Given the equilibrium reaction N2(g) + 3 H2(g) <---> 2NH3(g) + heat, an increase in temperature will:

Question

a) Increase the value of K
b) Decrease the value of K
c) Increase or decrease the value of K depending upon the concentration of reactants
d) Will have no effect on the value of K
e) Will increase or decrease the value of K depending on the concentration of ammonia

To change the value of K in the reaction H2(g) + I2(g) <---> 2HI(g), one would:
a) Increase pressure
b) Decrease the volume of the system
c) Decrease pressure
d) Increase volume
e) Change the temperature

Please help me understand how to figure out these types of problems!

Answer 1

Le Chatelier's Principle is all about which way the reaction will shift when we mess around with a system in equilibrium. In plain English it states that a system in equilibrium will try to undo what we do to it. For a system such as A + B ==> C + D + heat.
If we add A it will shift to the right because that we the system uses up what we added. If we add B the same thing. If we add C it shifts to the left to use up what we added. Same with D. If we add heat, it MUST shift to the left so as to use up the heat we added. For gaseous reactions just remember that increased causes a shift to the side with fewer moles (so as to occupy the smaller volume).

So for #1, write the K expression.
K = (NH3)^2/(N2)(H2)^3
For increase T it must shift to the left (see my discussion above). If it goes to the left then what happens to the substances? NH3 gets smaller while H2 and N2 get larger. If you substitute into K a smaller number for NH3 and larger numbers for N2 and H2, K must get smaller.
For #2 you are asked to change K. There is no reasoning involved here. Pressure, concentration, volume, all of those things SHIFT the equilibrium BUT they do NOT change the value of K. The only thing that can change K is T.

Answer 2

I understand now, thanks a lot!

Answer 3

I omitted a very important word in one of the last sentences. What I wrote was, "For gaseous reactions just remember that increased causes a shift to the side with fewer moles (so as to occupy the smaller volume)."
I should have written, "For gaseous reactions just remember that increased PRESSURE causes a shift to the side with fewer moles (so as to occupy the smaller volume)."