AP Physics B Review Session - Weeblychettyscience.weebly.com/uploads/8/5/2/7/8527773/ap-chem_eq_kin...FISD Equilibrium Review Session 2015 ... (g). The reaction was initiated by adding

AP Chemistry Review Session Saturday, April 25, 2015

8:30 am – 3:45 pm

Frisco ISD Transportation West Facility 10701 Dallas Pkwy Frisco, TX 75034

(just South of Wakeland HS on the Tollway frontage road) A light breakfast and pizza lunch will be provided.

Come review for the AP Chemistry Exam. You will get to hear from all the AP Chemistry

teachers in the district and work with students from other classes. This is a great way

to review what you already know and get a second perspective on what you already

know.

Format:

Short (20-ish minute) “crash course” review over each topic including common misconceptions and pitfalls

AP-style multiple choice questions from each unit along with multiple choice testing strategies

1-3 free response questions per topic to work as a group and finish at home

You will receive copies of all of the notes and questions to take home and study!

Bring a pencil, calculator, periodic table, notebook paper, and brain!

Time Topic/Presenter Topic/Presenter 8:30 – 9:00 Light breakfast

9:00 – 10:30 Bonding and IMFs Sternitzke

Equilibrium Smith

10:30 – 12:00 Kinetics McDonald

Thermodynamics Sadri/ Edwards-F

12:00 – 12:45 Lunch 12:45 – 2:15 Thermodynamics

Sadri/ Edwards-F Kinetics

McDonald

2:15 – 3:45 Equilibrium Smith

Bonding and IMFs Sternitzke

FISD Equilibrium Review Session 2015

General Equilibrium Any bond that can be formed can be broken. Chemical equilibrium is a dynamic, reversible state in which rates of

opposing processes are equal. At equilibrium, all species (reactants and products) are present and at a constant

concentration.

At equilibrium Keq = [products]z For equilibrium expressions, coefficients become exponents.

[reactants]x Pure solids and liquids are left out.

Magnitude of K can be used to determine whether the equilibrium lies toward the products or reactants.

K > 1 equilibrium position is to the right (products favored)

K < 1 equilibrium position is to the left (reactants favored)

K = 1 at equilibrium [reactants] [products]

Only temperature will change the value of K

Kc concentration in molarity (aqueous) Kp pressure (gases)

Kp = Kc (RT)∆n R = 0.0821 T = Kelvin

Q is the reaction quotient which provides a measure of the progress of the reaction Q < K shift right Q > K shift left

To calculate K, use a RICE table

R reaction

I initial concentration

C change in concentration (mole ratio!)

E equilibrium concentration

Le Chatelier’s Principle – an equilibrium system will shift to relieve a stress and re-establish equilibrium

Adding/subtracting a reactant/product; equilibrium position will shift away from an addition and toward a

removal (think about Q vs K)

Increasing/decreasing temperature; pay attention to sign on ΔH—shift away from temp increase and

toward temp decrease

Changing volume/pressure; increase in pressure causes decrease in volume—shift toward least moles of gas Solubility Equilibrium Even insoluble salts dissociate a little. Saturated salt solutions are at equilibrium.

The dissociation equation: MX3(s) M3+ + 3 X- becomes the Ksp expression: Ksp = [M3 ] [X-]3

A Ksp value is unique to a given salt and at a given temperature. Solubility indicates the amount of salt that dissociates to form a saturated solution. Basically, it indicates the equilibrium position. (You can have different solubilities with the same Ksp.)

When comparing solubilities of salts, always reference Ksp values (data) rather than solubility rules (observations).

Precipitation formation is dependent on concentrations and how they compare with Ksp. You use the given concentrations to calculate Q, the reaction coefficient and then compare with Ksp.

When Q = Ksp, the solution is saturated and the maximum amount of salt is in ion form.

When Q < Ksp, the solution is unsaturated and additional salt will dissociate Q = Ksp.

When Q > Ksp, the solution is supersaturated and precipitation will occur until Q = Ksp.

Insoluble salts with very small Ksp values will precipitate before salts with larger Ksp values.

Acid-Base Equilibrium

Acids donate protons Bases accept protons

Strong acids and bases dissociate 100% and form weak conjugates

HClO4, HClO3, HCl, HBr, HI, H2SO4, HNO3

Group I cations + OH-, Group II cations except Be2+ and Mg2+

Weak acids establish equilibrium in aqueous solution.

HC2H3O2 + H2O C2H3O2- + H3O

+ NH3 + H2O OH- + NH4+

Conjugate acid-base pair: HC2H3O2 and C2H3O - Conjugate acid-base pair: NH3 and NH4

+

Kw = 1.0 x 10-14 = Ka x Kb = [H+] [OH-] acidic [H+] > [OH-] basic [H+] < [OH-]

pH = -log [H+] pOH = -log [OH-] pH + pOH = 14

Ka and Kb—acid/base dissociation constant; larger value indicates more dissociation

Ka/Kb calculations—use a RICE table

R HA + H2O A- + H3O+ Ka = x2

I 0.50 0 0 0.50 – x ignore x

C -x +x +x

E 0.50 – x x x

Salt hydrolysis: salt can be acid, base, neutral depending on strength of acid/base reactants

Pull off weak end and add water. If end up w/ OH- then use Kb for calculation; if H3O+, then use Ka calculation

Buffers are solutions that contain a weak acid/conjugate base or weak base/conjugate acid. Because they have both constituents, they can neutralize added acid or base with the pH changing much. Optimal buffering occurs when the molarities of the constituents are equal. Useful formula for calculating the pH of a buffer solution:

[ ]acid form

H Kabase form

Titration—using a known concentration of a solution to determine an unknown concentration

strong acid/strong base pH = 7 at eq point

strong acid/weak base pH at equivalence point >7

weak acid/strong base pH at equivalence point < 7

4 areas of titration curve

1. Beginning—pH calculated using M of strong acid/base or Ka/Kb for weak acid/base

2. Buffer region—pH calculated using neutralization to determine new concentration for strong acid/base

Or buffer calculation for weak acid/base

3. Equivalence point—moles acid = moles base = moles salt

pH calculated using salt hydrolysis and Ka or Kb

4. End point—one drop past equivalence point

pH calculated using excess acid or base; determine new concentration using total volume

