A classmate asks for your help writing an equilibrium expression for the combustion of methane gas: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l). Explain to her why it is not meaningful to create an equilibrium expression or calculate an equilibrium constant for this reaction.

Question

A classmate asks for your help writing an equilibrium expression for the combustion of methane gas: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l). Explain to her why it is not meaningful to create an equilibrium expression or calculate an equilibrium constant for this reaction.

aaronq · Answer

do you understand what's going on in this reaction? and what type of reactions CAN have a Kc?

anonymous · Answer

Hello again, I understand what is going on in this reaction, but, what do mean by CAN have a Kc?

aaronq · Answer

well an equilibrium constant is only applicable to reactions that can establish a dynamic equilibrium, and by that i mean that the reaction can be forwards AND backwards..

in this reaction you're burning something, can the reaction go backwards?

anonymous · Answer

ooooohhhh, I see... So there is no equilibrium at all, because this equation is only one way. So the equilibrium expression and constant would be useless

aaronq · Answer

yeah exactly, it's not reversible, so there can't be an equilibrium constant.

anonymous · Answer

Thank you very much!

aaronq · Answer

no problem !