A student ran the following reaction in the laboratory at 431 K:PC15 (9) PC13 (g) + Cl₂ (g)When she introduced 2.98 moles ofPC15 (g) into a 1.00 liter container, she found the equilibrium concentration ofPC15 (9) to be 2.94 M.Calculate the equilibrium constant, Kc, she obtained for this reaction.Kc =Submit Answer[Review Topics][References]Use the References to access important values if needed for this question.Show HintRetry Entire Group9 more group attempts remaining2PreviousNextorneret

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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 431 K:
PC15 (9) PC13 (g) + Cl₂ (g)
When she introduced 2.98 moles of
PC15 (g) into a 1.00 liter container, she found the equilibrium concentration of
PC15 (9) to be 2.94 M.
Calculate the equilibrium constant, Kc, she obtained for this reaction.
Kc =
Submit Answer
[Review Topics]
[References]
Use the References to access important values if needed for this question.
Show Hint
Retry Entire Group
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Expert Solution

Step 1: Determine the equilibrium constant for the reaction using given data:

$G i v e n comma T h e space r e a c t i o n colon P C l subscript 5 space left parenthesis straight g right parenthesis end subscript space space space rightwards harpoon over leftwards harpoon space space space space space P C l subscript 3 space left parenthesis straight g right parenthesis end subscript space space space plus space space space space C l subscript 2 space left parenthesis straight g right parenthesis end subscript space space space space space space a t space 431 space K I n i t i a l comma space space space space m o l e s space o f space P C l subscript 5 space equals space 2.98 space L v o l u m e space o f space c o n t a i n e r space equals space 1.00 space L A t space e q u i l i b r i u m comma space M o l a r i t y space o f space P C l subscript 5 space equals space 2.94 space M$

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