1. Consider N2 (g) + O2 (g) ↔ 2 NO (g). The reaction was initiated by adding 15.0 moles of NO to a 1.0-L flask.

At equilibrium, 3.0 moles of oxygen are present in the 1.0-L flask. The value of Keq must be A. 0.33 B. 3.0 C. 5.0 D. 9.0

2. In the reaction 3 W + X ↔ 2 Y + Z, all substances are gases. The reaction is initiated by adding an equal

number of moles of W and of X. When equilibrium is reached, A. [Y] = [Z] B. [X] = [Y] C. [W] = [X] D. [X] > [W]

3. Consider the reaction system for the endothermic decomposition below at equilibrium in a 2.0 liter sealed rigid

flask at 298 K. PH3PCl3 (s) ↔ PH3 (g) + PCl3 (g) Keq = 3.5 x 10-2

When the vessel containing the system is immersed in an ice bath, all of the following occur EXCEPT A. The total pressure decreases. B. The volume of PH3 (g) decreases. C. The partial pressure of PCl3 (g) decreases. D. The number of molecules of PH3PCl3 (s) increases.

4. Consider the equilibrium H2 (g) + I2(g) ↔ 2 HI (g). In a closed system, the initial partial pressure of

hydrogen is 1.25 atm and the initial partial pressure of iodine is 1.75 atm. At equilibrium, the partial pressure of hydrogen is 1.00 atm. Which expression gives the value of Kp for this system?

A. (0.50)2 C. (0.50)2 (1.00) (1.75) (1.00) (1.50) B. (0.25)2 D. (0.25)2 (1.25) (1.75) (1.00) (1.50)

5. Consider a system at equilibrium according to the equation N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

If Ar (g) is added to such an equilibrium system at constant volume and temperature, the total pressure A. Increases and the number of NH3 molecules present remains the same. B. Decreases and the number of NH3 molecules present increases. C. Remains the same and the number of NH3 molecules present remains the same. D. Increases and the number of NH3 molecules present increases.

HgO(s) + 4 I- + H2O HgI42- + 2 OH- H < 0

6. Consider the equilibrium above. Which of the changes below will increase the concentration of HgI4

2-? A. Adding a catalyst B. Adding 6 M HNO3 C. Increasing the mass of HgO present D. Increasing the temperature

Questions 7 – 9 should be answered using the following responses for the reaction which follows. o

2 2C(s) + H O(g) H (g) + CO(g) ΔH = 131 kJ/mol

A. Some C(s) is added B. The temperature is increased C. The container is compressed D. Some H2O(g) is added

7. Which change will alter the value of the equilibrium constant, K? 8. Which change will not affect the position of the equilibrium? 9. Which change will decrease the quantity of H2(g) at equilibrium?

10. What is the value of the equilibrium constant for the reaction 2 2 2 4( ) 2 ( ) ( )N g O g N O g in terms of the

K values for the reactions:

2 2 1

2 2 4 2

1 1N (g) + O (g) NO(g)

2 2

2 NO(g)+ O (g) N O (g)

K

K

A. K1 + K2 B. K1

2 + K2 C. 2 K1 × K2 D. K1

2 × K2

11. What is the molar solubility in water of Ag2CrO4? (The Ksp for Ag2CrO4 is 8 x 10-12.) A. 2 x 10¯12 M

B. (4 x 10¯12 M)1/2

C. (4 x 10¯12 M)1/3

D. (2 x 10¯12 M)1/3

12. The Ksp for BaF2 is 2.4 10-5. When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M Ba(NO3)2, will a precipitate form?

A. No, because Q is 1 10-12 and since it is less than Ksp no precipitate will form.

B. Yes, because Q is 1 10-12 and since it is less than Ksp a precipitate will form.

C. No, because Q is 1.25 10-7 and since it is less than Ksp no precipitate will form.

D. Yes, because Q is 1.25 10-7 and since it is less than Ksp a precipitate will form.

13. The solubility of manganese (II) hydroxide, Mn(OH)2, is 2.2 10-5 M. What is the Ksp of Mn(OH)2?

A. 1.1 10-14

B. 4.3 10-14

C. 2.1 10-14

D. 4.8 10-10

14. Magnesium hydroxide, Mg(OH)2, has a Ksp = 6.2 × 10-10. The solubility of Mg(OH)2 will be lowest in 1.0 L of which of the following? A. 0.10 M HCl B. 0.10 M NaOH

C. 0.10 M MgCl2

D. Pure H2O

AgBr (Ksp = 5.4 × 10-13) Ag2CO3 (Ksp = 8.0 × 10-12) AgCl (Ksp = 1.8 × 10-10)

15. On the basis of the Ksp values above, what is the order of solubility from least soluble to most soluble for these

compounds? A. AgBr < Ag2CO3 < AgCl B. AgBr < AgCl < Ag2CO3 C. AgCl < Ag2CO3 < AgBr D. Ag2CO3 < AgBr < AgCl

16. Which quantity is the same for separate 25 mL portions of 1.0 M strong and weak acids (Ka 1 x 10-5)? A. initial pH B. pH at equivalence point of titration with 1.0 M NaOH C. percent ionization D. volume of 1.0 M NaOH to reach equivalence point

17. If the dissociation of water, + -

2H O H + OH is endothermic, and some room temperature water is heated

to boiling, which of the following is true? A. The pH will be 7; the solution will be acidic. B. The pH will be greater than7, but the solution will be neutral. C. The pH will be less than 7, but the solution will be neutral. D. The pH will be less than 7; the pOH will be less than7, and the solution will be acidic.

18. Solution 1 consists of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2. Solution 2 consists of 0.10 M HC2H3O2 and 0.10 M NaC2H3O2. A. Solution 1 is a buffer but solution 2 is not. B. Solution 2 is a buffer but solution 1 is not. C. Both solutions are buffers with the same buffering capacity. D. Both solutions are buffers with the same pH.

