The equilibrium constant, Ke, for the following reaction is 1.80x10-2 at 698K. 2HI(g) =H₂(g) + I2(g) If an equilibrium mixture of the three gases in a 15.7 L container at 698K contains 0.212 mol of HI(g) and 0.493 mol of H₂, the equilibrium concentration of 12 is M.
The equilibrium constant, Ke, for the following reaction is 1.80x10-2 at 698K. 2HI(g) =H₂(g) + I2(g) If an equilibrium mixture of the three gases in a 15.7 L container at 698K contains 0.212 mol of HI(g) and 0.493 mol of H₂, the equilibrium concentration of 12 is M.
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.36PAE: The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of...
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The equilibrium constant, Ke, for the following reaction is 1.80x10-2 at 698K.
2HI(g) =H₂(g) + I₂(g)
If an equilibrium mixture of the three gases in a 15.7 L container at 698K contains 0.212 mol of HI(g) and 0.493 mol of H2, the equilibrium concentration of 1₂ is
M.
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Transcribed Image Text:m/ilmn/takeAssignment/takeCovalentActivity.do?locator assignment-take
[References]
Use the References to access important values if needed for this question.
The equilibrium constant, Ke, for the following reaction is 1.80x10-2 at 698K.
2HI(g) =H₂(g) + I₂(g)
If an equilibrium mixture of the three gases in a 15.7 L container at 698K contains 0.212 mol of HI(g) and 0.493 mol of H2, the equilibrium concentration of 1₂ is
M.
Submit Answer
Try Another Version
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s
Previous
Cengage Learning Cengage Technical Support
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F
h
B
C
C
31°C Haze
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