Chemical Equilibrium 5.1 - Calculate the partial pressure of monoatomic hydrogen in hydrogen in hydrogen gas at 2000K and 1atm.
1
H 2 ( g ) H ( g ) 2 o H 29 8 217990 J
Para:
o
S 29 8
49.35 J / K
Então:
1
3
C p C P , H ( g ) C P , H 2
H
o 2000
H o 298
2
2
2000
29 8
1
8.314 31 3.035 2
o C P dT H 29 8 C P ( 2000 298)
217990 3.0351702 212824 J
S
o 2000
S
o 29 8
2000
29 8
o C P dT S 29 8 C P ln
49.35 3.035 ln
2000 298
2000 298
43.57 J
0 0 0 G2000 H 2000 T S 2000 212824 2000 43.57 125684 J o G2000 RT ln K RT ln
RT ln
P H 2
1/ 2
P H ( g )
125684 ln
P H ( g ) P H 2
P H 2
1/ 2
RT ln
1/ 2
P H ( g )
P H 2
P H ( g )
125684 8.314 2000
P H 2 ( g ) P H 2( g ) 1, P H 2 1 ln
1 P H ( g )
1/ 2
atm 7.562 P H 0.0005atm
7.56
5.2 - For the reaction : Co( s )
1 2
O2 ( g ) CoO( S )
G o 59850 19.6T where G o is in calories and T is in Kelvin. a) Calculate the oxygen equilibrium pressure (atm) over Co and CoO at 1000 C. b) What is the uncertainty in the value calculated in part a if the error in Ho term is estimated to be 500 cal? Solução: a) At a) At 1000C, Go =-59850+19.6T=-5985 =-59850+19.6T=-59850+19.6 0+19.6(1000+273) = -34899.2cal = -1458.79J/mol No equilibrio: G
o
RT ln K RT ln
ln P O2
P O2
1
P O2
1/ 2
1 2
RT ln P O2
27.6
1.07 10
12
atm
b) Incerteza b) Incerteza em Ho = 500cal/mol = 2090J/mol Então a incerteza em Go = 500cal/mol = 2090J/mol Portanto:
1
1 RT ln P O' 2 RT ln P O 2 2090 2 2
1 2
RT ln
ln
P O' 2 P O 2
P O' 2 P O 2
P O' 2 P O 2
2090
0.25
1.286
P P O 2
28.6%
145879 J
Similarly, Similarly, uncertainty in Ho =- 500cal/mol =- 2090J/mol
1
1 RT ln P O' 2 RT ln P O 2 2090 2 2 '
ln
P O 2 P O 2
0.25
'
P O 2 P O 2
0.779
P P O 2
22.1%
5.3 - Calculate the temperature at which silver oxide (Ag2O) begins to decompose into silver and oxygen upon heating: a) in pure oxygen at P = 1 atm; b) in air at Ptotal = 1 atm.
DATA
H f forA forAg g 2 O 7300 cal / mol
Standard En tr opy at at 298K
[cal/(mol.K)] Ag2O 29.1 O2 49.0 Ag 10.2
Assume that Cp = 0 for the decomposition reaction.
Solution: (a) Ag2O = 1/2O2 + 2Ag cal / mol 30514 H o H f o , AgO 7300cal 1
S o 2S Ag , 298 S O 2, 298 S Ag 2O, 298 2
1
2 10.2 49 29.1 66.044 J / mol . K 2
044T G o H o T S o 30514 T S o 30514 66.044
when Ag2O begins to decompose,
G G o RT ln J 0 ie 30514 66.044T RT ln P O 2 0 a) in pure oxygen at 1 atm, RTlnP O2 = 0 30514-66.044T = 0 T = 462K b) in air at P total = =1 atm , P O2 =0.21 ie. 30514- 66.044T + RTln0.21 = 0 T = 386K
5.4 - One step in the manufacture of specially purified nitrogen is the removal of small amounts of residual oxygen by passing the gas over copper gauze at approximately 500 C. The following reaction takes place: 2Cu( s)
1 2
O2 ( g ) Cu2 O( s)
a)Assuming that equilibrium is reached in this process, calculate the amount of oxygen present in the purified nitrogen; b) What would be the effect of raising the temperature to 800 C? Or lowering it to 300 C? What is the reason for using 500 C? c) What would be the effect of increasing i ncreasing the gas pressure?
For
2Cu( s)
1 2
O2 ( g )
o
Cu2 O( s) , G
(in calories ) is – 39850+15.06T. 39850+15.06T.