19. Which acid forms the strongest conjugate? HF Ka = 7.2 x 10-4 HNO2 Ka = 4.0 x 10-4 HCO2H Ka = 1.8 x 10-4 HC3H5O3 Ka = 1.38 x 10-4

A. HF B. HNO2 C. HCO2H D. HC3H5O3

20. A 0.50 M solution of an unknown acid has a pH = 4.0. Of the following, which is the acid in this solution? A. HBr (strong acid) B. HF (Ka = 6.8 x 10-4) C. HOCl (Ka = 2.0 x 10-8) D. C6H5OH (Ka = 1.0 x 10-10)

21. Which is the best description of the concentration of ions in solution when 0.050 mol OH- (aq) is added to 1.0

liter of 0.10 M solution of NaH2PO4? (Assume no change in volume.) [H2PO4

-] [HPO42-] [PO4

3-] A. negligibly small 0.050 negligibly small B. negligibly small negligibly small 0.050 C. 0.050 0.050 negligibly small D. 0.050 negligibly small 0.050

2   2 C s CO g CO g

Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are recorded in the table below.

(a) Write the expression for the equilibrium constant, Kp , for the reaction.

(b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume of the solid

carbon is negligible.)

(c) For the reaction mixture at equilibrium at 1,160 K , the partial pressure of the CO2(g) is 1.63 atm. Calculate

(i) the partial pressure of CO(g) , and (ii) the value of the equilibrium constant, Kp .

(d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at

equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.)

In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s) , plus CO(g) and CO2(g) , each at a partial pressure of 2.00 atm at 1,160 K. (e) Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches

equilibrium. Justify your prediction with a calculation.

Time (hours)

Total Pressure of Gases in Container at 1,160 K (atm)

0.0 5.00

2.0 6.26

4.0 7.09

6.0 7.75

8.0 8.37

10.0 8.37

4 3NH Cl(s) NH (g) + HCl(g) When solid ammonium chloride is heated, it decomposes as represented above. The value of Kp for the reaction is 0.0792 at 575 K. A 10.0 g sample of solid ammonium chloride is placed in a rigid, evacuated 3.0 L container that is sealed and heated to 575 K. The system comes to equilibrium with some solid NH4Cl remaining in the container.

(a) Write the expression for the equilibrium constant for the reaction in terms of partial pressures (i.e., Kp).

(b) Calculate the partial pressure of NH3(g), in atm, at equilibrium at 575 K.

(c) A small amount of NH3(g) is injected into the equilibrium mixture in the 3.0 L container at 575 K. (i) As the new equilibrium is being established at 575 K, does the amount of NH4Cl(s), in the container

increase, decrease, or remain the same? Justify your answer.

(ii) After the new is established at 575 K, is the value of Kp greater than, less than, or equal to the value before the NH3(g) was injected into the container? Justify your answer.

(d) When the temperature of the container is lowered to 500 K, the number of moles of NH3(g) in the container decreases. On the basis of this observation, is the decomposition of NH4Cl(s) endothermic or exothermic? Justify your answer.

In another experiment, 20.00 mL of 0.800 M NH4Cl(aq) is prepared. The ammonium ion reacts with water according to

the equation : + +

4 2 3 3NH (aq) + H O( ) NH aq) + H O (aq) .

(e) Calculate the value of the equilibrium constant for the reaction of the ammonium ion with water. (At 25C the value of Kb for NH3 is 1.8 × 10-5.)

(f) A solution is prepared by mixing 20.0 mL of 0.800 M NH3(aq) with 20.0 mL of 0.800 M NH4Cl(aq). Assume volumes are additive.

(i) Is the solution acidic, basic, or neutral? Justify your answer.

(ii) Calculate the pH of the solution that would result from adding 0.0200 mol of HCl to the solution. Assume that the addition of the HCl does not change the volume of the solution.

Several reactions are carried out using AgBr, a cream-colored silver salt for which the value of the solubility-product constant, Ksp, is 5.0 × 10-13 at 298 K. (a) Write the expression for the solubility-product constant, Ksp, of AgBr. (b) Calculate the value of [Ag+] in 50.0 mL of a saturated solution of AgBr at 298 K. (c) A 50.0 mL sample of distilled water is added to the solution described in part (b), which is in a beaker with some

solid AgBr at the bottom. The solution is stirred and equilibrium is reestablished. Some solid AgBr remains in the beaker. Is the value of [Ag+] greater than, less than, or equal to the value you calculated in part (b)? Justify your answer.

(d) Calculate the minimum volume of distilled water, in liters, necessary to completely dissolve a 5.0 g sample of

AgBr(s) at 298 K. The molar mass of AgBr is 188 g/mol. (e) A student mixes 10.0 mL of 1.5 × 10-4 M AgNO3 with 2.0 mL of 5.0 × 10-4 M NaBr and stirs the resulting mixture.

What will the student observe? Justify your answer with calculations. (f) The color of another salt of silver, AgI(s), is yellow. A student adds a solution of NaI to a test tube containing a

small amount of solid, cream-colored AgBr. After stirring the contents of the test tube, the student observes that the solid in the test tube changes color from cream to yellow.

(i) Write the chemical equation for the reaction that occurred in the test tube. (ii) Which salt has the greater value of Ksp: AgBr or AgI? Justify your answer.

1

Kinetics Review 2015

Vocabulary:

activated complex

activation energy, Ea

Arrhenius equation

bimolecular

catalyst (hetero / homo)

Collision Theory

elementary steps

first order

half-life

initial rate

instantaneous rate

integrated rate law

intermediate

kinetics

molecularity

molecular orientation

overall reaction order

rate constant, k

rate determining step

rate expression

rate of reaction

reaction mechanism

rate law

second order

termolecular

transition state

unimolecular

zero order

Key Concepts:

Collision Theory: The more effective the collisions, the faster the reaction rate

Reactions form products when effective collisions occur between reactants:

1) with the correct collision geometry (orientation) and

2) with sufficient energy to overcome activation energy

FACTOR AFFECT ON RATE COLLISION THEORY

Concentration

of Reactants

Increasing concentration increases the rate (only

reactants in the rate determining step affect the rate)

More molecules = more effective

collisions

Temperature Increasing temperature increases the rate (estimate

that 10°C increase doubles the rate)

