Equilibrium Process

Practice Exam Equilibrium Name (last)_____________________ (First)_______________________ Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper number of significant figures. By signing your signature above you agree that you have worked alone and neither give nor received help from any source. Keep your eyes on your own paper at all times

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 = 0.08206 L•atm / mol•K)

Quadratic Equation: ax2 + bx + c = 0

€

x =−b ± b2 −4ac

2a

1 What is the general equilibrium constant expression, Keq, for the reversible reaction: 2 A(aq) + 3 B(aq) D C(l)

a) Keq = [A]2[B]3/[C]0 b) Keq = [A]2[B]3/[C]1

c) Keq = [C]0 / [A]2 [B]3 d) Keq = [C]1 / [A]2[B]3 e) none of the above

2 At 100°C, Kp is 2.65 for the equilibrium: SO2Cl2(g) D SO2(g) + Cl2(g).

If ΔH°rxn = +93.1 kJ/mol, what is K’p at 200°C ?

a) K’p = 2.65

b) K’p < 2.65

c) K’p > 2.65

d) none of these

e) cannot be determine

3 In which reaction does Kp = Kc ?

a) NO (g) + O2 (g) D N2O3(g)

b) N2 (g) + O2 (g) D 2 NO(g)

c) CaCO3 (s) D CaO(s) + CO2 (g)

d) N2 (g) + H2O (g) D NO( g) + H2(g)

e) none

4 Which of the following will shift the reaction to the left (reactant)

C2H4 (g) + H2O(g D C2H5OH (g) ΔH = -47.8 kJ

a) Addition of steam b) removal of ethanol

c) increase in pressure d) addition of heat e) none

5 Consider the diagram to the right for: A(g) + B(g) D C(g) ΔH = ( + ) From this reaction coordinate diagram, Keq can be determined to be:

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

6 Concentration and solubility of gas The solubility of CO2 gas in water is 0.240 g per 100 ml at a pressure of 1.00 atm and 10.0°C. Consider a 100-mL

solution that is saturated with CO2 at 1.00 atm and 10°C. You can assume that the partial pressure of CO2 is 1.00atm. Answer the following question concerning this solution.

i) (Multiple choice) If the pressure is change to 760 mmHg: a) more CO2 can be dissolve in solution b) The solution is saturated with CO2 c) The solution will release CO2 gas d) the temperature of the solution will drop e) none

ii) (Multiple choice) If the temperature is raised to 25°C: a) The solution is saturated with CO2 b) more CO2 can be dissolve in solution c) The solution will release CO2 gas d) the pressure of the solution will drop e) none

iii) (Multiple choice) If an additional 50-mL of solvent is added : a) The solution is saturated with CO2 b) the mole fraction , χ, of CO2 will increase c) The solution will release CO2 gas d) more CO2 can be dissolve in solution e) none

iv) Calculate the molar concentration (M), molality (m), parts per million concentration (ppm), and mole fraction (χ) of CO2 at 10.0 °C. Assume the density of the solution = 1.00 g/cc

M = 0.240g∗ 1.0mol

44.0g∗

1100ml

∗1000ml

1L= 5.45• 10−2

m = 0.240g∗1.0mol

44.0g∗

1100ml

∗1ml1g

#

$ %

&

' ( *

1000g1kg

)

* +

,

- . = 5.45∑10−2

ppm =

0.240g100g + 0.240g

∗106 = 2394ppm

χ =

0.240g∗ 1mol44.0g

0.240g∗ 1mol44.0g

$

% &

'

( ) + 100g∗ 1mol

18.0g$

% &

'

( )

= 9.81 ⋅10−4

iv) Calculate Kp for the above reaction at 10°C. The Mass Action expression is Kc = [CO2(aq)] /[CO2(g)]

Kc =CO2 (aq)[ ]CO2 (g)[ ]

=

massMWt

PRT

=0.0545 M

1atm

0.08206 L •atmmol•K

∗283K#

$ %

&

' (

= 1.266 Kp = Kc (RT )−1 = 1.266 0.08206 L •atm

mol•K∗283K$

% &

'

( )

−1= 5.45∗10−2

7 Tooth enamel is composed of the mineral hydroxylapatite, Ca5(PO4)3OH (Ksp= 6.8•10-37). By adding fluoride to

drinking water, the fluoride can reacts with hydroxylapatite to form a more tooth decay-resistant mineral, called fluorapatitie, Ca5(PO4)3F (Ksp = 1.0•10-60) Calculate the solubility of both hydroxylapatite and fluorapatitie. How many times greater is this mineral more soluble ?