Solution: (a) When the equilibrium is reached, 1
G G o RT ln J G o－ RT ln PO 0
2 4.18 (39850 15.06T ) ln P O 2 1 RT 2
2
T = 500 C = 773K ln P O 2
4.18 (39850 15.06 773) 36.69 1 8.314 773 2
P O 2 1.14 10 26 atm
(a) at T=300C=573K, Although the equilibrium P O2 is very low, kinetically the reaction is not favoured and reaction speed is very slow. slow. So 300 C is not suitable
at -10
At T=800C=1073K, lnPO2 =-22.2, PO2 =2.2810 atm. b) At 800C, if the equilibrium is reached, nitrogen can be of high purity level. However, However, at this high temperature , particles of Cu will weld together to reduce effective work surface. So it is not suitable to use this high temperature in purification either. either. c) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level level of N 2 .
5.5. The solubility of hydrogen(PH2 = 1 atm ) in liquid copper at 1200 C 7.34cm3(STP) per 100g of copper. Hydrogen in copper exists in monatomic form. (a) Write the chemical equation for the dissolution of H2 in copper; (b) What level of vacuum(atm) must be drown over a copper melt at 1200 C to reduce its hydrogen content to 0.1 0. 1 cm3 (STP) per 100g? (c) A 100g melt of copper at 1200 C contains 0.5 cm3(STP) of H2. Argon is bubbled through the melt slowly so that each bubble equilibrates with the melt. How much argon must be bubbled through the melt to reduce the H2 content to 0.1 cm3(STP) per 100g ?
Note: STP means standard temperature temperature and pressure(298K and 1 atm). atm). Solution: (a) H2(g) = 2H 1
1
(b) H K a 2 P H 22
P H 2 1atm atm, H 7.34cm 3 / 10 100 0 gCu gCu 1
K a 2 is
[ H ] P H 1 /22
a constant, [ H ]' ( P H ' 2 )1 / 2
' 1/ 2 H 2
( P
)
P H 1 /22 [ H ]' [ H ]
0.1 7.34
0.0136
( P H ' 2 ) 0.00019atm 18.56 10130 18.8 Pa
(c ) The amount of H2 needed to be brought out by Ar is:
n
P V RT
10130 (0.5 0.1) 10
6
8.314 298
1.6 10 6 mol
This amount of H2 is in equilibrium with the melt in the bubble, ie. The partial pressure of H2 in the bubbles is 18.8Pa. nRT T 4.05 102 P H ' 2V bubble nR P H 2 4.05 10 / 18.8 0.00215m 2.15 L '
2
3
2.15L Ar is needed to be bubbled into the melt.
5.6 - The following equilibrium data have been determined for the reaction: NiO NiO( s) CO( g ) Ni( s) CO2 ( g ) -3
T( C) K 10 663 4.535 716 3.323 754 2.554 793 2.037 852 1.577
o
o
a)Plot the data using appropriate axes and find H , K and G at 1000K; b)Will b)Will an atmosphere atmosphere of 15%CO 2, 5%CO, and 80%N 2 N2 oxidize nickel at 1000K? Solution: (a)
d ln K a
H o
Plot ln K a ~ 1/ T
1 d ( ) R T
8.6
Kduishu Linear Fit of Data1_Kduishu
8.4
8.2
8.0 a
K n l 7.8
lnKa =2.01+6003(1/T)
7.6
7.4
7.2 0.88
0.90
0.92
0.94
0.96
0.98
1/T, 10
1.00
1.02
1.04
1.06
1.08
-3
. d ln K a dT
H o R
6003 H o R 6003 49909 J
At T=1000K, lnK a =8.01, K a = 3010 o G1000 RT ln K a 8.314 1000 8.01 66600 J 66.6kJ
(b)
G G o RT ln J RT ln K a RT ln J RT ln J
15% 5%
J K a
3 K a
So the atmosphere will oxidize Ni.
5.7 - At 1 atm pressure and 1750 C, 100 g of iron dissolve 35cm3 (STP) of nitrogen. Under the same conditions, 100 g of iron dissolves 35 cm3 of hydrogen. Argon Argon is insoluble in molten iron. How much gas will 100 g of iron dissolve at 1750 C and 760 mm pressure under an atmosphere that consists of:
(a) 50% nitrogen and 50% hydrogen? (b) 50% argon and 50% hydrogen? (c) 33% nitrogen, 33 hydrogen, and 34 argon? Solution: N2 =2N, H2 = 2H 1
1
N K a, N 2 P N 22
1
1
2 P H 22 , ， H K a, H
[ N ]
For N2 dissolving: P
1/ 2 N 2
[ N ]
[ H ]
For H2 dissolving： P
'
( P N 2 )
1/ 2 H 2
'
1/ 2
[ H ]' ( P H ' 2 )1 / 2
a)For dissolving N2, P N2 = 1 atm, [N]=35cm3/100g melt, [ N ] ‘
( P N ' 2 )1 / 2 [ N ] 1/ 2 N 2
P
35 (0.5)1 / 2 24.75cm3 / 100 g melt
3
’
similarly: [H] =24.75cm /100g melt 3 total gas : [H] ＋[N] = 49.5 cm /100g melt 3
b) [H] =24.75 cm /100g melt 1/2
1/2
c)[H]＋[N] = [N](0.33) /1+[H](0.33) /1=20.10+20.10 = 3 40.2cm /100g melt
5.8 - Solid silicon in contact with solid silicon dioxide is to be heated to a temperature of 1100 K in a vaccum furnace. The two solid phases are not soluble in each other, but is known that silicon and silicon dioxide can react to form gaseous silicon monoxide. For the reaction: Si( s) SiO2 2SiO SiO ( g )
the Gibbs free energy change (J) is. G o 667000 25.0T ln T 510T
(a) Calculate the equilibrium pressure of SiO gas at 1100K; (b) For the reaction above, calculate Ho and So at 1100K; (c) Using the Ellingham chart (Figure 5.7), estimate the pressure of oxygen (O2) in equilibrium with the materials in the furnace. 2 a) G o RT ln K RT ln P SiO 2 RT ln P SiO At 1100K, o G =667000+25.0TlnT-510T = 667000+25.01100ln1100-510 1100 =667000+192584-561000 =298584 -2RTlnPSiO =298584 lnPSiO =-16.32 -8 PSiO = 8.110 (atm)
o
b)G =667000+25.0TlnT-510T =-RTlnK ln K
667000 RT
d ln K （
H o R
667000 2
RT
25 R
ln T 510
25 RT
667000
)dT (
25 R
667000 R
T
H o 667000 25T o
T = 1100K, H = 639500J S o
H o G o T
-30
667000 298584
(c ) PO2 =10 atm
1100
334.9 J / K
25 R
T )d (1 / T )
H o R
d (1 / T )
5.9 - What is the pressure of uranium (gas) in equilibrium with uranium dicarbide o
DAT DATA: At 2263K,
for UC2 is – 82,000 82,000 cal/mol
G f
Vapor pressure of pure uranium is: ln P (atm atm, uranium ) 25.33
100000
T
(T in K )
Solution: ca l / mol 342760 J G f o 82000cal
U ( g ) C ( s ) UC 2( s )
G f o ,UC 2 RT ln K RT ln(
1
P u ( g )
) RT ln P u ( g )
RT ln P u ( g ) 342760 ln P u ( g ) 1.2 10 (atm) 8
vapor pressure of uranium: ln P u ( g ) (atm, uranium ) 25.33 P u ( g )
100000 T
25.35
100000 2263
18.89
0.6 10 8 (atm)
the vapor pressure is lower than the one determined by chemical reaction. It is the one in equilibrium with dicarbide.
5.10 - The direct reduction of iron oxide by hydrogen maybe represented by the following equation: Fe Fe2 O3 3 H 2 2 Fe Fe 3 H 2 O
What is the enthalpy change, in joules, for the reaction? Is it exothermic or endothermic? 2 Fe
3
O2 Fe2 O3 2 1 H 2 O2 H 2 O 2
G o 810250 254.0T G o 246000 54.8T
Solution: 2 Fe
3
O 2 Fe 2 O3
2 1 H 2 O 2 H 2 O 2
G3o
(1) ( 2)
G1o 810250 254.0T G 2o 246000 54.8T
Fe Fe2 O3 3 H 2 2 Fe Fe 3 H 2 O (3) o o 3G2 G1 3 24600 54.8T (810250 254.0T ) 72250 89.6T
H 72250 J o 3
The reaction is an endothermic one.