Higher kinetic energy of

molecules = more collisions and

greater effectiveness of collisions

Catalyst

Increase the rate of reaction by lowering the

activation energy needed; NOT CONSUMED

during the reaction

Less energy needed to begin

reacting = more effective

collisions

Surface Area

of Reactants Increasing surface area increases the rate

More area for collisions to occur

= more effective collisions

Reaction Rates: the Δ in concentration of a reagent per unit of time

reactant rates are always negative (disappearing/decreasing)

product rates are always positive (appearing/increasing)

2 NO2(g) O2(g) + 2 NO(g)

Relative Rates: Relates the products and reactions to each other based on the balanced equation

Rate of Reaction = -

= +

= +

Work MC # 1, 2, 3 FRQ: 1990 (a,b), 1998 (b)

Work MC # 4, 5, 6 FRQ: 1991 (c)

2

Rate Law/Expression: Rate = k [A]n [B]

m [C]

p where n, m, and p are rate orders (and can be fractions)

k = rate constant

Overall Rate Order = n + m + p

*** Orders must come from experimental data; data given in a table, graph, or mechanism –

NOT the coefficients from the balanced equation!!***

Integrated Rate Law:

Rate

Order Rate Law Integrated Rate Law

Straight

Line Plot Rate Constant

Units

for k

0 Rate = k[A]° = k [A] = –kt + [A]0 [A] vs. t

slope = -k k = | slope | M s

–1

1 Rate = k [A] ln[A] = –kt + ln[A]0 ln[A] vs. t

slope = -k k = | slope | s

–1

2 Rate = k [A]2

[A] –1

vs. t

slope = k k = | slope | M

–1 s

–1

Zero Order First Order Second Order

Half-Life: 1st Order: t1/2 = 0.693 / k MOST nuclear decay reactions are 1

st order

Mechanisms: sequence of elemental steps; rate dependent on slowest step; must meet 2 criteria:

1) elemental reactions up to the slowest step add up to the overall reaction

2) rate law must be consistent with mechanism

Work MC # 7, 8, 9, 10, 11, 12, 13, 14 FRQ: 1991 (a), 1998 (c)

Work MC # 15, 16, 17, 18, 19, 20 FRQ: 1991 (b), 1996 (d), 1998 (b)

Work MC # 21, 22, 23 FR: 1990 (c,d), 1991 (d), 1996 (a,b)

3

reaction intermediates – produced in 1 step and consumed in another

catalyst – goes in and comes out unchanged; NOT included in the final rxn

Arrhenius Equation: relates temperature and activation energy to concentration; represents the

fraction of collisions with sufficient energy to produce a reaction

Slope = -Ea/R ; Since this is solving for Activation Energy, be sure to use ENERGY “R”

Boltzman Distribution Graph:

- Distribution of energy at a given temperature

General Rule of Thumb:

- Every 10°C increase in temperature

will cause the number of collisions

with the required Ea to double.

T2 has a greater number of effective collisions because more particles have the required Ea.

T1 has a fewer number of effective collisions because fewer particles have the required Ea.

Work MC # 24, 25 FRQ: 1996 (d), 1998 (a)

4

Kinetics Multiple Choice AP Chemistry Review Day

1. As the temperature of a reaction is increased, the rate of the reaction increases because the A) reactant molecules collide less frequently B) reactant molecules collide more frequently and with greater energy per collision C) activation energy is lowered D) reactant molecules collide less frequently and with greater energy per collision E) reactant molecules collide more frequently with less energy per collision

2. A catalyst can increase the rate of a reaction __________. A) By changing the value of the frequency factor (A) B) By increasing the overall activation energy (Ea) of the reaction C) By lowering the activation energy of the reverse reaction D) By providing an alternative pathway with a lower activation energy E) All of these are ways that a catalyst might act to increase the rate of reaction.

3. According to collision theory, which of the following factors does NOT influence the rate of reaction? A) collision frequency B) collision energy C) collision orientation D) collision rebound direction E) none of these

__________________________________________________________________________________________

4. At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2N2O5 (g) 4NO2 (g) + O2 (g). When the rate of formation of NO2

is 5.0 x10-4M/s, the rate of decomposition of N2O5 is __________ M/s.

A) 2.2 x 10-3

B) 1.4 x 10-4 C) 1.01 x 10-4 D) 2.5 x 10-4 E) 5.5 x 10-4

5. Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O

A) 2[O ]1

7 t

B) 2[NO ]1

4 t

C) 2[H O]1

6 t

D) 3[NH ]1

4 t

E) All of the above are valid expressions of the reaction rate.

5

6. The reaction 3O2 2O

3 is proceeding with a rate of disappearance of O

2 equal to 0.60 mol/Ls. What is

the rate of appearance of O3, in mol/Ls?

A) 0.60 B) 0.40 C) 0.10 D) 0.90 E) 1.20

___________________________________________________________________________________________

7. If the rate law for the reaction, 2A + 3B products is first order in A and second order in B, then the rate law is R = ______________ A) k[A][B]

B) k[A]2[B]3

C) k[A][B]2

D) k[A]2[B] E) k[A]2[B]2

8. The overall order of a reaction is 2. The units of the rate constant for the reaction are __________. A) M/s B) M-1s-1 C) 1/s D) 1/M E) s/M2

9. The kinetics of the reaction below were studied and it was determined that the reaction rate increased by a factor of 9 when the concentration of B was tripled. The reaction is __________ order in B. A + B → P A) zero

B) first C) second D) third E) one-half

10. A reaction was found to be zero order in A. Increasing the concentration of A by a factor of 3 will cause the reaction rate to __________.

A) remain constant B) increase by a factor of 27 C) increase by a factor of 9 D) triple E) decrease by a factor of the cube root of 3

11. The rate law for a reaction is rate = k[A][B]2 Which one of the following statements is false? A) The reaction is first order in A. B) The reaction is second order in B. C) The reaction is second order overall. D) k is the reaction rate constant E) If [B] is doubled, the reaction rate will increase by a factor of 4.

6

12. The reaction I- + OCl- IO- + Cl- is first order with respect to I- and first order with respect to OCl-.

The rate constant is 8.0 x 10-2

L/mols. What is the rate of reaction when [I-] = 0.10M and [OCl-] = 0.20M?

