Consider the following reaction: H20(g) + CO(g) = H2 (g) + CO2 (g). Amounts of H20, CO, H2, CO2 are put on a flask such that the system is already at equilibrium. a) If the CO put originally in the flask has radioactive 14C but the CO2 has not, where will 14C be found after a long, long time? b) If the CO2 put originally in the flask has radioactive 14C but the CO has not, where will 14C be found after a long, long time?
Consider the following reaction: H20(g) + CO(g) = H2 (g) + CO2 (g). Amounts of H20, CO, H2, CO2 are put on a flask such that the system is already at equilibrium. a) If the CO put originally in the flask has radioactive 14C but the CO2 has not, where will 14C be found after a long, long time? b) If the CO2 put originally in the flask has radioactive 14C but the CO has not, where will 14C be found after a long, long time?
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 12.27PAE: The following data were collected for a system at equilibrium at 140°C. Calculate the equilibrium...
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Question
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Consider the following reaction: H20(g) + CO(g) = H2 (g) + CO2 (g).
Amounts of H20, CO, H2, CO2are put on a flask such that the system is already at equilibrium.
-
a) If the CO put originally in the flask has radioactive 14C but the CO2 has not, where will 14C be found
after a long, long time?
-
b) If the CO2 put originally in the flask has radioactive 14C but the CO has not, where will 14C be found
after a long, long time?
-
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