Standard State Thermodynamic DataformulaAH? (kJ/mol)s° (J/mol·K)AG° (kJ/mol)formulaAHº (kJ/mol)s° J/mol·K)AG° (kJ/mol)SF,(g)SO(g)SOCI,(g)SO2(g)Ni(g)429.7182.2375.4-1209291.8-1105.3Ni(s)29.9-8.96.3222.0-59.9NiCl2(s)-305.397.7-334.4-212.5309.8-304.9O(g)02(g)03(g)249.4161.0231.7-296.1248.5-300.1So,Cl>(g)SO3(g)Se(g)205.0-364.0311.9-457.0142.7237.7163.4-395.2256.2-370.4P(g)316.5163.2267.8227.1176.7174.4P(red)-18.429.312.1Se(s)42.4-12.6Si(g)Si(s)SiBr4(1)SiC(s)SİCI4(g)SiCl,(1)P(s)P(white)PBr3(g)41.1-12.3450.0168.0399.941.118.83-139.3348.1-243.1-457.3277.8-540.1PBr3(1)PCI3(g)PCI3(1)-184.5240.2-256.1-65.316.6-70.2-287.0311.78-267.8-657.0330.7-755.6-319.7217.1-384.4-687.0239.7-758.5PCI5(g)-374.9364.6-305.0SiO2(s)-910.941.84-856.6PH3(g)РОС:(g)5.4210.213.5Sn(g)301.2168.5251.0Sn(s)SNC14(1)Ti(g)-558.5325.5-655.551.2-15.3POC13(1)-597.1222.5-663.4-511.3258.6-588.4PĄ(g)58.9280.0-24.6473.0180.3419.2P,010(s)-2984.0228.86-2697.0Ti(s)TiCl>(s)TiCl;(s)TiClą(g)30.7-9.2Pb(g)195.2175.4142.9-513.887.4-539.9Pb(s)PbBr,(s)64.8-19.3-720.9139.7-762.6-278.7161.5-326.9-763.2353.2-868.5TiClą(1)TiO2(s)252.3PBCO3(s)PbCl2(s)-699.1131.0-738.2-804.2-879.4-359.4136.0-399.9-944.050.6-959.1PbO(s)Rb(g)Xe(g)Zn(g)-217.368.7-237.8169.680.9170.130.2130.4161.082.4Rb(s)Zn(s)ZnBr2(s)ZNCO3(s)76.8-22.941.63RbBr(s)-394.6110.0-427.4-328.7138.5-370.0RbCl(s)-435.495.9-464.0-812.882.4-837.4167.8ZnCl,(s)ZnO(s)S(g)277.2227.2-415.1111.5-448.3S(monoclinic)S(rhombic)S(s)0.3032.550.10-348.343.64-318.331.88ZnS(s)-202.957.74-198.332.1-9.6ZnSOĄ(s)-978.6124.9-871.6 Use the data in the supporting materials to calculate the equilibrium constant (Kp) for the following reaction at 25°C.2 SO2(g) + O2(g) =2 S03(g)Кра4.24e24Calculate Kp for the reaction at 49°C using the following equations.(a) the van't Hoff equation below.(P),A,H°RT?a In KКр= 1.138e22(b) the Gibbs-Helmholtz equations below to find A-G° at 49°C and hence, Kp at the same temperature.AGΔΗ(i)= -T2A-G2A;G11(ii)T2T1T1Kp= 1.138e22(c) A-G° = ArH° – TA¡S° to find A-G° at 49°C and hence, Kp at the same temperature.Кр 3D 1.7742е22State the approximations employed in each case.The assumption that A,H° is temperature independent and equal to the value calculated at 298.15 K is required by all three partsA second assumption that AS° istemperature independent and equal to the value calculated at 298.15 K is required by part (c)Compare your results.The answers calculated in parts (a) and (b) ▼in part (c)are very close as expected because the formulas used to calculate them are related. On the other hand, the answer calculatedmay differ from the actual value due to the additional assumption involved.

The solution is given in the following steps : The value of Kp (a) 1.11024 x 10^22 (b) 1.1351 x 10^22 (c) 1.756 x 10^22

Use the data in the supporting materials to calculate the equilibrium constant (Kp) for the following reaction at 25°c. 2 SO2(g) + O2(9)=2 S03(g) Kp = 4.24e24 Calculate Kp for the reaction at 49°C using the following equations. (a) the van't Hoff equation below. a In (*),- RT2 Kp = 1.138e22 (b) the Gibbs-Helmholtz equations below to find A,G° at 49°C and hence, Kp at the same temperature. AG ΔΗ (i) A,G2_ A,G1 (ii) - A,H(-) T2 T1 Кр = 1.138e22 (c) A,G° = A,H° = TA,S° to find A,G° at 49°C and hence, Kp at the same temperature. %3D Kp = 1.7742e22