A) 4.4 x 10-4

M/s

B) 1.60 x 10-3

M/s

C) 4.9 x 10-2

M/s

D) 2.2 x 10-4

M/s

E) 2.4 x 10-5

M/s

13. A reaction and its rate law are given below. When [C4H6] = 2.0 M, the rate is 0.2 M/s.

What is the rate when [C4H6] = 4.0 M?

2 C4H

6 C

8H

12 Rate = k[C4H6]

2

A) 0.050 M/s B) 0.212 M/s C) 0.106 M/s D) 0.800 M/s E) 0.022 M/s

14. Below is some rate data for the hypothetical reaction, 2A + B C. What is the rate law for this reaction?

Experiment [A]o [B]o Rate (M/s)

1 3.0 M 1.0 M 0.100

2 3.0 M 2.0 M 0.400

3 6.0 M 1.0 M 0.100

A) Rate = k[A][B]

B) Rate = k[A]2[B]

C) Rate = k[A][B]2

D) Rate = k[A]2[B]

2

E) Rate = k[B]2

__________________________________________________________________________________________

15. For a first-order reaction, a plot of __________ versus __________ is linear. A) ln [A]t, 1/t B) ln [A]t, t C) 1/[A]t, t D) [A]t, t E) t, 1/[A]t

7

16. The acid catalyzed decomposition of hydrogen peroxide is a first order reaction with the rate constant given below. For an experiment in which the starting concentration of hydrogen peroxide is 0.20 M, what is the concentration of H

2O

2, 50 minutes after the reaction begins?

2H2O

2 2H

2O + O

2 k=6.93 x 10

-2 min

-1

A) 0.061 M B) 0.00625 M C) 0.0010 M D) 0.000658 M E) 0.0125 M

17. The graph shown below depicts the relationship between concentration and time for the following chemical reaction.

The slope of this line is equal to __________.

A) k B) -1/k C) ln [A]0 D) -k E) 1/k

18. If 87.5 percent of a sample of pure X decays in 24 days, what is the half-life of X? (A) 6 days (B) 8 days (C) 12 days (D) 14 days (E) 21 days

19. The reaction A → B is first order in [A]. Consider the following data.

Time (s) [A] (M)

0 1.60

10 0.40

20 0.10

The half-life of this reaction is __________ s. A) 0.97 B) 7.1 C) 5.0 D) 3.0 E) 0.14

8

20. What is the rate constant for a first order reaction for which the half-life is 69.0 sec?

A) 0.010 sec-1

B) 3.45 sec-1

C) 0.170 sec-1

D) 0.0618 sec-1

E) 69.0 sec-1

__________________________________________________________________________________________

21. Which statement is true about a reactant that appears in the balanced equation for a reaction but does not appear in the rate equation?

A) It is a catalyst B) It is an inhibitor C) Its concentration is too low to be important D) It takes part in the reaction after the rate determining step E) It is an intermediate

22. Assume a reaction occurs by the mechanism given below. What is the rate law for the reaction?

A + B C fast

C D slow A) Rate = k[A][B][C]

B) Rate = k[A]2

C) Rate = k[A][B] D) Rate = k[A][B]/[D] E) Rate = k[A]

23. Which of the following is true? A) If we know that a reaction is an elementary reaction, then we know its rate law. B) The rate-determining step of a reaction is the rate of the fastest elementary step of its

mechanism. C) Since intermediate compounds can be formed, the chemical equations for the elementary

reactions in a multistep mechanism do not always have to add to give the chemical equation of the overall process.

D) In a reaction mechanism, an intermediate is identical to an activated complex. E) All of the above statements are true.

___________________________________________________________________________________________

24. Which of the following is a graph that describes the pathway of reaction that is endothermic and has high activation energy?

A B C. D E

9

25. What would cause the change in the kinetic energy diagrams as shown?

A) Decreasing the H B) Decreasing the temperature C) Increasing the surface area D) Addition of a catalyst E) Increasing reactant concentration

10

Kinetics Multiple Choice - KEY AP Chemistry Review Day

1. B 2. D 3. D

_________________

4. D 5. E 6. B

_________________ 7. C 8. B 9. C 10. A 11. C 12. B 13. D 14. E

_________________ 15. B 16. B 17. D 18. B 19. C 20. A

_________________

21. D 22. C 23. A

_________________ 24. B 25. D

11

KINETICS FREE RESPONSE PRACTICE

1990 Q7 (Concepts: collision theory, factors affecting rate, rate law expression using mechanisms)

Consider the following general equation for a chemical reaction.

A(g) + B(g) C(g) + D(g)

(a) Describe the two factors that determine whether a collision between molecules of A and B results in a reaction.

(b) How would a decrease in temperature affect the rate of the reaction shown above? Explain your answer.

(c) Write the rate law expression that would result if the reaction proceeded by the mechanism shown below.

A + B ↔ [AB] (fast) [AB] + B → C + D (slow)

(d) Explain why a catalyst increases the rate of a reaction but does not change the value of the equilibrium constant for that reaction.

12

1991 Q3 Concepts: rate law expression using table logic, calculation of rate constant, k, with units, relative rates, mechanisms

The following results were obtained when the reaction represented above was studied at 25°C.

2 ClO2(g) + F2(g) → 2 ClO2F(g)

Experiment Initial [ClO2], (mol.L−1)

Initial [F2], (mol.L−1)

Initial Rate of Increase of

[ClO2F], (mol.L−1.sec−

1)

1 0.010 0.10 2.4 × 10−3

2 0.010 0.40 9.6 × 10−3

3 0.020 0.20 9.6 × 10−3

(a) Write the rate law expression for the reaction above. (b) Calculate the numerical value of the rate constant and specify the units. (c) In experiment 2, what is the initial rate of decrease of [F2]? (d) Which of the following reaction mechanisms is consistent with the rate law developed in (a). Justify

your choice.

I. ClO2 + F2 ↔ ClO2F2 (fast) ClO2F2 → ClO2F + F (slow) ClO2 + F → ClO2F (fast)

II. F2 → 2 F (slow) 2 (ClO2 + F → ClO2F) (fast)

13

1996 Q8 Concepts: mechanisms, factors affecting the rate

The reaction between NO and H2 is believed to occur in the following three-step process.