Ca5(PO4 )3OH ! 5Ca+2 + 3PO4-3 + OH-

Ksp = 6.8 ⋅10-37 = [Ca+2]5 •[PO4-3]3 •[OH-]

6.8⋅10-37 = [5s]5 •[3s]3 •[s] 6.8⋅10-37 = 3125s5 •27s3 •s 6.8⋅10-37 = 84375s9

8.06⋅10-42 = s9

2.72⋅10-05 = s

Ca5(PO4 )3F ! 5Ca+2 + 3PO4-3 + F-

Ksp = 1.0 ⋅10-60 = [Ca+2]5 •[PO4-3]3 •[F-]

1.0 ⋅10-60 = [5s]5 •[3s]3 •[s] 1.0 ⋅10-60 = 3125s5 •27s3 •s 1.0 ⋅10-60 = 84375s9

1.19⋅10-65 = s9

6.11⋅10-08 = s

2.72•10-5M / 6.11•10-8 M = 445 Ca5(PO4)3OH is 445 more soluble than Ca5(PO4)3F

8 Keq relation to chemical equation

As an environmental chemist you are studying catalytic converters and urban smog.

(1) N2 (g) + O2 (g) D NO (g) K’c = 4.8 • 10-10

(2) NO2 (g) D NO (g) + O2 (g) K”c = 1.1 • 10-5

Using the information above, what is Kc for the following reaction:

(3) 2 NO2 (g) D N2 (g) + 2 O2 (g) Kc = ? , T = 35°C

Rev Rxn1 : 2NO(g) ↔ N2 (g) + O2 (g) K = 1K'c

Rxn2 : 2NO2(g) ↔ 2NO(g) + O2 (g) K = K"c

Rxn3 : 2NO2(g) ↔ 2N2(g) + 2O2 (g) Kc = K"cK'c

=1.1• 10−5

4.8 • 10−10 = 2.29 • 104

1) 2NO (g) D N2 (g) + O2 (g) Rev Rxn 1

2) 2NO2 (g) D 2 NO (g) + O2 (g) Keep Rxn 2

3) 2 NO2 (g) D N2 (g) + 2 O2 (g) Kc = K”c/ K’c

9 Equilibrium concentration and Keq When 1.00 mol of PCl5 is introduced into a 5.000 L container at 500K, 75.00 % of the PCl5 dissociates to give an

equilibrium mixture of PCl5, PCl3 and Cl2. PCl5 (g) D PCl3 (g) + Cl2 (g) Calculate [PCl5, [PCl3] and [Cl2] at equilibrium and Kc & Kp.

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

i 0.20 M 0 M 0MΔ - x M +x +x e (.75∗ .20)M x x

PCl5[ ]eq = .20 - x = .05, x = .15

e[ ] .05 M[ ] .15[ ] .15[ ]

Kc =Cl2[ ]∗ PCl3[ ]

PCl5[ ]=

.15[ ]∗ .15[ ].05[ ]

= .45

Kp = Kc RT( )1 = .45 .08206 L •atmmol•K

∗500K#

$ %

&

' (

1= 18.46

10 Determination of Q: Cyclohexane (CH) undergoes a molecular rearrangement in the presence of AlCl3 to form methycyclopentane (MCP)

according to the equation CH D MCP: Kc = 0.143 at 25°C for this reaction. Predict the direction the reaction will shift to reach equilibrium if the initial concentrations of the species are [CH] = 0.200 M and [MCP] = 0.100 M.

11 Equilibrium Problem and Q Exactly 0.800 mol of POCl3 is sealed in a 20-mL flask at 620 K: POCl3 (g ) D PCl3 (g) + O2 (g) The equilibrium constant Kc, is 0.090 for this reaction.

i) Which statement is true about the above reaction?

a) Kc = Kp b) This reaction is exothermic

c) Addition of N2 result in a shift of the reaction to the left d) none

ii) Calculate Kp for the above reaction?

Kp = Kc(RT)2−1 = .090(.08206 L •atm

mol•K∗620K)1 = 4.57

iii) If the concentration of each specie is [POCl3]=0.00013, [PCl3] =0.0087 [O2 ]= 0.0078

in which direction will the equilibrium shift?

