Given the equation: CH4 (g) + 2 O2 (g) ⇋ CO2 (g) + 2 H2O (g) What direction will the reaction shift when more H2O (g) is removed?

Question

Given the equation:

CH_{4 (g)} + 2 O_2 (g) ⇋ CO_2 (g) + 2 H₂O  _(g)

What direction will the reaction shift when more H₂O  _(g) is removed?

Answer 1

This question is related to Le Chatelier's Principle. This principle states that if a system in dynamic equilibrium is disturbed by changing the conditions, then the position of the equilibrium will shift in a direction to counteract that change in order to re-establish the equilibrium. In the present situation, we are dealing with the following system which is in equilibrium:

CH₄(g) + 2 O₂(g) <==> CO₂(g) + 2H₂O (g)

If you remove H₂O (a product), the reaction will shift to the right (product side) in order to replace the water that has been removed in an attempt to re-establish the equilibrium.

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Other examples:

If you add CH₄(g), the reaction will shift to the right in order to remove the added CH₄

If you add O₂(g), the reaction will shift to the right in order to remove the added O₂

If you remove CH₄(g), the reaction will shift to the left in order to replace the CH₄ that was removed

If you remove O₂(g), the reaction will shift to the left in in order to replace the O₂ that was removed

If you add CO₂(g), the reaction will shift to the left in order to remove the added CO₂

If you add H₂O(g), the reaction will shift to the left in order to remove the added H₂O

If you increase or decrease the pressure, there will be no effect on the equilibrium since the number of moles of gas are the same (3 moles) on both sides of the reaction.