NO + NO ↔ N2O

2 (fast)

N2O

2 + H

2 → N

2O + H

2O (slow)

N2O + H

2 → N

2 + H

2O (fast)

(a) Write a balanced equation for the overall reaction.

(b) Identify the intermediates in the reaction. Explain your reasoning.

(c) From the mechanism represented above, a student correctly deduces that the rate law for the reaction is rate = k[NO]2[H

2]. The student then concludes that (1) the reaction is third-order and (2) the

mechanism involves the simultaneous collision of two NO molecules and an H2 molecule. Are

conclusions (1) and (2) correct? Explain.

(d) Explain why an increase in temperature increases the rate constant, k, given the rate law in (c).

14

1998 Q6 Concepts: potential energy diagram, graphing with rate laws, theoretical calculation of rate constant, k, with units, factors affecting rate

Answer the following questions regarding the kinetics of chemical reactions.

(a) The diagram below at right shows the energy pathway for the reaction O3 + NO → NO

2 + O

2.

Clearly label the following directly on the diagram.

(i) The activation energy (Ea) for the forward reaction

(ii) The enthalpy change (∆H) for the reaction

(b) The reaction 2 N2O

5 → 4 NO

2 + O

2 is first order with respect to N

2O

5.

(i) Using the axes at right, complete the graph that represents the change in [N

2O

5] over time as the reaction

proceeds.

(ii) Describe how the graph in (i) could be used to find the reaction rate at a given time, t.

(iii) Considering the rate law and the graph in (i), describe how the value of the rate constant, k, could be determined.

(iv) If more N

2O

5 were added to the reaction mixture at constant temperature, what would be the

effect on the rate constant, k? Explain.

15

(c) Data for the chemical reaction 2A → B + C were collected by measuring the concentration of A at 10-minute intervals for 80 minutes. The following graphs were generated from analysis of data.

Use the information in the graphs above to answer the following.

(i) Write the rate-law expression for the reaction. Justify your answer.

(ii) Describe how to determine the value of the rate constant for the reaction.

1

THERMOCHEMISTRY REVIEW AP Chemistry 2014-15

Vocabulary: System

Surroundings

Work

Temperature

Absolute Zero

Kinetic Energy

Potential Energy

Chemical Energy

Bond Dissociation Energy

Calorie (kilocalorie)

Joule (kilojoule)

Calorimeter

Calorimetry

Conduction

Convection

Endothermic

Exothermic

Enthalpy

Entropy

Free Energy

Specific Heat Capacity

Spontaneous

Non-Spontaneous

Standard State

Quick Summary:

Thermodynamics is the study of heat and its transformations. Energy can be described as either Kinetic Energy,

the energy of motion, or Potential Energy, stored energy. Chemical Energy is the potential energy contained in

bonds. Energy can be converted from one form to another, but according to the Law of Conservation of Energy,

cannot be created or destroyed.

Energy is defined as the capacity to do work. Energy exchanges occur between a system and the surroundings.

The system is the part of the universe that we are studying. The surroundings is the rest of the universe.

Heat is the transfer of thermal energy. Temperature is a measure of the thermal energy in a system, and is directly

proportional to the average kinetic energy of the molecules in the system.

The SI unit for energy is the Joule (J). The calorie (cal) is a unit of energy defined as the amount of energy

needed to raise 1 gram of water 10C. The relationship between a Joule and a calorie is: 1 cal = 4.184 J

** A calorie should not be confused with a nutritional Calorie which is really a kilocalorie: 1 Calorie = 1000 cal

When a reaction results in heat energy being given off, it is Exothermic. When a reaction absorbs heat energy, it

is Endothermic. These heat changes that occur with chemical reactions can be expressed using an Energy

Diagram, that plots the potential energy of the Reactants before the reaction, the Products after the reaction, and

the Activation Energy, or the amount of energy that must be overcome in order to produce a chemical reaction.

Work MC # 1, 2, 3, 4

2

Thermochemical Relationships:

Calculating Enthalpy:

Enthalpy is a measure of energy that can be released as heat in a constant pressure condition. Enthalpy changes

are associated with physical and chemical changes. Enthalpy changes for a reaction (H) are given in kJ/mol.

There are four ways to calculate H for a chemical reaction:

1) BOND ENERGIES – Energy in needed to break bonds in chemical processes (Endothermic) and Energy

is released when new bonds are formed in chemical processes (Exothermic)

Draw the Lewis structure of all reactants and products

Count all bonds broken and all bonds formed in the process

H = bonds broken) – (bonds formed)

2) CALORIMETRY – use temperature change and Specific Heat Capacity to measure heat absorbed by water

that is released from a chemical process

Specific Heat Capacity is the energy needed to change one gram of a substance one degree (J/g0C)

Q = mCT

Q released = Q absorbed by water

Sometimes the calorimeter absorbs some of the heat and this must be subtracted

3) HESS’ LAW – the overall enthalpy of a reaction is the sum of the reaction enthalpies of the steps into

which that reaction can be divided

manipulate the given equations for the steps of the reactions so they add up to the overall equation

When reversing a reaction, the new H is of opposite sign

When a reaction is multiplied or divided, the new H is also multiplied or divided

4) ENTHALPY OF FORMATION – given as the enthalpy change observed in one mole of a compound

when formed form its constituent elements

H0

f of elements in their standard state is ZERO

H = H0fproducts) – H

0f (reactants)

Enthalpy Entropy

Work MC # 5, 6, 7, 8, 9

EQUILIBRIUM ELECTROCHEMISTRY

Gibb’s Free Energy

Spontaneity

3

Calculating Entropy:

Entropy is a measure of randomness (or disorder). Increases in entropy occur when 1) a pure solid or liquid

dissolves in a solvent, 2) when a gas molecule escapes from a solvent, 3) when molecules become more complex,

4) when phase changes occur from solid to liquid to gas, and 5) when the number of particles increases.

Unlike enthalpy, S is only zero for a perfect crystal at absolute zero.

S = S0

fproducts) – S0f (reactants)

Calculating Free Energy:

Gibb’s Free Energy (or simply Free Energy) is the amount of energy that is available to do work. Free Energy is

the driving force of a reaction.

