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

Jul 6, 2018

All other things equal, there would be a decline in $\left[{H}_{2} O\right]$.

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:

${H}_{2} O \left(g\right) r i g h t \le f t h a r p \infty n s {H}_{2} \left(g\right) + \frac{1}{2} \textcolor{w h i t e}{l} {O}_{2} \left(g\right)$

Removing ${H}_{2} \left(g\right)$ 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.

$n \left({H}_{2} O\right)$ declines whereas $V$ stays the same (assumed.)
$c = \frac{n}{V}$, what would be the direction of the change in $c \left({H}_{2} O\right)$?

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