Consider the following reaction:PCl5(g) PCl3(g) + Cl2(g)If 4.46×10-3 moles of PCl5(g), 0.286 moles of PCl3, and 0.333 moles of Cl2 are at equilibrium in a 17.4 L container at 650 K, the value of the equilibrium constant, Kc, is

Number of moles of PCl5 = 4.46 x 10-3 moles Number of moles of PCl3 = 0.286 moles Number of moles of…

Answered: Consider the following reaction:…

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 14.2QE: Describe a nonchemical system that is in equilibrium, and explain how the principles of equilibrium...

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Question

Consider the following reaction:

PCl₅(g) PCl₃(g) + Cl₂(g)

If 4.46×10^-3 moles of PCl₅(g), 0.286 moles of PCl₃, and 0.333 moles of Cl₂ are at equilibrium in a 17.4 L container at 650 K, the value of the equilibrium constant, K_c, is

Expert Solution

Step 1

Number of moles of PCl₅ = 4.46 x 10^-3 moles

Number of moles of PCl₃ = 0.286 moles

Number of moles of Cl₂ = 0.333 moles

Given reaction :

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$

$C o n c e n t r a t i o n o f P C l_{5} = \frac{N u m b e r o f m o l e s}{V o l u m e o f s o l u t i o n (i n L)} = \frac{4.46 x 10^{- 3}}{17.4} = 2.56 \times 10^{- 4} M C o n c e n t r a t i o n o f P C l_{3} = \frac{N u m b e r o f m o l e s}{V o l u m e o f s o l u t i o n (i n L)} = \frac{0.286}{17.4} = 0.016 M C o n c e n t r a t i o n o f C l_{2} = \frac{N u m b e r o f m o l e s}{V o l u m e o f s o l u t i o n (i n L)} = \frac{0.333}{17.4} = 0.019 M$

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