P4(s) + 5 02(g)P4010(s) Kp = 0.0112 A reaction vessel initially contains 2.5 mol of P4010(s). What is the expression for Kp that would be derived from the ICE table. A. Kp C. = Kp = 2.5 - x x(5x)5 1 (5x)5 B. D. Kp = 2.5 + x x(5x)5 Kp = (5x)5

P4(s) + 5 02(g)P4010(s) Kp = 0.0112 A reaction vessel initially contains 2.5 mol of P4010(s). What is the expression for Kp that would be derived from the ICE table. A. Kp C. = Kp = 2.5 - x x(5x)5 1 (5x)5 B. D. Kp = 2.5 + x x(5x)5 Kp = (5x)5

Chemistry & Chemical Reactivity

9th Edition

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter15: Principles Of Chemical Reactivity: Equilibria

Section: Chapter Questions

Problem 3PS: Kc = 5.6 1012 at 500 K for the dissociation of iodine molecules to iodine atoms. I2(g) 2 I(g) A...

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Question

P4(s) + 5 02(g)P4010(s)
Kp = 0.0112
A reaction vessel initially contains 2.5 mol of P4010(s).
What is the expression for Kp that would be derived from the ICE table.
A. Kp
C.
=
Kp =
2.5 - x
x(5x)5
1
(5x)5
B.
D.
Kp
=
2.5 + x
x(5x)5
Kp = (5x)5

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