Aclassmate asks for your 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.

Explanation:

When discussing equilibrium of a chemical reaction, the reaction must be a reversible reaction which can be altered to either produce more of reactants or more of the products. Also, the enthalpy change of the reaction determines whether the forward reaction/backward reaction is endothermic or exothermic.

Below is the equation for the combustion of methane

CH₄ + 2O₂ ⇒ CO₂ + 2H₂O

The combustion of methane is an irreversible reaction (meaning nothing can be done to it to get more reactants) which will give off heat (exothermic reaction). hence, the equilibrium expression is not needed here as the type of reaction it is gives us all the information we need to know already.

odds are decent that your handwriting right now looks a little bit different than it did five years ago, and will look different again in five more. when i look back through my notebooks from middle school to now, i see a long, slow progression from meticulous and teeny letters to bigger, sloppier ones, plus a few side detours: sometime in high school, for example, i decided those loopy cursive lowercase f’s were just the coolest, adopting them as the lone rep for cursive in my otherwise all-print handwriting.

explanation: