CHAPTER TOPIC NO. OF PROBLEMS SOLVED

PAGE NOS.

8 Thermochemistry 2011 Rates of Reactions 2012 Chemical

Equilibrium20

13 Acids and Bases 2018 Electrochemistry 20

1) A 340-g sample of water was heated from 5.10° to 67.50° C. Calculate the amount of heat absorbed by water. The specific heat of water is 4.184 J/g°C.

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q = m s Δtq = (340)(4.184)(67.5 – 5.10)q = 8.9 x 104 J

2) When one mole of caffeine, C8H10O2 is burned in air, 2.98x103 KJ of heat is evolved. 3.00g of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by 9.63°C. What is the heat capacity of the calorimeter?

3.00g x 1mol C8H10N4O2 x 2.98 x 10 3 k J x 1000J = 4.60 x 104J 194.22 g 1mol C8H10N4O2

qrxn = -Ccal Δt4.60 x 104J = -Ccal (9.63°C)Ccal = 4.78 x 103 J/°C

3) The specific heat aluminum is 0.902 J/g°C. How much heat is absorbed by an aluminum pie tin with a mass of 352g to raise the temperature from room temp.(25.00°C) to oven temperature(190.6°C)?

q = m s Δtq = (352)(0.902)(190.6 – 25.00)q = 52.6 kJ

4) Magnesium sulfate is often used in first-aid hot packs, giving off heat when dissolved in water. When 5.00g MgSO4 dissolves in 25.0mL water (d=1.00g/mL) at 25.0°C, 2.25 KJ of heat is evolved.

a. Is the process exothermic? Yes

b. What is qH2O?qH2O = - qrxn

qH2O = 2.25 kJc. What is the final temperature of the solution?

q = m s (t2 – t1)-2250 J = (25)(4.184)(t2 – 25.0)t2 = 46.5°C

5) Urea (NH2)2CO, is used in the manufacture of resins and glues. When 15.0g of urea is dissolved in 450mL H2O (d=1.00g/mL) at 27.0°C, 31.4 KJ of heat is absorbed.

a. Is the solution exothermic? No

b. What is qH2O?qH2O = - qrxn

qH2O = -31.4 kJ

c. What is the final temperature of the solution?q = m s (t2 – t1)

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-2250 J = (25)(4.184)(t2 – 25.0)t2 = 10.3°C

6) In earlier times, ethyl ether was commonly used as anesthetic. It is, however, highly flammable. When ten milliliters of ethyl ether C4H10O(l), (d = 0.714g/mL)is burned in a bomb calorimeter, the temperature rises from 21.1°C to 40.3°C. The calorimeter heat capacity is 9.56 kJ/°C.

a. What is the q for the calorimeter?qcal = Ccal Δtqcal = (9.56 kJ/°C)(19.1°C)qcal = 184 kJ

b. What is q when 10.0 mL of ethyl ether is being burned?q = -184 kJ

7) Acetylene, C2H2, is used in wedding torches. It releases a lot of energy when burned in oxygen. One gram of acetylene releases 39.7kJ. When 0.560 g of acetylene is burned in a bomb calorimeter (heat capacity = 1.117kJ/°C), the final temperature of the bomb is 56.7°C. What is the initial temperature of the calorimeter?

qcal = Ccal Δtqcal = (1.117kJ/°C)(56.7 – t1)t1 = 36.8 °C

8) In photosynthesis, the following reaction takes place:6CO(g) + 6H2O(l) → 6O2(g) + C6H12O6(s) ΔH = 3901 kJ

a. Calculate ΔH when one mole CO2 reacts.1mol CO2 x 3901 k J = 650.2 kJ 6mol CO2

b. How many kilojoules of energy are liberated when 30.00 g of glucose, C6H12O6, is burned in oxygen?

30.00g C6H12O6 x 1mol C6H12O6 x 3901 k J = 649.5 kJ 180.18g 1mol C6H12O6

9) Given the following thermochemical equations,C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1354.5 kJC(s) + O2(g) → CO2(g) ΔH = -297.5 kJH2(g) + ½ O2(g) → H2O(l) ΔH = -267.6 kJ

Calculate ΔH for the decomposition of one mole of acetylene, C2H2(g), to its elements in their stable state at 25°C and 1 atm.

2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ΔH = -1354.5 kJCO2(g) → 2C(s) + 2O2(g) ΔH = 595 kJ2O(l) → H 2(g) + ½ O 2(g) Δ H = 267.6 kJ 2H2(g) → 2C(s) + H2(g) ΔH = -491.1 kJ

10) Given the following thermochemical equations,2H2(g) + O2(g) → 2H2O(l) ΔH = -571.6 kJN2O(g) + H2O(l) → 2HNO3(l) ΔH = -73.7 kJ

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½ N2(g) + 3/2 O2(g) + ½ H2→ HNO3(l) ΔH = -174.1 kJCalculate ΔH for the formation of one mole of dinitrogen pentoxide from its elements in their stable state at 25°C and 1 atm.

2O(l) → H2(g) + ½ O2(g) ΔH = 285.8 kJHNO3(l) → N2O5(g) + H2O(l) ΔH = 73.7 kJ2(g) + 3O2(g) + H2(g)→ 2HNO3(l) Δ H = -348.2 kJ 2(g) + 5O2(g) → N2O5(g) ΔH = 11.3 kJ

11) Given 2CuO(s) → 2Cu(s) + O2(g) ΔH = 314.6 kJa. Determine the heat of formation of CuO.

1 mol CuO x 314.6 kJ = 157.3 kJ 2 mol CuO But since the equation is reversed, ΔH°f = -157.3 kJ

b. Calculate ΔH° for the formation of 17.60 g of CuO.17.60g x 1mol CuO x 314.6 k J = 34.80 kJ 79.55g 2 mol CuO

But since the equation is reversed, ΔH°f = -34.80 kJ

12) Obtain ΔH° from the equationZn(s) + 2H+

(aq) → Zn2+(aq) + H2(g)

ΔH° = Σ ΔH°f prod - Σ ΔH°f react

13) Obtain ΔH° for the following thermochemical equations:a. 2H2S (g) + 3O2(g) → 2SO2(g) + 2H2O(g)

b. 3Ni(s) + 2NO3-(aq) + 8H+

(aq) → 3Ni2+(aq) + 2NO(g) + 4H2O(l)

14) Calculate ΔH° fora. The reaction between copper (II) oxide and carbon monoxide to give copper metal

and carbon dioxide.b. The decomposition of one mole of methyl alcohol (CH3OH) to methane and

oxygen gas.

15) When ammonia reacts with dinitrogen oxide gas (ΔH°f = 82.05 kJ/mol), liquid water and nitrogen gas are formed. How much heat is liberated or absorbed by the reaction that produces 345 mL of nitrogen gas at 25°C and 717mm Hg?

16) Finda. ΔE when a gas absorbs 18 J of heat and has 13 J of work dome on it.b. q when 72 J of work is done on a system and its energy is increased by 61 J.

17) Consider the combustion of one mole of acetylene, C2H2, which yields carbon dioxide and liquid water.

a. Calculate ΔH.b. Calculate ΔE at 25°C

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18) Given the following reactions,N2H4(l) + O2(g) → N2(g) + 2H2O(g) ΔH = -534.2 kJH2(g) + ½ O2(g) → H2O(g) ΔH = -241.8 kJ

Calculate the heat of formation of hydrazine.

19) Brass has a density 8.25 g/mL and a specific heat of 0.362 J/g°C. A cube of brass 22.00 mm on an edge is heated in a Bunsen burner flame to a temperature of 95.0°C. It is then immersed in 20.0 mL of water (d = 1.00 g/mL, c = 4.18 J/g°C) at 22.0°C in an insulated container. Assuming no heat loss, what is the final temperature of the water?

20) Microwave ovens convert radiation to energy. A microwave oven uses radiation with a wavelength 12.5 cm. Assuming that all the energy form the radiation is converted to heat without loss, how many moles of photons are required to raise the temperature of a cup of water (350.0 g, specific heat = 4.18 J/g°C) from 23.0°C to 99.0°C?

RATES OF REACTIONS

1) Radium spontaneously decomposes by a reaction of the first order. If it takes 25 1/3 years for a 1% decomposition of radium, what is its half-life?

[A] = 0.99[A]o

ln 0.99 = -k(25 1/3 years)k = 3.97 x 10-4/year

t ½ = 1.75 x 103 years

2) If the first order reaction A → B + C is started at noon and 80% of A has been converted to its products at 2:00 pm of the same day, at what time was 10% of A converted?

[A] = 0.20[A]o

ln 0.20 = -k(2h)k = 0.805/h

[A] = 0.90[A]o

ln 0.90 = -(0.805/h)tt = 0.13 hours

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3) In the first order reaction of type A → B + C, it takes 20 minutes to bring about decomposition of 30% of the initial substance, what total time would be required to decompose 60% of it?

ln 0.70 = -k(20 min)k = 0.018/min

for a 60% decomposition

ln 0.40 = -(0.018)tt = 51 min

4) The effect of concentration of reactants on the following reaction2H2(g) + 2NO(g) → 2H2O(g) + N2(g)

is summarized by the following experimental data obtained at 800°C

Experiment NO (M) H2 (M) Rate (M/min)I 6.00x10-3 1.00x10-3 1.5x10-3

II 6.00x10-3 2.00x10-3 3.0x10-3

III 6.00x10-3 3.00x10-3 4.5x10-3

IV 1.00x10-3 6.00x10-3 2.2x10-3

V 2.00x10-3 6.00x10-3 8.8x10-3

Determine a. the individual and overall orders

r = k[NO]m[H2]n

r1 = 1.5 x 10 -3 = (6.00 x 10 -3 ) m (1.00 x 10 -3 ) n r4 = 2.2 x 10 -3 = (1.00 x 10 -3 ) m (6.00 x 10 -3 ) n r2 = 3.0 x 10-3 = (6.00 x 10-3)m(2.00 x 10-3)n r5 = 8.8 x 10-3 = (2.00 x 10-3)m(6.00 x 10-3)n

0.5 = 0.5n ¼ = (½ )m

n = 1 m = 2

overall order = 3

b. the rate constant

k = 4.2 x 104/M2min

c. the rate law

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rate = k[NO]2[H2]

5) Using the same data in No.4, answer the following:rate = k[NO]2[H2]let [NO] = A

[H2] = B

a. What happens to the rate of reaction if [NO] is doubled?r = [2A]2[B]the rate is quadrupled

b. Tripling [H2] increases the rate by what factor?r = [A]2[3B] the rate increases thrice

6) For the following hypothetical reactionA(g) + 2B(g) + 3C(g) → 4D(g)

the rate law was found to beRate of formation of D = k[A]2[B]

a. Doubling [A] would increase the rate by what factor? 4b. Doubling [B] would increase the rate by what factor? 2c. Tripling [C] would increase the rate by what factor? No effectd. Calculate k from the following data:

[A] = 2.0x10-2 [B] = 3.0x10-2 [C] = 5.0x10-2

Rate of formation D = 4.0x10-3M/min

k = 3.3 x 102 /M2min

7) The data show the effect of the concentration of reactants on the rate of formation of NOCl gas by the equation

Cl2(g) + 2NO(s) → 2NOCl(g) at 450°CExpt [Cl]

(M)[NO](M)

Rate of NOCl f ormed(M/sec)

1 1.0 x 10-3 2.5 x 10-4 2.6 x 10-10

2 1.0 x 10-3 5.0 x 10-4 5.2 x 10-10

3 1.0 x 10-3 10.0 x 10-4 1.04 x 10-9

4 2.0 x 10-3 10.0 x 10-4 4.16 x 10-9

5 3.0 x 10-3 10.0 x 10-4 9.36 x 10-9

6 4.0 x 10-3 10.0 x 10-4 1.664 x 10-8

Determinea. The rate law

r = k[Cl2]m[NO]n

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b. The individual and overall ordersr1 = 2.6 x 10 -10 = (1.0 x 10 -3 ) m (2.5 x 10 -4 ) n r3 = 1.04 x 10 -9 = (1.0 x 10 -3 ) m (10.0 x 10 -4 ) n r2 = 5.2 x 10-10 = (1.0 x 10-3)m(5.0 x 10-4)n r4 = 4.16 x 10-9 = (2.0 x 10-3)m(10.0 x 10-4)n

0.5 = 0.5n ¼ = (½ )m

n = 1 m = 2

overall order = 3

c. The rate constant

k = 1.0/M2sec

8) Using the same data in No.7, how would the reaction rate be affected bya. Doubling the concentration of NO

double the rateb. Increasing the concentration of NO by 4 times

increase the rate by 4 timesc. Doubling the concentration of Cl2

quadruple the rated. Increasing the concentration of Cl2 by 3

increase the rate by 9 times

9) For a reaction of A and B to form C, the following data were obtained from the following experiments

Expt [A] [B] Rate1 0.60 0.15 6.3 x 10-3

2 0.20 0.60 2.8 x 10-3

3 0.60 0.15 7.0 x 10-4

Calculatea. The rate law

r1 = 6.3 x 10 -3 = (0.60) m (0.15) n r2 = 2.8 x 10 -3 = (0.20) m (0.60) n r3 = 7.0 x 10-4 = (0.20)m(0.15)n r3 = 7.0 x 10-4 = (0.20)m(0.15)n

9 = 3m 4 = 4n

m = 2 n = 1

overall order = 3

b. The numerical value for k

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k = 0.12/ M2min

10) Consider the reaction3A + B → 2C + D

If the rate of formation of D is 2.0M/sec, calculatea. Rate of disappearance of A

-1 Δ[A] = 2.0 M/sec 3 Δt

Δ[A] = -6.0 M/sec Δt

b. Rate of formation of C 1 Δ[C] = 2.0 M/sec 3 Δt

Δ[C] = 4.0 M/sec Δt

c. Rate of oxidation of B-Δ[B] = 2.0 M/sec Δt

Δ[B] = -2.0 M/sec Δt

11) The decomposition of ammonia in tungsten at 1100°C is zero-order with a rate constant of 2.5x10-4M/min.

a. Write the rate lawr = k

b. Calculate the rate when [NH3] = 0.750Mr = 2.5x10-4M/min

c. At what concentration of NH3 is the rate equal to the rate constant?Any value since the rate is independent of the concentration.

