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Exam 2 General Chemistry 201. May 12, 2009

No credit will be given for correct numerical answers without a clear indication of how they were obtained. Show all work and provide detailed explanation. Report all numerical answers to the proper Significant Figures.

Show all work on the exam pages and not on scratch paper. Please keep your eyes on your paper!

System Pressure: LENGTH: VOLUME MASS Temperature English: 760 mmHg = 14.7 psi

1 atm = 101.3 KPa 1 ft = 12 in 1 mile = 5280 ft 1 yd = 3 ft

1 gal = 4 qt 1 qt = 57.75 in3

1 lb = 16 oz 1 ton = 2000lb

T°F = 1.8 (T°C) + 32

SI- English:

1 atm = 760 torr 1atm = 760 mmHg

1 in = 2.54 cm 1 mi = 1.609 km

1 L = 1.057 qt 1 qt = 0.946 L

1 lb = 453.6 g 1 oz = 28.35 g

T°C = (T°F - 32) / 1.8

Misc. info 1 J = 1 kg m2 / s2 1 mole = 6.02•1023 Density H2O: 1.0 g/ml

Equilibrium constant Kp and Kc: Kp = Kc(RT)Δn

(R = .08206 L•atm / mol•K)

Quadratic Equation: ax2 + bx + c = 0

€

x =−b ± b2 −4ac

2a

4 pts each 1 For the combination reaction below, which will cause the reaction to shift to the right (product)? A(s) + B(s) AB(g) ΔH (+)

a) double the container volume b) decrease the temperature c) add more A(s) d) double the concentration of the product

2 Consider the reaction: 2H2(g) + CO(g) CH3OH(g) A reaction vessel contains the three gases at equilibrium with a total pressure

of 1.00 atm. What will happen to the partial pressure of hydrogen if enough argon is added to raise the total pressure to 1.4 atm? a) decrease b) increase c) increase then decrease d) unchanged 3 A mixture of 0.600 mol of Br2 and 1.600 mol of I2 is placed into a rigid 1.000-L container at 350°C: Br2(g) + I2 (g) 2IBr(g)

When the mixture has come to equilibrium, the concentration of IBr is 1.190 M. What is the equilibrium constant at 350°C? a) 1.48 b) 282 c) 237 d) 1.18 4 For the reaction 3 H2(g) + N 2(g) 2 NH3 (g), the relationship between Kc and Kp at a given temperature T is:

a) Kc = Kp b) Kp = Kc(RT) 2 c) Kc = Kp(RT)2 d) none

5 Choose the correct concentration vs. time curve for the decomposition of calcium carbonate: CaCO3 (s) CaO (s) + CO2 (g)

a)

CaCO3

CO2

Concentration

Time b)

Concentration

Time

CaCO3

CO2 c)

Concentration

Time

CaCO3

CO2 d) none

6 Consider the diagram to the right for: 2C(g) A(g) + 2B(g) ΔH = ( + )

From the reaction coordinate diagram, the equilibrium constant is determined to be:

a) Keq < 1 b) Keq = 1 c) Keq > 1 d) Keq cannot be determined.

7 A 10.0 gram sample of solid NH4Cl(s) is heated in a 5.00L container to 900°C. The reaction: NH4Cl (s) NH3(g) + HCl(g)

At equilibrium the pressure of NH3 (g) is 1.20 atm. What is the equilibrium constant, Kp ?

a) 1.44 b) 1.20 c) 2.40 d) none 8

Consider the following theoretical reaction: A(g) + 2B (g) 2P(g). At the 33-minute mark, what is the value of Q? a) Q < Keq b) Q = Keq

c) Q > Keq d) Q cannot be determined.

9 Refer to the graph from question 8. At what time is Q = Keq ?

a) 33 min b) 18 min c) 28 min d) no correct choice 10 Which of the following compounds has the lowest molar solubility in water ?

a) Al(OH)3 Ksp = 2•10-32 b) BaSO4 Ksp = 1.1 × 10-10

c) PbSO4 Ksp = 1.3•10-8 d) MgC2O4 Ksp = 1.0•10-5

11 (20 pts) Consider the metabolic breakdown of glucose- C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l) ΔH = - 2802 KJ After the following stress (see below) are imposed to the reaction, describe and explain how each specie concentration in the reaction

is affected as a new equilibrium is attained. i) State if Q is less than (<), greater than (>) or equal to (=) Keq

ii) State if Q increase, decrease or does not change as equilibrium is re attained and explain your answer in details

Concentration: I, D, N or ? CO2 H2O C6H12O6 O2 Energy

Absorb or produce

i) Decrease the pressure of O2.

Shift to Left Q > Keq

Reaction reverse. Remnoval of reactant means that there is a void in the reactant. The reaction will respond by replacing the reactant and converting products bact to reactants. This happens when there is an increase in Q, meaning that Q > Keq. For equilibrium to be established again, Q will need to decrease so that equilibrium is re-establish.