G3o
5.11 - Calcium carbonate decomposes into calcium oxide and carbon dioxide according to the reaction CaCO3 CaO CO2
DAT DATA for the pressure of carbon dioxide in equilibrium with CaO and CaCO3: Temperature (K) 1030 921
Pressure (atm) 0.10 0.01
a) What is the heat effect ( H) of the decomposition of one mole of CaCO3 ? Is the reaction endothermic or exothermic? (b) At what temperature will the equilibrium pressure of CO 2 equal one atmosphere? Solution: (a) H o 1 d ln K d , K P CO R T H o 1 d d ln P CO 2 R T P CO 2, 2 H o 1 1 ln P CO 2,1 R T 2 T 1 0.01 1 H o 1 ln 0.1 R 921 1030 H o 166528 J
2
the reaction is endothermic b) PCO2 =1atm ln
1 0.1
H o 1 R
(
T
1 1030
)
T 1168 K
At 1168K, the equilibrium pressure of CO 2 equals one atmosphere.
5.12 - In the carbothermic reduction of magnesium oxide, briquettes of MgO and and carbon are heated at high temperature in a vacuum furnace to form magnesium (gas) and carbon monoxide(gas). (a) write the chemical reaction for the process; (b) What can you say abou the relationship between the pressure of magnesium gas and the pressure of o f carbon monoxide? (c) Calculate the temperature at which the sum of the pressures of Mg(gas) and CO reaches on atmosphere. With With T in Kelvin, the free energies of formation, in calories, of the relevant compounds are: MgO
G f o 174000 48.7T
CO
G f o 28000 20.2T
(a). The reaction is: (b).
G G o
o f ,CO
G
MgO Mg O( s) C ( s) CO( g ) Mg ( g )
o f , MgO
146000 68.9T
G o RT ln K RT ln( P CO( g ) P Mg ( g ) ) (146000 68.7T ) P CO P Mg
(c) Ptotal = 1 atm, P CO = 0.5 atm, P Mg =0.5 atm RT ln(0.5 0.5) (146000 68.7T ) 4.18
T = 2037 K
5.13 - Metallic silicon is to be heated to 1000 C. To prevent the formation of silicon dioxide (SiO 2), it is proposed that a hydrogen atmosphere be used. Water Water vapor, which is present as an impurity in the hydrogen, can oxidize the silicon.
(a) Write Write the chemical equation for the oxidation of silicon to dioxide by water vapor; (b) Using the accompanying data, where Go is in joules, determine the equilibrium constant fro the reaction at 1000 C (1273K); (c) What is the maximum content of water in the hydrogen (ppm) that is permitted if the oxidation at 1000 C is to be prevented ? (d) Check the answer to part c on the Ellingham diagram (Figure 5.7) DATA
H 2 ( g )
1
O2 ( g ) H 2 O ( g )
G o 246000 54.8T
2 Si O2 SiO2 ( s )
G o 902000 174T
Solution: (a) Si( s) 2 H O( g ) SiO SiO ( s) 2 H ( g ) 2
(b) H ( g ) 12 O 2
2
2
2
( g ) H 2 O( g )
Si O2 SiO2 ( s )
(1) ( 2)
G(o1) 246000 54.8T G(o2) 902000 174T
Si( s) 2 H 2 O( g ) SiO2 ( s) 2 H 2 ( g )
(3)
G3o
G3o G2o 2G1o 902000 174T (246000 54.8T ) 2 410000 64.4T G3o RT ln K 410000 64.4T 410000 64.4 1273 At T 1273 K , ln K 31 8.314 1273 K 2.9 1013
(c) 2
P H 2( g ) 13 K 2.9 10 P H 2O ( g ) P H 2( g ) P H 2O( g ) P H 2O( g ) P H 2( g )
5.38 106
1 5.38 10
6
0.186 10 6 0.186 ppm
5.14 - Solid barium oxide(BaO) is to be prepared by the decomposition of the mineral witherite (BaCO 3) in a furnace open to the atmosphere (P = 1 atm).