12) The reactionICl(g) + ½ H2(g) → ½ I2(g) + HCl(g)

is first order in both reactants. The reaction rate is 4.89x10-5M/s when [ICl] = 0.0500M and [H2] = 0.0150M.

a. What is the value of k?

k = 0.0652/Msec

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b. At what concentration of H2 is the rate equal to 2.50 x 10-4M/s and the concentration of ICL is 0.117?

[H2] = 0.0328 M

13) In the first order decomposition of acetone at 500°CCH3-CO-CH3(g) → products

It is found that the concentration is 0.0300M after 200 minutes and 0.0200M after 400 minutes. Find:

a. The rate constant

ln [0.0200] = -k(200 min) [0.0300]k = 2.03 x 10-3 /min

b. The half-life

t ½ = 342 min

c. The initial concentrationln [A] – ln [A]o = ktln [A]o = ln(0.03) + (2.03 x 10-3)(200)

[A]o = e -3.10

[A]o = 0.0450 M

14) The decomposition of dimethyl ether (CH3)2O at 510°C is first order with the rate constant equal to 6.8x10-4/sec. If the initial pressure is 135 torr, what is the partial pressure after 1420 seconds?

ln [A] = ln [A]o – ktln [A] = ln(135) – (6.8x10-4/sec)(1420 sec)ln [A] = 3.94ln [A] = 51.4 torr

15) For the reaction2N2O(g) → 2N2(g) + O2(g)

the rate constant is 0.033M/min at 838K AND 11.4M/min at 1001K. What is the activation energy for the reaction?

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Ea = 250 kJ/mol

16) For the reaction: 2NO(g) + O2(g) → 2NO2(g), the rate constant is 0.0396M/min at 673K and 13.68M.min at 821K.

a. What is the activation energy?

Ea = 181 kJ/mol

17) The decomposition of N2O5 to N2 and NO3 is a first order reaction. At 55°C, the reaction has a half-life of 1.67s and at 73°C, the reaction has a half-life of 0.203s. What is its activation energy?

k1 = 0.415/sk2 = 3.41/s

Ea = 111 kJ/mol

18) The decomposition of NO2 is a second order reaction. At 550K, a 0.250M sample decomposes at a rate of 1.17M/min.

a. What is the rate law?r = k[NO2]2

b. What is the rate constant at 550K?

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k = 1.17M/min (0.250M)2

k = 18.7 M/min

c. What is the rate of decomposition when [NO2] = 0.800M?r = k[NO2]2

r = (18.7 M/min)(0.800M)2

r = 12.0 M/min

19) Complete the table for the reaction 2R(s) + 3S(g) → products. The reaction is first order in R and second order in S.

[R] [S] k (M2/min)

Rate (M/min)

a 0.200 0.200 1.49 0.0119b 4.95 0.633 0.42 0.833c 0.100 2.33 0.298 0.162d 0.0500 0.0911 15.0 0.00624

a. Rate = (0.149)(0.200)(0.200)2

b.

c.

d.

20) Complete the table for the reaction 2A(g) + 3B(g) → products which is first order in both

reactants.

[A] [B] K (/Mh) Rate (M/h)a 0.0600 0.240 1.13 0.0163b 0.0360 0.500 0.530 0.0954c 0.238 0.0150 8.04 0.0287d 0.355 0.140 0.0135 0.000672

a. Rate = (1.13)(0.0600)(0.240)

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b.

c.

d.

CHEMICAL EQUILIBRIUM

1) At a certain temperature, the rate constant for the reaction 3C2H2 ↔ C6H6 is 4.0. If the equilibrium concentration of C2H2 is 0.60M, what is the concentration of C6H6?

Kc = [C6H6] [C2H2]3

4.0 = [C6H6] (0.60)3

[C6H6] = 0.86 M

2) Consider the gaseous equilibrium 2A ↔ 2B + C. Four moles of A are put in a 1-L flask. At equilibrium, the flask is found to contain 1.0 mole of C. Calculate the equilibrium constant.