D

N

N

I

Absorb

ii) Add CO2 by decomposition of CaCO3 (s)

Shift to Left Q > Keq

Reaction reverse. Adding product cause the reaction to shift left to remove the excess products. This happens when there is an increase in Q, meaning that Q > Keq. For equilibrium to be re-establish. Q will need to decrease so that equilibrium is re-establish.

D

N

N

I

Absorb

iii) Double the amount of C6H12O6 (s)

No net shift Q = Keq

No effect since C6H12O6 is a solid, the concentration does

not change for the remaining chemicals. Q = Keq.

N

N

N

N

N

iv) Decreasing the temperature for rxn.

Shift to right Q < Keq

Removal of energy to an exothermic reaction means that product is being affected or that keq at the new

temperature is higher than Keq at the old temperature.

The means Q < Keq. For equilibrium to establish itself again, Q must increase to match Keq, the reaction shift towards product.

I

N

N

D

Produce

v) Decrease volume by half

No net shift Q = Keq

No net effect since the moles of gas in the products are the same of the moles of gas in the reactants. Q does not change since Q = Keq.

N

N

N

N

N

12 (18 pts) Consider the dissociation of iodine: I2 (g) 2I (g)

A 1.000 g sample of I2 is heated to 1200.°C in a 500-mL flask. At equilibrium the total pressure is 1.510 atm.

i) What is the partial pressure of I2 (g) and I (g) at equilibrium.

ii) What is the mole fraction of I2 and I at equilibrium. ii) Calculate Kp for the reaction.

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1.00 g I2 : mol I2 = 1.00 g 1 mol253.8g

= 3.94× 10-3 mol

pI2 : pI2 = nRTV

= 3.94× 10-3 mol x 0.08206 x 1473K.500L

= 0.9525 atm

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I2(g) ↔ 2I(g)

i 0.9525 atm 0Δ - x + 2xe[ ] 0.9525 - x atm 2x

€

PT = 1.51 atm = pI2 + pI = 1.51 atm = 0.9525 - x + 2x

x = 1.51 - 0.9525 = 0.5575 atm

€

PI2= 0.9525 - x = 0.3950atm

PI = 2x = 1.115 atm

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χI2

= 0.3950atm0.3950atm + 1.115 atm

= 0.2616

χI

= 1.115 atm1.115 atm + 0.3950atm

= 0.7384

€

Kp = 1.115( )

2

0.3950 = 3.15

13 Check mark if you agree___ or disagree___ with the statements below. Write a complete justification for your answer. (12 pts) i) Solids will always have their solubility increase with increasing temperature because at higher temperature molecular collisions

between solute and solvent are more effective resulting in more stable solute-solvent interaction. Agree ___ or disagree ____ Note all dissolution process are endothermic, some are exothermic. See solubility vs. temperature graph

ii) Consider a reaction with a Keq ~ 1.0 * 10 + 65, at equilibrium, all reactants converts to product and no reactant remains.

Agree ___ or disagree ____ At equilibrium, products and reactants are still exchanging except that the majority of the chemicals in this mixture is the product

iii) Initially when a reaction begins, and reactant converts to product but before the reaction reaches equilibrium, the reaction

quotient, Q, is decreasing. Agree ___ or disagree ____ If the reaction is moving in the forward direction, the Qc < Keq, Qc is therefore increasing to catch-up with Keq iv) Solubility of solid in aqueous solution increases with a raise of pressure because at higher pressure, the solid solute will be force in

to the aqueous solvent more effectively. Agree ____ or disagree ___ Solubility of a solid is unaffected by pressure.

14 (15 pts) At 31.65°C in a 0.1000 L vessel, Kp = 7.110• 10-5 for the reaction; N2(g) + O2(g) N2O2(g) i) If 1.000 mol N2(g) and O2(g) are placed in the vessel, what is the total pressure at equilibrium? Since Kp is so small the amount of N2O2 formed is negligible.

€

Kp = kc(RT)Δn; Kc = kp(RT)−Δn; Δn = 1 - 2 = -1

Kc = kp(RT)−Δn = 7.11• 10-5(0.08206×304.65)1

Kc = 1.78• 10-3

N2 (g) + O2 (g) ↔ N2O2(g)

i 10.0M 10.0M 0Δ - x - x + x[e] 10.0 - x 10.0 - x x

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Kc = x

10.0 - x( )2≈

x

10.0( )2

= 1.78• 10-3

0.178M = x; N2[ ] = O2[ ] = 10.0M − 0.178M ≈ 9.82M

PTotal = PN2+ PO2

+ PN2O2=

(nN2+ nO2

+ nN2O2)

VRT

PTotal = 19.82 M 0.08206×304.65( ) PTotal = 4.95• 102 atm

ii) What mass (mg) of N2O2 is produced at equilibrium?

