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Hi. It's Mr. Andersen and this chemistry essentials video 67. It's on equilibrium disturbances.
In this reversible reaction I have two aqueous solutions of cobalt. And so if I push it to
the left, in other words if I have more of these reactants it's going to be a pink color.
If I push it to the right it's going to be more of the products and so I'm going to have
this blue color. And so what I can do is I can vary which way the reaction goes to the
left pink, to the right blue and I can change color. Now what's a good way to do that? Temperature.
Because as I change the temperature I'm actually changing the equilibrium constant so I can
move it back and forth. So in a reversible reaction where we have reactants and products,
over time it will eventually establish an equilibrium. And we can measure that using
the equilibrium constant which is essentially the concentration of products divided by the
concentration of reactants. And that's going to stay the same over time. Now before we
get there, the reaction is going to be at Q, where we have the products and reactants
over time. And eventually those two are going to be the same when we reach equilibrium.
But what can happen is you can have disturbances, disturbances that change our Q and we have
to reestablish K. And some disturbances that actually change that K value and so Q has
to approach a new K. Now let's dig in a little bit deeper and actually look at real K values.
And so this that Haber process. What we have here is our initial equilibrium. And then
we have some kind of a disturbance. And then we establish equilibrium at the end. Equilibrium
constant, remember to figure that out, this is going to be our equation for that. We're
going to write the concentration of ammonia raised to the second power. Where am I getting
that from? Remember we have two moles of that. So that's going to be raised to the second
power over our reactants which is going to be concentration of nitrogen times the concentration
of hydrogen raised to the third power. Why raised to the third power? It's because we
have 3 moles right in front of here. And so let's figure out what our K value is going
to be initially. Well to figure that out all we do is plug in the values. We plug in the
concentrations of our H, in this case it's going to be 4. Our NH3 which is going to be
3. And then our N which is going to be 1. So if I plug in those values I get 9 over
64. It's around 0.14. Now if we move to that point where the disturbance occurs, let's
figure it out there. What's our new hydrogen value? Our new hydrogen value is going to
be 6. So I'm going to plug in a 6 here. What's going to be our ammonia and nitrogen values?
Those are going to be the same. So if I figure out what my new Q value is, again there's
been some kind of a disturbance, our value is going to be 0.042. So our Q value now is
lower than our K. And so where is this reaction going to go? It's going to move towards the
right. And so we're going to move it to the right and establish a new equilibrium. So
how does that occur? Well we're going to decrease the amount of hydrogen. And as we decrease
the amount of hydrogen, it goes from 6 down to 5, we're going to have to adjust our other
values as well. So here is our new value. It's again K equals 0.14 and that reaction
moved to the right to reestablish that. Again that's Le Chatelier's Principle since our
K value is going to be lower that Q value the whole thing moved over to the right. Let's
look at temperature now and see how temperature affects K. And so in the reaction we have
colorless dinitrogen tetraoxide and there's a reversible reaction where that can become
nitrogen dioxide which is going to be reddish brown. Now this is an endothermic reaction.
What does that mean? As we move from the left to the right side it's actually consuming
energy. And so let's put that heat over on left side. So what happens if we increase
our temperature? If we increase our temperature it's really like adding more reactants on
the left side and so it's going to push that reaction over to the right side. And so we're
going to get a movement to the right. What's going to happen to our K value? Our K value
is actually going to increase because we're going to have more products. And that whole
thing is going to shift to the right. Now what would this appear as, as we increase
temperature? It's going to be darker in color because we're getting more of this reddish
brown gas on the right side. And so in an endothermic reaction what happens as you increase
the temperature? You increase your K value. Let's look at what happens if we decrease
temperature. As we decrease temperature we're actually going to shift that to the left side.
What's going to happen to our K value? Now our K value is actually going to go down where
it's going to decrease. And so so there's a direction relationship between temperature
and K as long as it's an endothermic reaction. Now let's look at an exothermic reaction.
So in the Haber process it's an exothermic. So we could put heat on the right side. What
happens if we increase temperature now? That's actually going to push it to the left side.
What's going to happen to our K value? It's going to go down. Likewise if we were to increase
it, it's going to go in the opposite direction. Here are some actually temperature and equilibrium
constants for the Haber process. And you can see as we increase temperature our K value
is actually decreasing in an exothermic reaction. So if you wanted to increase yield, if we
wanted to increase the amount of ammonia we would actually lower the temperature. It's
more complex than that because we want to speed up the reaction as well. And so did
you learn to really apply LeChatelier's Principle to values of K and Q? I hope so. And I hope
that was helpful.