2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)You have 0.4 moles C2H2 and 0.2 moles O2 present initially in a 5L flask. You allow the reaction to run until it reaches equilibrium. Which of the following is the correct ICE table for this process? Question 5 options: Conc.C2H4O2CO2 H2O Initial0.4 0.2 0 0 Change -x-x +x +x Equil. 0.4-x0.2-x x x Conc.C2H4O2CO2 H2O Initial0.08 0.040 0 Change -2x-5x +4x +2x Equil. 0.08-2x0.04-5x 4x 2x Conc.C2H4O2CO2 H2O Initial0.4 0.2 0 0 Change -2x-5x +4x +2x Equil. 0.4-2x0.2-5x 4x 2x Conc.C2H4O2CO2 H2O Initial0.080.040 0 Change -20x-20x +20x +20x Equil. 0.08-20x0.04-20x 20x 20x

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Answered: 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)…

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 12.22PAE

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2C₂H₂(g) + 5O₂(g) 4CO₂(g) + 2H₂O(g)

You have 0.4 moles C₂H₂ and 0.2 moles O₂ present initially in a 5L flask. You allow the reaction to run until it reaches equilibrium. Which of the following is the correct ICE table for this process?

Question 5 options:

Conc.	C₂H₄	O₂	CO₂	H₂O
Initial	0.4	0.2	0	0
Change	-x	-x	+x	+x
Equil.	0.4-x	0.2-x	x	x

Conc.	C₂H₄	O₂	CO₂	H₂O
Initial	0.08	0.04	0	0
Change	-2x	-5x	+4x	+2x
Equil.	0.08-2x	0.04-5x	4x	2x

Conc.	C₂H₄	O₂	CO₂	H₂O
Initial	0.4	0.2	0	0
Change	-2x	-5x	+4x	+2x
Equil.	0.4-2x	0.2-5x	4x	2x

Conc.	C₂H₄	O₂	CO₂	H₂O
Initial	0.08	0.04	0	0
Change	-20x	-20x	+20x	+20x
Equil.	0.08-20x	0.04-20x	20x	20x

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