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

1 Answer
Jul 6, 2018

Answer:

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