Use the data in the supporting materials to calculate the equilibrium constant (Kp) for the following reaction at 25°c. 2 SO2(g) + O2(9)=2 S03(g) Kp = 4.24e24 Calculate Kp for the reaction at 49°C using the following equations. (a) the van't Hoff equation below. a In (*),- RT2 Kp = 1.138e22 (b) the Gibbs-Helmholtz equations below to find A,G° at 49°C and hence, Kp at the same temperature. AG ΔΗ (i) A,G2_ A,G1 (ii) - A,H(-) T2 T1 Кр = 1.138e22 (c) A,G° = A,H° = TA,S° to find A,G° at 49°C and hence, Kp at the same temperature. %3D Kp = 1.7742e22

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Standard State Thermodynamic Data
formula
AH? (kJ/mol)
s° (J/mol·K)
AG° (kJ/mol)
formula
AHº (kJ/mol)
s° J/mol·K)
AG° (kJ/mol)
SF,(g)
SO(g)
SOCI,(g)
SO2(g)
Ni(g)
429.7
182.2
375.4
-1209
291.8
-1105.3
Ni(s)
29.9
-8.9
6.3
222.0
-59.9
NiCl2(s)
-305.3
97.7
-334.4
-212.5
309.8
-304.9
O(g)
02(g)
03(g)
249.4
161.0
231.7
-296.1
248.5
-300.1
So,Cl>(g)
SO3(g)
Se(g)
205.0
-364.0
311.9
-457.0
142.7
237.7
163.4
-395.2
256.2
-370.4
P(g)
316.5
163.2
267.8
227.1
176.7
174.4
P(red)
-18.4
29.3
12.1
Se(s)
42.4
-12.6
Si(g)
Si(s)
SiBr4(1)
SiC(s)
SİCI4(g)
SiCl,(1)
P(s)
P(white)
PBr3(g)
41.1
-12.3
450.0
168.0
399.9
41.1
18.83
-139.3
348.1
-243.1
-457.3
277.8
-540.1
PBr3(1)
PCI3(g)
PCI3(1)
-184.5
240.2
-256.1
-65.3
16.6
-70.2
-287.0
311.78
-267.8
-657.0
330.7
-755.6
-319.7
217.1
-384.4
-687.0
239.7
-758.5
PCI5(g)
-374.9
364.6
-305.0
SiO2(s)
-910.9
41.84
-856.6
PH3(g)
РОС:(g)
5.4
210.2
13.5
Sn(g)
301.2
168.5
251.0
Sn(s)
SNC14(1)
Ti(g)
-558.5
325.5
-655.5
51.2
-15.3
POC13(1)
-597.1
222.5
-663.4
-511.3
258.6
-588.4
PĄ(g)
58.9
280.0
-24.6
473.0
180.3
419.2
P,010(s)
-2984.0
228.86
-2697.0
Ti(s)
TiCl>(s)
TiCl;(s)
TiClą(g)
30.7
-9.2
Pb(g)
195.2
175.4
142.9
-513.8
87.4
-539.9
Pb(s)
PbBr,(s)
64.8
-19.3
-720.9
139.7
-762.6
-278.7
161.5
-326.9
-763.2
353.2
-868.5
TiClą(1)
TiO2(s)
252.3
PBCO3(s)
PbCl2(s)
-699.1
131.0
-738.2
-804.2
-879.4
-359.4
136.0
-399.9
-944.0
50.6
-959.1
PbO(s)
Rb(g)
Xe(g)
Zn(g)
-217.3
68.7
-237.8
169.6
80.9
170.1
30.2
130.4
161.0
82.4
Rb(s)
Zn(s)
ZnBr2(s)
ZNCO3(s)
76.8
-22.9
41.63
RbBr(s)
-394.6
110.0
-427.4
-328.7
138.5
-370.0
RbCl(s)
-435.4
95.9
-464.0
-812.8
82.4
-837.4
167.8
ZnCl,(s)
ZnO(s)
S(g)
277.2
227.2
-415.1
111.5
-448.3
S(monoclinic)
S(rhombic)
S(s)
0.30
32.55
0.10
-348.3
43.64
-318.3
31.88
ZnS(s)
-202.9
57.74
-198.3
32.1
-9.6
ZnSOĄ(s)
-978.6
124.9
-871.6

Use the data in the supporting materials to calculate the equilibrium constant (Kp) for the following reaction at 25°C.
2 SO2(g) + O2(g) =2 S03(g)
Кра
4.24e24
Calculate Kp for the reaction at 49°C using the following equations.
(a) the van't Hoff equation below.
(P),
A,H°
RT?
a In K
Кр
= 1.138e22
(b) the Gibbs-Helmholtz equations below to find A-G° at 49°C and hence, Kp at the same temperature.
AG
ΔΗ
(i)
= -
T2
A-G2
A;G1
1
(ii)
T2
T1
T1
Kp
= 1.138e22
(c) A-G° = ArH° – TA¡S° to find A-G° at 49°C and hence, Kp at the same temperature.
Кр 3D 1.7742е22
State the approximations employed in each case.
The assumption that A,H° is temperature independent and equal to the value calculated at 298.15 K is required by all three parts
A second assumption that AS° is
temperature independent and equal to the value calculated at 298.15 K is required by part (c)
Compare your results.
The answers calculated in parts (a) and (b) ▼
in part (c)
are very close as expected because the formulas used to calculate them are related. On the other hand, the answer calculated
may differ from the actual value due to the additional assumption involved.

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