2A ↔ 2B + CI 4.0 0 0C - 2x + 2x + x E 4.0 – 2x 2x x

x = 1.0 mole

Kc = [B] 2 [C] [A]2

Kc = (2.0) 2 (1.0) (4.0 – 2x)2

Kc = 1.0

3) The equilibrium constant for the gaseous equilibrium X ↔3Y is 2.0. A 2-L flask contains 3.0 moles of X at equilibrium. How many moles of X does the flask contain at equilibrium?

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X ↔ 3YI a 0C - x + 3x E a – x 3x

3x = 1.5x = 0.5Kc = [Y] 3 [X]2.0 = (3x) 3 (y-x)2y = 3x3 + 2xy = 2.2y – x = 1.7 Mmoles of X = 3.4

4) Consider the following gaseous equilibrium 3A ↔ 2B + C. Three moles of A were put into a 1-L flask. At equilibrium it was found that 0.5 moles of C was present. What is the value of the equilibrium constant?

3A ↔ 2B + CI 3.0 0 0C - 3x + 2x + x E 3.0 – 3x 2x x

x = 0.5 M

Kc = [B] 2 [C] [A]3

Kc = (1.0) 2 (0.5) (3.0-1.5)3

Kc = 0.155) 4.0 moles of PCl are placed in a 2-L flask.

PCl5 ↔ PCl3 + Cl2

When equilibrium is established, the flask is found to contain 0.40 moles of Cl 2.

Determine the equilibrium constant for this reaction.

PCl5 ↔ PCl3 + Cl2

I 2.0 0 0C - x + x + x E 2.0 – x x x

x = 0.2M

Kc = [PCl3][Cl2] [PCl5]Kc = (0.2)(0.2) (2.0 – 0.2)Kc = 0.02

6) If HI is 25% dissociated according to the equation 2HI ↔ H2 + I2 at a given temperature.

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a. Calculate the equilibrium concentrations of HI, H2 and I2 if we start with 10 moles of HI in a 1.0-L container

2HI ↔ H2 + I2

I 10 0 0C - 2x + x + x E 10 – 2x x x

10-2x = 7.5x = 1.25[HI] = 7.50 M[H2] = 1.25 M[I2] = 1.25 M

b. Calculate the value of the equilibrium constantKc = [H2] 2 [I 2] [HI]2

Kc = (1.25) 2 (1.25) (7.50)2

Kc = 0.02787) A mixture containing 1.000 moles of H2(g) and 1.000 moles of CO2(g) is placed in a 1.000L

container at 800°C. At equilibrium, 0.491 moles of CO2(g ) and 0.491 moles of H2O(g) are present.

H2(g) + CO2(g) ↔ H2O(g) + CO(g)

Calculate a. The equilibrium concentrations of H2(g) and CO2(g)

H2(g) + CO2(g) ↔ H2O(g) + CO(g)

I 1.000 1.000 0 0C - x - x + x + x E 1.000 – x 1.000 – x 0.491 0.491

x = 0.491[H2], [CO2] = 0.509 M

b. The equilibrium constantKc = [H2O] [CO] [H2][CO2]Kc = (0.491) 2 (0.590)2

Kc = 0.931

8) A mixture containing 0.075M of HCl(g) and 0.033M O2(g) is allowed to come to equilibrium at 480°C

4HCl(g) + O2(g) ↔ 2Cl2(g) + 2H2O(g)

At equilibrium, the concentration of Cl2 is 0.030Ma. What are the equilibrium concentrations of HCl, O2 and H2O?

4HCl(g) + O2(g) ↔ 2Cl2(g) + 2H2O(g)

I 0.075 0.033 0 0C - 4x - x + 2x + 2x

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E 0.075 – 4x 0.033 – x 2x 2x2x = 0.030 Mx = 0.015 M[HCl] = 0.015 M[O2] = 0.018 M[H2O] = 0.030 M

b. What is the rate constant at 480°C?Kc = [H2O] 2 [Cl 2] 2 [HCl]4[O2]Kc = [0.030] 2 [0.030] 2 [0.015]4[0.018]Kc = 890

9) At 100°C the equilibrium constant for the reaction CO(g) + Cl2(g) ↔ COCL2(g) is 4.57x109/M. If 1.00 mole COCl2 is confined in a 1.00-L container, find the concentration of CO after equilibrium.

COCL2(g) ↔ CO(g) + Cl2(g) I 1.00 0 0C - x + x + x E 1.00 – x x x

1/Kc = 2.19 x 10-10 Test equation:

Kc = x 2 __ (1.00-x)x2 = 2.19 x 10-10

x = 1.48 x 10-5 M[CO] = 1.48 x 10-5 M

10) Analysis of an equilibrium mixture of N2 + 3H2 ↔ 2NH3 showed concentrations of H2

and NH3 to be 0.60M and 0.20M respectively. If the equilibrium concentration is 0.37, what is the concentration of N2?

