A student ran the following reaction in the laboratory at 427 K:PCl5(g) PCl3(g) + Cl2(g)When she introduced 4.21 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of PCl5(g) to be 4.17 M.Calculate the equilibrium constant, Kc, she obtained for this reaction.

PCl5(g) → PCl3(g) + Cl2(g)Initial 4.21 M 0 0 Initial…

Answered: A student ran the following reaction in…

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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A student ran the following reaction in the laboratory at 427 K:

PCl₅(g) PCl₃(g) + Cl₂(g)

When she introduced 4.21 moles of PCl₅(g) into a 1.00 liter container, she found the equilibrium concentration of PCl₅(g) to be 4.17 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

Expert Solution

Step 1

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g) I n i t i a l 4.21 M 0 0 I n i t i a l c o n c e n t r a t i o n o f a f t e r d i s s o c i a t i o n - x x x P C l_{5} = 4.21 m o l e s / 1 l i t r e a t e q u i l i b r i u m 4.21 x - x x x = 4.21 M G i v e n a t e q u i l i b r i u m [P C l_{5}] = 4.17 M 4.21 - x = 4.17 M x = 4.21 - 4.17 x = 0.04 M [P C l_{3}] = 0.04 M [C l_{2}] = 0.04 M K_{C} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{0.04 M x 0.04 M}{4.17 M} K_{C} = 3.837 x 10^{- 4} M$

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