2 POCl3 (g ) D 2 PCl3 (g) + O2 (g) Balance the Equation !!!!

Q =

PCl3[ ]2 ∗ O2[ ]POCl3[ ]2

=.0087[ ]2 ∗ .0078[ ]

.00013[ ]2= 34.9 ≈35

Q = 35 > Kc Reaction shifts or moves to the left (Reactant) 12 LeChatelier Principle Please complete the following table for the hypothetical reaction: R(g) D 2P(g) ΔH = (- )

Disturbance Affecting …. Net direction of Rxn. (Forward, Reverse or none)

Effect on K eq. (Increase, decrease or no change)

Concentration Increase [reactant] Forward no change

Increase [product] Reverse no change

Pressure (Vol) Increase P (volume decrease)

Reverse no change

Decrease P (increase volume)

Forward no change

Temperature Increase Temp Reverse decrease

Decrease Temp Forward increase

Catalyst Add catalyst to reactant

none no change

13 Fe(NO3)3 (0.00100 mol) and KSCN (0.200 mol) are added to water to make 1-liter of solution. The red complex ion

FeSCN2+ is produced. Calculate the conc. of Fe3+(aq) & FeSCN2+(aq) at equilibrium, if Kf of the FeSCN2+ is 8.9 × 102.

Fe+3 + SCN- ! FeSCN+2

i[ ] 0.00100 0.200 0Δ[ ] - x - x +xeq[ ] 0.00100 - x 0.200 - x +x

kf =8.9•102 = FeSCN+2[ ]

Fe+3[ ] • SCN-[ ]

kf =8.9•102 = x[ ]0.00100 - x[ ] • 0.200 - x[ ]

kf =8.9•102 = x[ ]2.00 •10−4 − .201x + x2

8.9•102 2.00•10−4 − .201x + x2( ) = x[ ]

0.178−179.89x +890x2 = 0

8.9•102 2.00•10−4 − .201x + x2( ) = x[ ]

0.178−179.89x +890x2 = 0

a =890, b = -179.9, c = 0.178

x = 179.89 ± -179.89( )2 −4 890( ) 0.178( )

2 890( )

x = 179.89 ± 31726.73212 890( )

x = 179.89 ± 178.121780

= 0.20113 or 0.000994386

x = 9.94 •10-4 MFeSCN+2[ ] = 9.94 •10-4 M

Fe+3[ ] =5.614 •10−6 M

14 Consider the table below: Analysis for KCl

i) What is the temperature in which K2Cr2O7 and KCl both

have the same solubility?

K2Cr2O7 T = 59° s = 38 g per 100 ml H2O

K2Cr2O7 T = 71° s = 49 g per 100 ml H2O

ii) What is the solubility (g per 100g H2O) at this temperature.

See chart below

iii) What is the molar solubility for each of these compounds at

this temperature

See chart below

iv) Calculate the Ksp for each of these chemicals at this temperature

Ksp = [K+]{Cr2O7-}= (2s)2 • s = 4s3

Chem MWt s (g/100ml) [s] M T °C Ksp

K2Cr2O7 294.18 49.00 1.67 71 18.48

15 Solution is for Ca3(PO4)2

Look up the Ksp value for the salt Ca3(PO4)2 i) What is the molar solubility? ii) What is the solubility express in g per 100 ml solution? iii) What is the molar solubility in 0.010 M K3PO4 solution?

Compound Ksp Solubility 1 Ca3(PO4)2 2.0 • 10-29 2 Mg3(PO4)2 6.3 • 10-26

iv) iv) Given 1 •10-5 M solution of Ca(NO3)2, how much H3PO4 is necessary to cause a precipitate?

Ca3(PO4 )2 ! 3Ca+2 + 2PO4-3

MW = 310.18 gmol

Ksp = 2.0 ⋅10-29 = [Ca+2]3 •[PO4-3]2

2.0 ⋅10-29 = [3s]3 •[2s]2

2.0 ⋅10-29 = 108s5

1.85 ⋅10-31 = s5

7.14 ⋅10-07M = s

solubility g per 100ml H2O

s = 7.14 ⋅10-07 molL

∗310.18gmol

∗.100L100ml

s = 2.21 ⋅10-05g per 100cc H2O

Ksp = 2.0 ⋅10-29 = [Ca+2]3 •[PO4-3]2

2.0 ⋅10-29 = [3s]3 •[0.010M]2

2.0 ⋅10-29 = .0027s3

7.41⋅10-27 = s3

1.95 ⋅10-09 = s

Ksp

= 2.0 ⋅10-29= [Ca+2]3 •[PO4-3]2

2.0 ⋅10-29= [1 ⋅10-5]3 •[2s]2

2.0 ⋅10-29= 1.0 ⋅10-15 •4s2

2.0 ⋅10-29= 4.0 ⋅10-15 •s2

5.00 ⋅10-15= s2

7.07-08M= s

at Equilb, [PO42-] =2s = 1.414e-7M,

H3PO

4 =1.414e-7M

16 Check mark if you agree___ or disagree___ with the statements below and justify for your answer completely. i) Consider a reaction with an Keq ~ 1.0 * 10-55, when the reaction reaches equilibrium, then no product is ever formed.

