CHEM 162 - General Chemistry II
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exam_3_review.pdf
CHEM 162 Exam 3 Review page 1 of 4
CHEM 162: Exam 3 Study Guide CHAPTER 14: Chemical Kinetics chemical kinetics: study of the factors that influence reaction rates
reaction rate: a positive quantity expressing the concentration change with time
Use experimental data given to – determine a general reaction rate given
concentrations of reactants/products over time – determine the rate of disappearance/ consumption of a reactant or rate of
appearance/production of a product given the rate of disappearance/appearance of another reactant/product in the reaction
Determine reaction rate – Given experimental data of concentrations and
time.
Know the terms: rate law, rate constant (k)
Reaction Order: – zero-‐order, first-‐order, second-‐order – Determine overall order for a reaction given rate
law.
Distinguish between the instantaneous reaction rate and the average reaction rate.
Determine rate law using initial rates method – Given data of concentrations and rates. – Cannot be determined given only the balanced
chemical equation. half-‐life (t1/2): the time required for the concentration of a reactant to decrease by half Determine the reaction order given experimental data of reactant concentration over time. – Recognize the plots giving a straight line for
zero-‐order, first-‐order, and second-‐order reactions.
Determine the reaction order given experimental data of reactant concentration over time (Continued) – Recognize that the half-‐life is only constant for
first-‐order reactions. – Compare slopes for the first and last sets of data
to see if slope changes for each reaction order. Do calculations given integrated rate laws for zero-‐order, first-‐order, and second-‐order reactions. – Solve for concentration at a given time given
initial concentration and rate constant. – Carry out natural log (ln) calculations for 1st-‐
order reactions—review your algebra! – Solve for the time required for the
concentration to decrease to a given amount. – Solve for half-‐life given rate constant, k, or vice
versa.
COLLISION MODEL/THEORY: reactant molecules must collide to react
Activation Energy (Ea): minimum energy needed for chemical reaction
Reaction Rate and Temperature – As T↑, reaction rate ↑ since molecules move
faster and more molecules have activation energy.
Three Factors Affecting Reaction Rate 1. Concentration 2. Orientation of molecules 3. Temperature and Kinetic Energy
– Molecules move faster at higher temps – Molecules must have activation energy (Ea)
to react – Molecules must vibrate strongly
enough and collide with enough force to make and break bonds
CHEM 162 Exam 3 Review page 2 of 4
CHAPTER 14: Chemical Kinetics (Continued) TRANSITION STATE MODEL
Reaction Energy Profiles – Indicate the transition state, activation
energy (Ea) for reactants and products, ΔH for a reaction given a reaction energy profile
– Distinguish between the activated complex and the transition state
– Determine if a reaction is endothermic or exothermic
– Given a reaction energy profile for a multi-‐ step mechanism, determine the rate-‐ determining step based on largest Ea
Catalyst: substance added to a system that lowers the activation energy of a reaction. – Know catalysts provide an alternative
pathway that eases the collision geometry requirement
– Recognize that catalysts increase reaction rate without being consumed in reaction
– Know homogeneous versus heterogeneous catalysts
– Enzymes: catalysts with unique active sites that speed up biochemical reactions
Arrhenius Equation: k = A e–Ea/RT where A=frequency factor, Ea=activation energy, R=8.3145 J/mol·∙K, T=temperature in K – Know that A reflects the collision geometry
requirement – Know how temperature affects k – Be able to solve for any variable given the other variables and/or graphical data
Two-‐Point Equation of Arrhenius Equation: – Solve for the rate constants or activation
energy at two different temperatures
⎟ ⎠
⎞ ⎜ ⎝
⎛ −=⎟
⎠
⎞ ⎜ ⎝
⎛
21
a
1
2 T 1
T 1
R E
k k
ln
