1. Consider the following reaction: PCl3(g)+ Cl2(g) ⇌ PCl5(g) K = 4.30 x 10-6Initially, 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 6P Cl33)P Cls13)+I2.50M1.40M-x+xE2.50 – x1.40 -x[PCI]Rule of 100:K2.504.30x 10= 581395581395 » 100Therefore, 2.50-x = 2.50Rule of 100:[CI,]1.404.30x10= 325581325581 » 100Therefore, 1.40 – x = 1.40PCI,K =(PCI, XCI,)4.30 x 10 -6(2.50)(1.40)-64.30 x 103.5x= 1.51 x 10 -M[PCI,]= 2.50 – x2.50 – 1.51 × 10-5= 2,50M[C] = 1.40 – x1.40 – 1.51 × 10 s1.40M[PC!3] =x = 1.51 x 10 MTherefore, the equilibrium concentration of [PCI3] is 2.50M, [Cl] is 1.40M, and [PCI3] is1.51 x 10 M.

Reaction: PCl3 + Cl2 ⇌PCl5 K = 4.30 × 10-6 Initial : 2.50M 1.40…

Answered: . Consider the following reaction:…

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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1. Consider the following reaction:

PCl_3(g)+ Cl_2(g) ⇌ PCl_5(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

= 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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