Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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1. Consider the following reaction:
PCl3(g)+ Cl2(g) ⇌ PCl5(g) K = 4.30 x 10-6
Initially, 2.50 M PCl3 and 1.40 M Cl2 are placed into a container and allowed to reach equilibrium. Determine the equilibrium concentrations of all the species.
I attached my answer.
![1. PCl + Cl2«) ÷ PCI5)
3(g)
K = 4.30 x 10 6
P Cl33)
P Cls13)
+
I
2.50M
1.40M
-x
+x
E
2.50 – x
1.40 -x
[PCI]
Rule of 100:
K
2.50
4.30x 10
= 581395
581395 » 100
Therefore, 2.50-x = 2.50
Rule of 100:
[CI,]
1.40
4.30x10
= 325581
325581 » 100
Therefore, 1.40 – x = 1.40
PCI,
K =
(PCI, XCI,)
4.30 x 10 -6
(2.50)(1.40)
-6
4.30 x 10
3.5
x= 1.51 x 10 -M
[PCI,]= 2.50 – x
2.50 – 1.51 × 10
-5
= 2,50M
[C] = 1.40 – x
1.40 – 1.51 × 10 s
1.40M
[PC!3] =x](https://content.bartleby.com/qna-images/question/9ff59a79-09f3-4968-a1fd-b58139a18392/a2c6bbc0-e716-4c11-90a9-53e1926634fb/u8qm1mc_processed.png)
Transcribed Image Text:1. PCl + Cl2«) ÷ PCI5)
3(g)
K = 4.30 x 10 6
P Cl33)
P Cls13)
+
I
2.50M
1.40M
-x
+x
E
2.50 – x
1.40 -x
[PCI]
Rule of 100:
K
2.50
4.30x 10
= 581395
581395 » 100
Therefore, 2.50-x = 2.50
Rule of 100:
[CI,]
1.40
4.30x10
= 325581
325581 » 100
Therefore, 1.40 – x = 1.40
PCI,
K =
(PCI, XCI,)
4.30 x 10 -6
(2.50)(1.40)
-6
4.30 x 10
3.5
x= 1.51 x 10 -M
[PCI,]= 2.50 – x
2.50 – 1.51 × 10
-5
= 2,50M
[C] = 1.40 – x
1.40 – 1.51 × 10 s
1.40M
[PC!3] =x
![= 1.51 x 10 M
Therefore, the equilibrium concentration of [PCI3] is 2.50M, [Cl] is 1.40M, and [PCI3] is
1.51 x 10 M.](https://content.bartleby.com/qna-images/question/9ff59a79-09f3-4968-a1fd-b58139a18392/a2c6bbc0-e716-4c11-90a9-53e1926634fb/3zsooyv_processed.png)
Transcribed Image Text:= 1.51 x 10 M
Therefore, the equilibrium concentration of [PCI3] is 2.50M, [Cl] is 1.40M, and [PCI3] is
1.51 x 10 M.
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