Reaction Mechanisms – sequence of steps by which a reaction occurs at the molecular level – The slowest step in a mechanism is the rate-‐
determining step – Given the reaction mechanism for a reaction,
determine the rate law – If a fast step is followed by a slow step,
express the rate law with respect to only the reactants (excluding intermediates)
– Given the experimentally determined rate law, determine the correct reaction mechanism given possible mechanisms
molecularity of a reaction – unimolecular, bimolecular, termolecular – Explain why unimolecular and bimolecular
steps are common but termolecular steps are very rare
– Determine the corresponding rate law for a given elementary step intermediate: species produced in an earlier step and consumed in later step of mechanism Distinguish between a catalyst and an intermediate – Be able to identify the correct mechanism for a reaction given information on any catalysts, intermediates, rate laws, and/or reaction orders
CHEM 162 Exam 3 Review page 3 of 4
CHAPTER 18: THERMODYNAMICS system: that part of the universe being studied surroundings: the rest of the universe outside the system 1st Law of Thermodynamics: Energy is neither created nor destroyed. Enthalpy change, ΔH = qreaction at constant pressure (e.g atmospheric pressure) – endothermic reaction: ΔH = +
– energy of reactants < energy of products; surroundings feel cooler after reaction;
– for physical changes, products have higher kinetic energy than reactants
– exothermic reaction: ΔH = – – energy of reactants > energy of products;
surroundings feel hotter after reaction; – for physical changes, products have lower
kinetic energy than reactants
spontaneous process: occurs without external intervention or stimulus
nonspontaneous process: only occurs with external intervention or stimulus
Entropy, S: measure of randomness factor – Ssolid < Sliquid < Sgas – S=0 only for a perfect crystalline solid at 0K (3rd Law of Thermodynamics) – S > 0 for all other substances, even naturally
occurring elements; the more complex the molecule the greater its absolute entropy, S°.
– Recognize ΔS is positive for a reaction that increases the # of moles of gas particles.
– Recognize if Ssys increases or decreases based on increased kinetic energy or changes in physical state.
Calculate ΔS° given S° data 2nd Law of Thermodynamics: For any
spontaneous process, the Suniv increases.
ΔSuniv = ΔSsys + ΔSsurr
– Recognize that the Ssurr= T
HsysΔ
Definitions of standard state 1. A gaseous substance with P=1 atm 2. An aqueous solution with a concentration of
1M at a pressure of 1 atm 3. Pure liquids and solids 4. The most stable form of an element at 1 atm
and 25°C G=Gibbs free energy – Be able to calculate ΔG° given ΔGf° data
Gibbs’ Equation: ΔG = ΔH -‐ T ΔS
If ΔG < 0 → a spontaneous reaction If ΔG > 0 → a nonspontaneous reaction;
reverse reaction is spontaneous. If ΔG = 0 → reaction is at equilibrium
– Given ΔH and ΔS indicate if a reaction is always spontaneous, never spontaneous, only spontaneous at high temperatures or at low temperatures.
Two Driving Forces for a chemical reaction – Exothermicity: a reaction occurs to form more
stable compounds with stronger bonds – Increased Entropy: a reaction occurs to
increase the number of available energy states (higher entropy)
Standard state conditions: ΔG° = ΔH° -‐ T ΔS° – Be able to calculate ΔG°, ΔH°, T in Kelvins,
and/or ΔS° given the other variables – Know standard state conditions
CHEM 162 Exam 3 Review page 4 of 4
BE PREPARED TO SOLVE PROBLEMS COMBINING
CONCEPTS FROM VARIOUS CHAPTERS.
THE USUAL GENERAL CHEMISTRY PERIODIC TABLE WILL BE PROVIDED, ALONG WITH THE EQUATIONS ON THE FOLLOWING PAGE.
THE FOLLOWING EQUATIONS WILL ALSO BE PROVIDED:
[A]t = −kt + [A]0 t1/2 k 2
[A]0 =
ln [A]t = −kt + ln [A]0 0
t
[A] [A]
ln = −kt t1/2 k
0.693 =
t[A] 1
− 0[A]
1 = kt t1/2 =
0[A] k 1
⎟ ⎠
⎞ ⎜ ⎝
⎛ −=⎟
⎠
⎞ ⎜ ⎝
⎛
21
a
1
2 T 1
T 1
R E
k k
ln
ΔSuniv = ΔSsys + ΔSsurr Ssurr= – T
HsysΔ
ΔG = ΔG° + RT ln Q ΔG° = – RT ln K
⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ −=
121
2
T 1
T 1
R ΔH-
K K
ln !
and ⎟⎟ ⎠
⎞ ⎜⎜ ⎝
⎛ − =
21
21
1
2
TT TT
R ΔH-
K K
ln !