Kc = [NH3] 2 [N2][H2]3

[N2] = (0.20) 2 __ (0.37)(0.60)3

[N2] = 0.50 M11) Consider the reaction CO + H2O ↔ CO2. Into a 1-L vessel are placed 5.0 moles CO and

3.0 moles H2O and the system is heated to a reacting temperature. At equilibrium, analysis shows the presence of 1.5 mole CO2. What is the value of the equilibrium constant?

CO + H2O ↔ CO2

I 5.0 3.0 0C - x - x + x E 5.0 – x 3.0 – x x = 1.5

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[CO] = 3.5 M[H2O] = 1.5 M

Kc = [CO2] 2 [CO][H2O]Kc = (1.5) 2 (3.5)(1.5)Kc = 0.29

12) Consider the reaction 2A + B ↔ 2C + D. Initially, 3.0 moles of A and 5.0 moles of B are placed in a vessel and reacted. At equilibrium, 0.5 moles of D is found. What is Kc?

2A + B ↔ 2C + DI 3.0 5.0 0 0C - 2x - x + 2x + x E 3.0 – 2x 5.0 – x 2x xx = 0.50

2.0 4.5 1.0 0.5Kc = [C] 2 [D] [A]2[B]Kc = (1.0) 2 (0.5) (2.0)2(4.5)Kc = 0.028

13) Consider the following equilibrium at high temperature 2CH4(g) + 3O2(g) ↔ 2CO(g) + 4H2O(g). The equilibrium constant is 0.045. The concentrations of O2, CO and H2O are as follows: [O2] = 2.00, [CO] = 3.00 [H2O] = 1.00. what would be the concentration of CH4

at equilibrium?Kc = [CO] 2 [H 2O] 4 [CH4]2[O2]3

[CH4]2 = [CO] 2 [H 2O] 4 Kc [O2]3

[CH4]2 = (3.00) 2 (1.00) 4 0.045 (2.00)3

[CH4] = 5.0 M14) Carbon monoxide and water react at 1000°C to give the ff equation:

CO + H2O ↔ CO2 + H2

Into a 1-L vessel is placed initially 0.750 moles CO and 0.500 moles H2O. Analysis at equilibrium shows the presence of 0.200 moles H2. Find the value of Kc.

CO + H2O ↔ CO2 + H2

I 0.750 0.500 0 0C - x - x + x + x E 0.750 – x 0.500 – x x xx = 0.200

0.550 0.300 0.200 0.200Kc = [CO2][ H2] [CO][ H2O]

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Kc = (0.200)(0.200) (0.550)(0.300)Kc = 0.242

15) Consider the following reaction at 600°CN2(g) + 3H2(g) ↔ 2NH3(g)

Initially, 4.0 moles NH3 is placed in a 1.0-L container. At equilibrium, 3.0 mole H2 are found in the container. Determine Kc.

` 2NH3(g) ↔ N2(g) + 3H2(g)

I 4.0 0 0C - x + x + 3x E 4.0 – 2x x 3x x = 1.0

2.0 1.0 3.0Kc = [N2][H2] 3 [NH3] 2

Kc = (1.0)(3.0) 3 (2.0) 2

Kc = 6.75 but the equation is reversed, therefore:Kc`= 1/ Kc

Kc` = 0.15

16) State the direction in which each of the following equilibrium systems would be shifted upon the application of the stress listed beside the equation.

a. N2(g) + O2(g) ↔ 2NO(g) (endothermic) T↓ ←b. 2NOBr(g) ↔ 2NO(g) + Br2(g) P↓ →c. 3Fe(s) + 4H2O(g)↔ Fe3O4(g) + 4H2(g) P↑ ←d. C(s) + CO2(g) ↔ 2CO(g) P↓ →e. C(s) + CO2(g) ↔ 2CO(g) + C(s) no effectf. N2O4(g) ↔ 2NO2(g) [NO2]↑ ←g. N2(g) + 3H2(g) ↔ 2NH3(g) + catalyst no effecth. BaCl2 H2O(s) ↔ BaCl2(s) + H2O(g) - H2O(g)→

17) For the reaction 2NO2(g) ↔ N2O4(g)

At a given temperature, 1.0 mole NO2 is placed in a 10.0-L container. After equilibrium is established, there are 0.20 moles N2O4 present. Calculate Kc at this temperature.

2NO2 ↔ N2O4

I 0.1 0C - 2x + x E 0.1 – 2x x

0.06 0.02Kc = [N2O4] [NO2] 2

Kc = (0.02)

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(0.06)2

Kc = 5.6

18) For the reaction 2HI(g) ↔ H2(g) + I2(g) Kc = 0.016 at 500°CIf 1.0 mole HI is placed in a 2.0-L container and allowed to come to equilibrium at 500°C, calculate the equilibrium concentrations of HI, H2 and I2.