a) Write the equation of the decomposition (witherite and BaO are immiscible). b) Based on the accompanying accompanying data, data, what is the heat effect effect of the decomposition of the witherite(J/mol). Specify whether heat is to be added (endothermic) or evolved (exothermic). c )How high must the temperature be raised to raise the carbon dioxide pressure above the mineral to one atmosphere? DATA Thermodynamic Properties
G f o ( 298)
[KCAL/(g.mol)]
H f o ( 298)
CO2 -94 -94 BaO -126 -133 BaCO3 -272 -291 (Assuming that CPCO2+ CP BaO = CP BaCO3 BaCO3)
Solution: BaCO O3 ( s) BaO BaO( s) CO2 ( g ) (a) BaC
(b) CP = 0 H H f o ,CO2( 298) H f o , BaO( 298) H f o , BaCO3(298) 94 133 (219) 64kcal 267.52kJ
the reaction is endothermic (c ) At 298K, o G298 G f o ,CO2, 298 G f o ,CaO, 298 G f o ,CaCO3, 298
94 126 272 52kcal 217.36kJ o o o G298 H 298 T S 298 o S 298
o o H 298 G298
T
GT o H T o T S T o
when PCO2=1 atm,
(267.52 217.36) 298
168 J / mol . K
GT o 0, ie 2675201 168T T 1592 K
indicated, Mg has a very stable oxide. 5.15 - As the Elligham diagram indicated, Therefore Mg metal can be obtained from the oxide ore by a two-step process. First the oxide is converted converted to a chloride. chloride. In the second step the chloride is converted to metal Mg by passing H 2 gas over liquid MgCl2 at 1200 C. The reaction in this last step is: MgCl 2 (l ) H 2 ( g ) Mg ( g ) 2 HCl ( g )
a) Calculate the equilibrium pressure of H2(g), Mg(g) and HCl(g) if the total pressure is maintained constant at 1 atm. b)Calculate the maximum vapor vapor pressure of H2O that can be tolerated tolerated in the hydrogen without causing the oxidation of the Mg vapor. DATA Reaction
G
Mg(g)+Cl2(g) = MgCl (l) H2 (g) + Cl2(g) = 2HCl(g) Mg(g) +1/2O2(g) = MgO(s) H2 (g) + 1/2O2(g) = H2O(g)
o
at 1200C
－425484 J －207856 J －437185 J －165280J
Solution: (a) MgCl 2 (l ) H 2 ( g ) Mg ( g ) 2 HCl ( g ) (1) G1o Mg(g)+Cl2(g) = MgCl (l) (2) H2 (g) + Cl2(g) = 2HCl(g)
o
G 2
(3)
－425484 J o
G 3
－207856 J
G1o G3o G2o 207856 425484 217628 J 2
G RT ln K RT ln o 1
P HCl ( g ) P Mg ( g ) P H 2( g )
2
8.314 1473 ln
217628 2
ln
P HCl ( g ) P Mg ( g ) P H 2( g )
＝17.78
2
P HCl ( g ) P Mg ( g ) P H 2( g )
5.27 10 7
P H 2 ( g ) P Mg ( g ) P HCl 1, P HCl 2 P Mg ( g )
let PMg(g) = x, PHCl = 2 x, PH2 = 1-3 x
P HCl ( g ) P Mg ( g ) P H 2( g )
1 3 x x(2 x)
2
5.27 107
x 1.6 10 3 (atm) o
Mg(g) + H2O(g) = MgO(s)+ H2 (g) (4) G 4 o Mg(g) +1/2O2(g) = MgO(s) (5) G 5 －437185 J o －165280J H2 (g) + 1/2O2(g) = H2O(g) (6) G 6 G4o G5o G6o 437185 (165280) 271905 J G4o RT ln K RT ln
P H 2( g ) P H 2O( g ) P Mg ( g )
271905 ln
P H 2( g ) P H 2O( g ) P Mg ( g )
＝22.2
1.6 10 3 P H 2O( g ) 1.6 10
3
22.2
ln P H 2O ( g ) 22.2 P H 2O ( g ) 2.28 10 10 (atm)
8.314 1473 ln
P H 2( g ) P H 2O( g ) P Mg ( g )
5.16 - A common reaction for the gasification of coal is: H 2 O( g ) C ( s) H 2 ( g ) CO( g )
a) Write Write the equilibrium constant for this reaction and compute its value at 1100K; b) If the total gas gas pressure is kept constant at 10 atm, calculate calculate the fraction of H2O that reacts; c) If the reaction temperature is increased, will the fraction of water reacted increase or decrease? Explain your answer. answer. Use the data in Table 5.1. 5 .1. Solution: H 2 O( g ) C ( s) H 2 ( g ) CO( g ) a) K
P CO( g ) P H 2( g ) P HO 2( g ) G o G f ,CO G f , H 2O 111710 87.65 1100 (246740 54.811100) 21676 J
G o RT ln K ln K 2.3 K 9.97 let let P H 2( g ) x atm, P CO( g ) x atm, P H 2O ( g ) (10 2 x) atm
b)
K
x
2
1 2 x x 4.14 atm
9.97
c) if the temperature is increased, the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.
