How would the equilibrium concentration of H_2O be affected by removing H_2 from the mixture?

1 Answer
Jul 6, 2018

All other things equal, there would be a decline in [H_2 O].

Explanation:

The question is likely to be talking about the thermal decomposition of H_2 O, a nonspontaneous process that becomes reversible at ridiculously-high temperature. The chemical equation for this reaction- a reversible process- is:

H_2 O (g) rightleftharpoons H_2 (g) + 1/2 color(white)(l) O_2 (g)

Removing H_2 (g) reduces its concentration in the mixture (assuming that volume stays unchanged.)

The Le Chatlier's Principle predicts that the system would respond to this change in a way as if it's trying to minimize the change's impact on the equilibrium conditions. H_2 O would be consumed as the system try to produce H_2 to (effortlessly) make up for the removal.

n(H_2O) declines whereas V stays the same (assumed.)
c = n / V, what would be the direction of the change in c(H_2O)?

Reference
"Water splitting." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 15 Jun. 2018. Web. 6 Jul. 2018, https://en.wikipedia.org/wiki/Water_splitting#Thermal_decomposition_of_water

"Thermochemical cycle." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 18 Mar. 2018. Web. 6 Jul. 2018, https://en.wikipedia.org/wiki/Thermochemical_cycle