€

mass N2O2 = 0.178 molL

∗ 0.100L ∗ 60.0gmol

∗1000mg

1gmass N2O2 = 1.1• 103 mg

15 (20 pts) Consider the following theoretical reaction: S2(g) + 2Cl2 (g) 2SCl2 (g). Initially 10 mol of S and 10 mol of Cl2 is placed in a

1.0 L vessel at 25°C. The reaction begins and equilibrium is achieved after 50 minutes. At equilibrium, [SCl2] = 8.0 M.

6pts) i) Calculate Kc and Kp for this problem at 25°C

6pts) ii) Calculate partial pressure of S2, Cl2 and SCl2.

4pts) iii) Sketch the concentration change for S2, Cl2 & SCl2 from t = 0 min. to t = 50 min. in the graph below and label your graph.

4pts) iv) If the volume of the vessel is doubled at the 60 min time, What is the new concentrations of S2, Cl2 & SCl2 at the instance this

volume change occur. Sketch the concentration changes for S2, Cl2 & SCl2 after 50 min to equilibrium on the graph. (Use dash)

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S (g) + 2Cl2 (g) 2SCl2(g) Δn = -1 T = 298, RT = 24.454i 10 M 10M 0Δ - x -2x +2x[e] 10 - x 10 -2x 8

x = 4 [S] = 6 M, [Cl2] = 2 M, SCl2 = 8 M pS = 146.7 atm; pCl2 = 48.9 atm; pSCl2 = 195.6 atm

Kc = SCl22

(Cl2)2(S)

= 82

(2)2(6) = 2.67

Kp = Kc (RT)Δn = 2.67 (24.454)-1 = 0.10905

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If volume is doubled, Concentration is decreased to half its value

[S]New = [S] * 1.0 L / 2.0 L =3 M[Cl2]New = [Cl2] * 1.0 L / 2.0 L = 1 M[SCl2]New = [SCl2] * 1.0 L / 2.0 L = 4 M

New Equilibrium S (g) + 2Cl2 (g) 2SCl2(g) RT = 24.454i 3 M 1M 4Δ - x -2x +2x[e] 3- x 1 -2x 4 + 2x

Kc = (4 −2x)2

(1 - 2x)2(3 - x) →

10.68 x3 + 38.72 x2 - 18.7 x - 7.99 = 0solve using polynomial.

A = S, B = Cl2, P = SCl2

16 (15pts) The solubility product for BaSO4, is Ksp = 1.1 × 10-10 Calculate the molar solubility and the solubility (per 100g H2O) of BaSO4 under the following condition-

i) in water. ii) in 0.100 M H2SO4

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MW BaSO4 (s )= 233.39 g

mol BaSO4 (s) Ba+2

(aq) + SO4-2

(aq) i[ ] Inxs 0 0Δ[ ] -s +s +se[ ] - s s

Ksp = 1.1 • 10-10 = s2

s =1.049• 10-5 M,

s g100ml H2O

=

s =1.049• 10-5 molL

•1L

1000ml•233.39 g

mol

= 2.45 •10-6 gml

•100ml

s =2.45 •10-4g100ml H2O

=0.245 mg

100ml H2O

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MW BaSO4 (s )= 233.39 g

mol BaSO4 (s) Ba+2

(aq) + SO4-2

(aq) i[ ] Inxs 0 0.100Δ[ ] -s +s +se[ ] - s s

Ksp = 1.1 • 10-10 = s (0.100)

s =1.10• 10-9 M,

s g100ml H2O

=

s =1.10• 10-9 molL

•1L

1000ml•233.39 g

mol

= 2.57 •10-10 gml

•100ml

s =2.57 •10-8g100ml H2O

=0.257 ng

100ml H2O

17 (10 pts) Magnesium ammonium phosphate has the formula MgNH4PO4. It is only soluble in water (Ksp = 2.3•10-13). Calculate the

solubility for this salt.

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MgNH4PO4 (s) Mg+2 (aq) + NH4+ (aq) + PO4

3- (aq)

i Solid 0 0 0 Δ -s +s +s +s[e] Excess s s s

ksp = s3 ; 2.3•10-13 = s3; s = 2.3•10-133

s = 6.13•10-05

18 (10 pts Bonus) Ethene (C2H6) can be form from the following reaction: 2CH3OH(g) + H2 (g) C2H6 (g) + 2H2O (g).

Use the following data to calculate Kp for the above reaction.

2 CH4 (g) C2H6 (g) + H2 (g) K’c = 9.5 x 10-13

CH4 (g) + 2H2O(g) CH3OH(g) + H2 (g) Kc = 2.8 x 10-21

Keq = Rxn1 * [Rev (Rxn2 x 2)] = 9.5 • 10-13 * [1 / 2.8 • 10-21 ]2 = 1.21 • 1029 = Kc = Kp