31. Listed below are AH; and S° values for PCI 3, Cl2, and PCl5. Substance AH (kJ mol¹) So (J.mol-1.K-1) PC13 (g) -287 311.8 C12 (9) 0 223.1 PCl5 (g) -374.9 364.6 Use these data to calculate the equilibrium constant (Keq) for the formation of PCl5 (g) from PC|3 (g) and Cl2 (g) at 200.0°C, as shown below. A. 0.137 B. 0.155 C. 3.98 D. 6.43 E. 7.32 PC13 (g) + Cl2 (g) = PCl5 (g)
31. Listed below are AH; and S° values for PCI 3, Cl2, and PCl5. Substance AH (kJ mol¹) So (J.mol-1.K-1) PC13 (g) -287 311.8 C12 (9) 0 223.1 PCl5 (g) -374.9 364.6 Use these data to calculate the equilibrium constant (Keq) for the formation of PCl5 (g) from PC|3 (g) and Cl2 (g) at 200.0°C, as shown below. A. 0.137 B. 0.155 C. 3.98 D. 6.43 E. 7.32 PC13 (g) + Cl2 (g) = PCl5 (g)
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter16: Spontaneity Of Reaction
Section: Chapter Questions
Problem 49QAP: Red phosphorus is formed by heating white phosphorus. Calculate the temperature at which the two...
Question
![31. Listed below are AH; and S° values for PCI 3, Cl2, and PCl5.
Substance
AH (kJ mol¹)
So (J.mol-1.K-1)
PC13 (g)
-287
311.8
C12 (9)
0
223.1
PCl5 (g)
-374.9
364.6
Use these data to calculate the equilibrium constant (Keq) for the formation of PCl5 (g) from PC|3
(g) and Cl2 (g) at 200.0°C, as shown below.
A.
0.137
B.
0.155
C.
3.98
D.
6.43
E.
7.32
PC13 (g) + Cl2 (g) = PCl5 (g)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49ba810f-9613-4237-89f5-bb23fd83addd%2Fb3a5fccf-9a6f-4896-bd4b-50c3233dad29%2Fctbyu9r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:31. Listed below are AH; and S° values for PCI 3, Cl2, and PCl5.
Substance
AH (kJ mol¹)
So (J.mol-1.K-1)
PC13 (g)
-287
311.8
C12 (9)
0
223.1
PCl5 (g)
-374.9
364.6
Use these data to calculate the equilibrium constant (Keq) for the formation of PCl5 (g) from PC|3
(g) and Cl2 (g) at 200.0°C, as shown below.
A.
0.137
B.
0.155
C.
3.98
D.
6.43
E.
7.32
PC13 (g) + Cl2 (g) = PCl5 (g)
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