rate for the second step includes and intermediate Rate = k [NO3][NO]

The rate of disappearance and formation are opposite and equal, which gives the equation

kf [NO][O2] = kr [NO3]

kf is formation and kr is reverse

Solve for NO3

[NO3] = (kf/kr)[NO][O2]

Substituting this into the elementary step rate law

Rate = k (kf/kr)[NO][O2][NO]

Combining the rates and rate constants

Rate = k[NO]^2[O2]

5.0(2)

Other AP Chemistry unit study guides

AP Chemistry Ultimate Guide

Unit 1: Atomic Structure and Properties

Unit 2: Molecular and Ionic Compound Structure and Properties

Unit 3: Intermolecular Forces and Properties

Unit 4: Chemical Reactions

Unit 5: Kinetics

Unit 6: Thermodynamics

Unit 7: Equilibrium

Unit 8: Acids and Bases

Unit 9: Applications of Thermodynamics

Studying for another AP Exam?

Check out our other AP study guides

Tags