Question

1. Consider the following reaction occurring in a closed chemical system. Assume that this reaction is at equilibrium and that in general the reaction to the right is favored.
CH3CH2OH + 3O2 2CO2 + 3H2O ∆H = –1,235 kJ/mol
• What type of chemical reaction is this?
• If more CH3CH2OH is added to the system, how will the reactions shift to reach equilibrium again?
• If water is extracted from the system, how will the reactions shift to reach equilibrium again?
• If heat is removed from the system, how will the reactions shift to reach equilibrium again?
• What is the name of the principle

Answer 1

\(CH_3CH_2OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\)

This is a combustion reaction. Note the \(\sf hydrocarbon\) framework in the presence of \(\sf oxygen\), the \(\sf production ~of~ water~ and ~carbon~ dioxide\). These are trademarks of every combustion reaction. (You may have carbon monoxide in the products sometimes as a result of incomplete combustion which happens when there isn't enough oxygen present)

Also note that \(CH_3CH_2OH \) (ethanol) it's oxidized by \(O_2\) in the process.

These questions deal with equilibrium concentration shifts due to stresses on the system.
The principle that addresses these is called \(\sf Le~Chatelier's ~Principle\).

If you remove any of the reactants, the system will shift to the \(\leftarrow \)
If you remove any of the products, the system will shift to the \(\rightarrow \)

You can treat the thermal energy produced as a product in an exothermic reaction.
So for example, if you supply heat to an exothermic reaction, the equilibrium concentrations will shift to the \(\leftarrow \).

Lastly, \(\sf \color{red}{this~ principle~ only ~pertains~ to ~reversible~ reactions}\), a combustion reaction \(\sf IS~NOT\) a reversible reaction. (you can't really unburn something, can you?) So there wont be any shifts.