Question #acd19

1 Answer
May 4, 2016


The concentration of nitrous oxide will decrease.


You are right, the position of the equilibrium will shift as a result of an increase in temperature.

As you know, equilibrium reactions are governed by Le Chatelier's Principle, which states that when a stress is placed on a system at equilibrium, the equilibrium will shift in a such as way as to reduce that stress.

Your starting equilibrium looks like this

#2"NO"_ ((g)) + "H"_ (2(g)) rightleftharpoons "N"_ 2"O"_ ((g)) + "H"_ 2"O"_((g)) + color(red)("heat")#

The fact that heat is located on the products' side tells you that you're dealing with an exothermic reaction, i.e. a reaction that gives off heat to the surroundings.

Now, you add a stress to the position of this equilibrium by adding heat.

How would the system counteract this added heat?

The position of the equilibrium will shift in such a way as to consume heat, since this will allow the equilibrium to be reestablished.

We've already established that the forward reaction

#2"NO"_ ((g)) + "H"_ (2(g)) -> "N"_ 2"O"_ ((g)) + "H"_ 2"O"_((g)) + color(red)("heat")#

is exothermic and thus produces heat, so the only way for the system to counteract the added heat is to favor the reverse reaction

#"N"_ 2"O"_ ((g)) + "H"_ 2"O"_ ((g)) + color(red)("heat") -> 2"NO"_ ((g)) + "H"_ (2(g))#

since this reaction consumes heat, i.e .is endothermic.

This means that the equilibrium will shift to the left


#2"NO"_ ((g)) + "H"_ (2(g)) rightleftharpoons "N"_ 2"O"_ ((g)) + "H"_ 2"O"_((g)) + color(red)("heat")#

#stackrel(color(blue)(larr))(color(white)(aaaaacolor(green)("shift to the left")aaaaaaaa))#

So, what does this mean?

Nitrous oxide, #"N"_2"O"#, will react with water, #"H"_2"O"#, and take in some of that added heat to form nitric oxide, #"NO"#, and hydrogen gas, #"H"_2#.

Since some of the nitrous oxide will be consumed, its concentration will decrease.