2HI(g) ↔ H2(g) + I2(g)

I 0.50 0 0C - 2x + x + x E 0.50 – 2x x x

Kc = [I2][H2] [HI] 2

0.016 = x 2 ___ 0.50 – 2x x2 + 0.032x – 8 x 10-3 = 0a = 1; b = 0.032; c = -8 x 10-3

x = 0.075

[HI] = 0.326 M[H2], [I2] = 0.072 M

19) Using the value given for Kc in the previous number, calculatea. Kc for the reaction H2(g) + I2(g) ↔ 2HI(g)

Kc` = 1/ Kc

Kc` = 63b. Kc for the reaction HI(g) ↔ ½ H2(g) + ½ I2(g)

Kc` = √ Kc

Kc` = 0.13c. Kc for the reaction ½ H2(g) + ½ I2(g) ↔ HI(g)

Kc` = 1/√ Kc

Kc` = 2.8d. Kc for the reaction 4HI(g) ↔ 2H2(g) + 2I2(g)

Kc` = Kc2

Kc` = 0.00026

20) For the reaction 2A(g) + B(g) ↔ 2C(g) Kc = 0.18 at 200°CCalculate Kc for

a. 2C(g) ↔ 2A(g) + B(g)

Kc` = 1/ Kc

Kc` = 5.6b. A(g) + ½ B(g) ↔ C(g)

Kc` = √ Kc

Kc` = 0.13c. C(g) ↔ A(g) + ½ B(g)

Kc` = 1/√ Kc

Kc` = 2.8

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d. 4A(g) + 2B(g) ↔ 4C(g)

Kc` = Kc2

Kc` = 0.032

ACIDS AND BASES

1) What is the pH value of

a. 0.010M HCl (100% effective ionization)

pH = 2.0

b. 0.30M NaOH (90% effective ionization)

[OH-] = (0.30M) (0.90)[OH-] = 0.27M

pOH = 0.57

pH = 14.00 – pOHpH = 13.43

c. solution of HCl in which the H+ concentration is 8.0M?

pH = -0.90

2) a. Given pH = 10.46; calculate [H+], [OH-], and pOH.

[H+] = 3.5 x 10-11 M

[OH-] = 2.9 x 10-4 M

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pOH = 14.00 – pHpOH = 14.00 – 10.46pOH = 3.54

3) What is the pH value of

a. 0.050M HNO3 (100% ionized)

pH = 1.30

b. 0.80M KOH (85% effective ionization)

[OH-] = (0.85) (0.80M)[OH-] = 0.68M

pOH = 0.17

pH = 14.00 – pOHpH = 14.00 – 0.17pH = 13.83

c. solution of HCL in which the hydrogen-ion concentration is 5.0M?

pH = -0.70

4) What is the hydroxyl-ion concentration of solution in which pOH = -0.27?

[OH-] = 1.9M

[H+] = 5.3 x 10-15M

5) Given pOH = 5.80; calculate [H+], [OH-], and pH.

[OH-] = 1.6 x 10-6 M

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pH = 14.00 – 5.80pH = 8.20

[H+] = 6.3 x 10-9 M

6) A certain organic acid has one replaceable hydrogen and in 0.010M aqueous solution is 0.18% ionized. What is the ionization constant of the acid?

[H+] = (0.0018) (0.010M)[H+] = 1.8 x 10-5 Mx = 1.8 x 10-5 M

Ka = 3.2 x 10-8 M

7) Lactic acid is a monobasic acid with an ionization constant of 1.6x10 -4. What is the lactate-ion concentration in 0.5.M solution of the acid?

HC3H5O3 → H+ + C3H5O3-

I 0.50 0 0C - x + x + xE 0.50 – x x x

Test equation:

x = 8.9 x 10-3M[C3H5O3

-] = 8.9 x 10-3M

8) Formic acid is a monobasic acid that at a certain temperature is 3.2% ionized in 0.20M solution. What is the ionization constant of formic acid at that temperature?