agree___ or disagree___

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

ii) 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

iii) For an exothermic reaction, a decrease in temperature leads to an increase in K eq for the reaction at the new

temperature. agree___ or disagree___ A g P + E, If the temperature decreases, the reaction shifts to the right so that

Keq (new) is bigger than Keq (old) 17 Equilibrium Kp & Kc:

Nitric oxide and bromine, at initial

partial pressures of 98.40 and 41.30

torr, respectively, were allowed to

react at 300 K in a seal vessel. At

equilibrium, the partial pressure of

Br2 = 22.10 torr.

2NO(g) + Br2(g) D 2 NOBr(g)

a) Calculate the pressure (in torr) of

each species at equilibrium.

b) Calculate Kp

18 Equilibrium:

Exactly 0.200 mol of PCl5 is sealed in a 2.0-L flask at 620 K.: PCl5 (g) D PCl3 (g) + Cl2 (g) The equilibrium constant Kc, is 0.600 for this reaction. (i) What is a, b and c in the quadratic equation that will give information on the concentration of all species at

equilibrium. (ii) What direction will the rxn proceed if the concentrations are: [PCl5]=0.00013, [PCl3] =0.0087 [Cl2 ]= 0.0078 ? (iii) If the equilibrium constant changes to 1.622 at 800 K, is this an exothermic or endothermic reaction? Explain.

19 LeChatelier The Haber process is used for the production of ammonia according to the reaction:

N2(g) + H2(g) NH3(g) at 127°C, the Kc =3.8•104 L2 mol-2

at 500°C, the Kc =6.0•10-2 L2 mol-2 i) The Haber reaction is:

a) An exothermic process b) An endothermic process c) is an adiabatic process d) none ii) The production of ammonia can be increase by:

a) increasing the temperature but decreasing the pressure b) decreasing the temperature and pressure c) increasing the pressure but decreasing the temperature d) increasing the pressure and temperature

Consider the graph for the Haber reaction. The reaction takes place in a

1-Liter vessel. The initial concentration of [H2] is 12.0 molar and that of [N2] is 4.0 molar.

iii) Sketch the change of the concentration of N2 from 0 to 3.5 min. See Diagram

iv) At 2 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine v) At 5 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine

vi) Calculate the equilibrium constant at the 4 .0 minute mark.

Q =

NH3[ ]2

H2[ ]3 ∗ N2[ ]=

8.0[ ]2

10.0[ ]3 ∗ 2.4[ ]= 0.027 ≈ .03

v) Calculate the equilibrium constant at the 8.0 minute mark.

Q =

NH3[ ]2

H2[ ]3 ∗ N2[ ]=

9.0[ ]2

11.0[ ]3 ∗ 2.0[ ]= 0.0304 ≈ .03

20 Heterogeneous Equilibrium: 10.0 ml of 7.40•10-4M HCl is added to 10.0 ml silver nitrate solution according to the equation below. For a

precipitation to occur, what is the initial concentration of silver nitrate ( or Ag+ ) ? AgNO3(aq) + HCl(aq) D AgCl(s) + HNO3(aq) Keq = 5.56 • 109

First write the net ionic equation for the reaction above, then remember (Ksp = 1/Keq )

[Cl-] = 7.40 ∗ 10−4 10.0ml20.0ml

= 3.7 ∗ 10−4 ; Ksp = 1Keq

= 1.80 ∗ 10−10

AgCl(s) ↔ Ag+(aq) + Cl-(aq) Ksp = 1.80∗ 10−10

[i] Excess 0 3.7 ∗ 10−4

Δ - s + s + s[e] Excess + s + 3.7 ∗ 10−4 + s

1.80∗ 10−10 = s (3.7 ∗ 10−4 ); s = 4.86 * 10-7

[Ag+] = 4.86 * 10-7 = [AgNO3]