Question #e0f51

1 Answer
Aug 15, 2015

Here's how you can approach this type of problems.


Once again, you're dealing with an equilibrium reaction

#2"NH"_text(3(g]) rightleftharpoons "N"_text(2(g]) + 3"H"_text(2(g])#, #DeltaH = +"92.4 kJ"#

The first important thing to realize is that this equlibrium is endothermic for the forward reaction. This means that when ammonia, #"NH"_3#, is converted to nitrogen gas and hydrogen gas, energy is needed, not released.

This is why #DeltaH# is positive. Automatically, the reverse reaction will be exothermic, meaning that when nitrogen gas and hydrogen gas react to form ammonia, heat is given off.

So, you start by adding a catalyst. Catalysts have absolutely no influence on the position of the equilibrium, all they do is increase the rate of the reaction.

In simple words, a catalyst will make sure that the equilibrium is reached faster, but it will not force the equilibrium to shift in one direction or the other.

Everything proceeds the same, only faster. So, a catalyst will have no effect on the equilibrium.

Next, you add hydrogen. Notice that hydrogen is present on the right side of the equilibrium. Le Chatelier's principle tells you that when you disrupt a chemical equilibrium, the equilibrium will react in such away as to counteract the changes and establish a new equilibrium.

So, how would the equilibrium counteract an increased concentration of a product? It will favor the reaction that will reduce the concentration of said product.

In your case, when you increase the concetration of hydrogen gas, the equilibrium will shift to the left, i.e. it will favor the reverse reaction

#"N"_text(2(g]) + 3"H"_text(2(g]) stackrel(color(red)("FAVORED"))(->) 2"NH"_text(3(g])#

This reaction consumes hydrogen gas and produces ammonia, which allows a new equilibrium to be established.

Finally, you increase the temperature. Increasing the temperature is equivalent to providing more energy to the reaction. Since you know that the forward reaction is endothermic, i.e. it requires energy to take place, adding more energy will of course favor this reaction.

The reverse reaction gives off energy, which is exactly the opposite of what the equilibrium "wants" to do when confornted with extra energy.

This means that the equilibrium will shift right, favoring the energy-consuming reaction. By using the excees energy to produce more hydrogen gas and nitrogen gas, a new equilibrium is established.