A change in Free Energy (G) is observed when reactants are converted into products:

G = G0fproducts) – G

0f (reactants)

The relationship between Enthalpy and Entropy is shown by: G = H - TS

Pay attention to UNITS when using this equation!

Enthalpies are usually measured in kJ

Entropies are usually measured in J

Determining Spontaneity:

G is the best indicator chemists have as to whether or not a reaction is spontaneous:

If G > 0, reaction is NOT spontaneous because energy must be added to cause reaction to occur

If G < 0, reaction IS spontaneous

If G = 0, reaction is at equilibrium

H S G

- (exo) + - Spontaneous at all temperatures

+ (endo) - + Non-spontaneous at all temperatures

- (exo) - -

+

At low temperatures, spontaneous

At high temperatures, non-spontaneous

+ (endo) + +

-

At low temperatures, non-spontaneous

At high temperatures, spontaneous

** Exothermic processes favor spontaneity, but at high temperatures, will be non-spontaneous

** Positive entropy processes favor spontaneity, but at low temperatures, will be non-spontaneous

Work MC # 10, 11, 12

Work MC # 13, 14, 15, 16

Work MC # 17, 18, 19, 20, 21

4

Relating Free Energy to Equilibrium:

The equilibrium point occurs at the lowest value of Free Energy available to the reaction system, so the value for

K can be calculated using G -

G0 = -RT ln K = -0.059/n log K ** R = 8.31 J/mol K

Relating Free Energy to Electrochemistry:

The maximum cell potential of a galvanic cell is directly related to the Free Energy difference between the

reactant and products in the cell –

G0 = -nFE

0

**When E0 is positive, G is positive, and the process is spontaneous

Work MC # 22, 23, 24, 25

5

Thermochemistry Multiple Choice:

1. Which of the following is the minimum energy required to initiate a reaction?

a. Free energy

b. Lattice energy

c. Kinetic energy

d. Activation energy

e. Ionization energy

2. The average _____________ is the same for any ideal gas at a given temperature.

a. Free energy

b. Lattice energy

c. Kinetic energy

d. Activation energy

e. Ionization energy

3. Consider the following PE diagram:

The forward reaction can be described as:

4. Consider the reaction

H2(g) + O2(g) H2O(l) H° = –286 kJ

Which of the following is true?

a. The reaction is exothermic.

b. The reaction is endothermic.

c. The enthalpy of the products is less than that of the reactants.

d. Heat is absorbed by the system.

e. Both A and C are true.

6

5. For the reaction:

AgI(s) + Br2(g) AgBr(s) + I2(s), H° = –54.0 kJ

Given: Hf° for AgBr(s) = –100.4 kJ/mol and Hf° for Br2(g) = +30.9 kJ/mol,

The value of Hf° for AgI(s) is:

a. –123.5 kJ/mol

b. +77.3 kJ/mol

c. +61.8 kJ/mol

d. –77.3 kJ/mol

e. –61.8 kJ/mol

6. The heat of formation of Fe2O3(s) is –826.0 kJ/mol. Calculate the heat change of the reaction

when a 53.99-g sample of iron is reacted.

a. –199.6 kJ

b. –399.2 kJ

c. –798.5 kJ

d. –1597 kJ

e. –2.230 104 kJ

7. A 4.4-g sample of Colorado oil shale is burned in a bomb calorimeter, which causes the temperature

of the calorimeter to increase by 5.0°C. The calorimeter contains 1.00 kg of water (heat capacity of

H2O = 4.184 J/g°C) and the heat capacity of the empty calorimeter is 0.10 kJ/°C. How much heat is

released per gram of oil shale when it is burned?

a. 21 kJ/g

b. 42 kJ/g

c. 0 kJ/g

d. 4.9 kJ/g

e. 0.21 kJ/g

8. Given the following information:

C (s) + O2 (g) CO2 (g) H = -393.5 kJ

H2 (g) + ½ O2 (g) H2O (l) H = -285.8 kJ

C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l) H = -1299.8 kJ

Find the enthalpy change for: 2 C (s) + H2 (g) C2H2 (g)

a. + 454.0 kJ

b. + 227.0 kJ

c. 0.0 kJ

d. – 227.0 kJ

e. – 454.0 kJ

9. 2 H2 (g) + O2 (g) 2 H2O (g)

From the table below, determine the enthalpy change for the above reaction.

BOND AVERAGE BOND ENERGY (kJ/mol)

H—H 436

O==O 499

H—O 464

a. 0 kJ

b. + 485 kJ

c. – 485 kJ

d. + 464 kJ

e. – 464 kJ

7

10. Choose the reaction expected to have the greatest increase in entropy?

a. H2O (g) H2O (l)

b. C (s) + O2 (g) CO2 (g)

c. Ca (s) + H2 (g) CaH2 (s)

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

e. 2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

11. Which of the following reactions has the largest positive value of S per mole of Cl2:

a. H2 (g) + Cl2 (g) 2 HCl (g)

b. Cl2 (g) + ½ O2 (g) Cl2O (g)

c. Mg (s) + Cl2 (g) MgCl2 (s)

d. 2 NH4Cl (s) N2 (g) + 4 H2 (g) + Cl2 (g)

e. Cl2 (g) 2 Cl (g)

12. For which of the following processes would S have a negative value?

I. 2 Fe2O3 (s) 4 Fe (s) + 3 O2 (g)

II. Mg2+

+ 2 OH1-

Mg(OH)2 (s)

III. H2 (g) + C2H4 (g) C2H6 (g)

a. I only

b. I and II only

c. I and III only

d. II and III only

e. I, II, and III

13. Given that Hvap is 60.3 kJ/mol, and the boiling point is 83.4°C, 1 atm, if one mole of this substance is

vaporized at 1 atm, calculate S.

a. –169 J/K mol

b. 169 J/K mol

c. 723 J/K mol

d. –723 J/K mol

e. 0

14. Which of the following combinations is true when sodium chloride melts?

a. H < 0, S > 0

b. H < 0, S < 0

c. H > 0, S > 0

d. H > 0, S < 0

e. H = 0, S = 0

15. Calculate G0 for the reaction given the following information:

2 SO2 (g) + O2 (g) 2 SO3 (g)

G0

f for SO2 (g) = - 300.4 kJ/mol

G0

f for SO3 (g) = - 370.4 kJ/mol

a. – 70.0 kJ

b. + 70.0 kJ

c. – 670.8 kJ

d. – 140.0 kJ

e. + 140.0 kJ

8

16. For the following reaction at 250C, H

0 = + 115 kJ and S

0 = + 125 J/K. Calculate G

0 for the reaction at

250C.

SBr4 (g) S (g) + 2 Br2 (l)

a. + 152 kJ

b. – 56.7 kJ

c. + 77.8 kJ

d. +37.1 kJ

e. – 86.2 kJ

17. For the reaction C2H6 (g) C2H4 (g) + H2 (g) H = +137 kJ/mol and S = +120 J/mol K.

This reaction is: a. Nonspontaneous at all temperatures b. Unreliable

c. Spontaneous only at high temperatures d. Spontaneous at all temperatures e. Spontaneous only at low temperatures

18. A reaction that is not spontaneous at high temperatures can become spontaneous at low temperatures if

H is __________ and S is __________. a. +, + b. -, - c. +, - d. -, +

e. +, 0

19. Which of the following must be true for a reaction that proceeds spontaneously from initial standard state

conditions?

a. G0 > 0 and Keq > 1

b. G0 > 0 and Keq < 1

c. G0 < 0 and Keq > 1

d. G0 < 0 and Keq < 1

e. G0 = 0 and Keq = 1

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

The reaction indicated above is thermodynamically spontaneous at 298 K, but becomes nonspontaneous at

higher temperatures. Which of the following is true at 298 K?

a. G, H, and S are all positive

b. G, H, and S are all negative

c. G and H are negative, but S is positive

d. G and S are negative, but H is positive

e. G and H are positive, but S is negative

21. The following reaction takes place at 120°C: H2O(l) H2O(g) H = +44.0 kJ/mol S = +119 J/mol K

Which of the following must be true?

a. The reaction is not spontaneous.

b. The reaction is spontaneous.

c. G = 0

d. G < 0

e. Two of these.

9

22. What is G° for the following electrochemical equation?

(°red(Ag+/Ag) = 0.800 V, °red(Cd

2+/Cd) = –0.403 V)

2Ag(s) + Cd2+

(aq) 2Ag+(aq) + Cd(s)

a. –232 kJ/mol

b. 116 kJ/mol

c. 232 kJ/mol

d. 464 kJ/mol

e. –464 kJ/mol

23. Determine G° for the weak acid, HF, at 25°C. (Ka = 7.16 10–4

)

a. 1.5 kJ

b. 177 kJ

c. 7.79 kJ

d. 1.77 kJ

e. 17.9 kJ

24. Which of the following statements is true about a voltaic cell for which °cell = 1.00 V?

a. It has G° > 0.

b. The system is at equilibrium.

c. It has K = 1.

d. The cathode is at a higher energy than the anode.

e. The reaction is spontaneous.

25. Given that Gf° for NH3 = –16.668 kJ/mol, calculate the equilibrium constant for the following reaction

at 298 K:

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

a. 6.97 105

b. 8.35 102

c. 1.01

d. 4.51 1069

e. 5.82 108

MC KEY: 1. D 2. C 3. A 4. E 5. E 6. B 7. D 8. B 9. C 10. E 11. D 12. D 13. B 14. C 15. D 16. A 17. C 18. B 19. C 20. B 21. E 22. C 23. E 24. E 25. A

10

FREE RESPONSE QUESTIONS:

#1 - Xe (g) + 3 F2 (g) XeF2 (g)

Under standard conditions, the enthalpy change for the reaction above going from left to right

(forward reaction) is H0 = - 294 kJ.

i. Is the value of S0, for the above reaction, positive or negative? Justify your conclusion.

ii. The above reaction is spontaneous under standard conditions. Predict what will happen to G for

this reaction as the temperature is increased. Justify your prediction.

iii. Will the value of K remain the same, increase, or decrease as the temperature increases? Justify

your prediction.

iv. Show how the temperature at which the reaction changes from spontaneous to non-spontaneous

can be predicted. What additional information is needed?

11

# 2 - 2 NO (g) + O2 (g) 2 NO2 (g) H0 = -114.1 kJ, S

0 = -146.5 J K

-1

The reaction represented above is one that contributes significantly to the formation of

photochemical smog.

a) Calculate the quantity of heat released when 73.1 g of NO (g) is converted to NO2 (g).

b) For the reaction at 250C, the value of the standard free-energy change, G

0, is -70.4 kJ.

i. Calculate the value of the equilibrium constant, Keq , for the reaction at 25

0.

ii. Indicate whether the value of G0 would become more negative, less negative, or remain

unchanged as the temperature is increased. Justify your answer.

c) Use the data in the table below to calculate the value of the standard molar entropy, S0, for

O2 (g) at 250C.

d) Use the data in the table below to calculate the bond energy, in kJ mol-1

, of the nitrogen-oxygen

bond in NO2. Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same

energy)

Standard Molar Entropy, S0 (J K

-1 mol

-1)

NO (g) 210.8

NO2 (g) 240.1

Bond Energy (kJ mol-1

)

Nitrogen- Oxygen bond in NO 607

Oxygen-Oxygen bond in O2 495

Nitrogen-Oxygen bond in NO2 ?

12

#3 - CO (g) + ½ O2 (g) CO2 (g)

The combustion of carbon monoxide is represented by the equation above.

a) Determine the value of the standard enthalpy change, H0

rxn, for the combustion of CO (g) at

298 K using the following information.

C (s) + ½ O2 (g) CO (g) H0

298 = -110.5 kJ mol-1

C (s) + O2 (g) CO2 (g) H0

298 = -393.5 kJ mol-1

b) Determine the value of the standard entropy change, S0

rxn, for the combustion of CO (g) at

298 K using the information in the following table.

c) Determine the standard free energy change, G0

rxn, for the reaction at 298 K. Include units

with your answer.

d) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.

e) Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.

Substance S0298 (J mol

-1 K

-1)

CO (g) 197.7

CO2 (g) 213.7

O2 (g) 205.1