HCHO2 → H+ + CHO2-

I 0.20 0 0C - x + x + x

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E 0.20 – x x x

[H+] = (0.032) (0.20M)[H+] = 6.4 x 10-3 Mx = 6.4 x 10-3 M

Ka = 2.1 x 10-4

9) A certain organic amine acts as a mono-acidic base in aqueous solution. A 0.050 soln. has a hydroxyl-ion concentration of 7.5x10-5M. What is the pH value of the solution, and the ionization constant of the base?

pOH = 4.12

pH = 14.00 – 4.12pH = 9.88

Ka = 1.1 x 10-7

10) Ethylamine is a derivative of ammonia and in aq. soln. is a mono-acidic base. At a certain temperature, a 0.30M solution of ethylamine has a pH value of 12.11. What is the ionization constant of this base at that temperature?

pOH = 14.00 – 12.11pOH = 1.89

[OH-] = 10-1.89

[OH-] = 0.013 M

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Kb = 5.9 x 10-4

11) Calculate the pH of each of the following aqueous solutions:

a. 2.17x10-3M HCl

pH = -log [2.17 x 10-3M]pH = 2.66

b. 0.100M NaHSO4

pH = -log [0.100M]pH = 1.00

c. 0.150M HNO2

pH = -log [0.150M]pH = 0.82

d. 7.50 x 10-2M NaOH

pH = 14.00 – (-log[OH-])pH = 14.00 + log (7.50x10-2M)pH = 12.88

12) Calculate the pH of an aqueous solution containing acetic acid at an analytical concentration of

a. 3.00 x 10-1M

pH = -log [3.00 x 10-1M]pH = 0.52

b. 3.00 x 10-10M

pH = -log [3.00 x 10-10M]pH = 9.52

13) A 0.0722M solution of a weak acid has a pH of 3.11. Calculate the Ka of this acid.

[H+] = 10-3.11

[H+] = 7.8 x 10-4 M

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Ka = 8.5 x 10-6

14) A 0.250M solution of a weak base has a pH of 9.91. Calculate Kb.

pOH = 14.00 – 9.91pOH = 4.09

[OH-] = 10-4.09

[OH-] = 8.1 x 10-5 M

Kb = 2.6 x 10-8

15)a. What is the pH value of a solution of which the hydrogen-ion concentration is

2.8x10-3M? Is the solution acidic or alkaline?

pH = -log [2.8 x 10-3M]pH = 2.55, acidic

b. What is the pH value of a solution of with a pOH value of 4.17? Is the solution acidic or alkaline?

pH = 14.00 – pOHpH = 9.83, alkaline

16) What is the hydrogen-ion concentration of a solution of which the pH value is -0.55?

[H+] = 100.55

[H+] = 3.5 M

17)a. What is the pOH value of a solution which the hydrogen-ion concentration is

5.2x10-4M? Is the solution acid or alkaline?

pOH = 14.00 – (-log[H+])pOH = 14.00 + log (5.3 x 10-4M)

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pOH = 10.72, acidic

b. What is the hydroxyl-ion concentration of a solution of which the pH value is 9.27? Is the solution acid or alkaline?

pOH = 14.00 – pHpOH = 4.73, alkaline

[OH-] = 10-4.73

[OH-] = 1.9 x 10-5 M

18) The ionization constant for ascorbic acid HC6H7O8 is 8.1 x 10-5. Calculate [H+] of 0.10M ascorbic acid.

HC6H7O8 ↔ H+ + C6H7O8-

I 0.010 0 0C - x + x + x E 0.010 – x x x

Test equation:

x2 = 8.1 x 10-5

x = 2.8 x 10-3

[H+] = 2.8 x 10-3

19) Propanoic acid, HC3H5O2 is 0.72% ionized in 0.25M solution. What is its ionization constant?

[H+] = (0.25) (0.0072M)[H+] = 1.8 x 10-3 M

Ka = 1.3 x 10-5

20) What are [H+], [C7H5O2-] and [HC7H5O2] in a 0.50M solution of benzoic acid if the

ionization constant is 6.4 x 10-5.?HC3H5O2 ↔ H+ + C3H5O2

-

I 0.50 0 0C - x + x + x E 0.50 – x x x

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Test equation:

x = 5.7 x 10-3

[H+] = 5.7 x 10-3 M[C7H5O2

-] = 5.7 x 10-3 M[HC7H5O2] = 0.49

ELECTROCHEMISTRY

1. For the cell diagramZn | ZnSO4(1.00M) || CuSO4(1.00F) | Cu

a. Write the anode reactionb. Write the cathode reactionc. What is the overall equation?

2. Calculate the equilibrium constant for the reactionMnO4

- + 5Fe2+ + 8H+ ↔ Mn2+ + 5Fe3+ 4H2O

3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.

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BIBLIOGRAPHY

Calculations of Analytic Chemistry 6th EditionLeicester Hamilton & Stephen Simpson

Analytic Chemistry 6th EditionAn Introduction

Douglas Skoog, Donald West & James Holler Analytic Chemistry

Principles and TechniquesLarry G. Harris

Analytical ChemistryAn Introduction

Douglas Skoog & Donald West College Chemistry 7th Edition

G. Brooks King, William Caldwell & Max Williams Fundamentals of Chemistry

Fred Redmore Chemistry

Principles and ReactionsWilliam Masterton & Cecile Hurley

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