Hey guys, in this brand new video, we're going to take a look at the equilibrium constant. Now, we're going to say that the variable we use to illustrate our equilibrium constant is the variable k. We're going to say k is a number equal to the ratio of products to reactants at a given temperature. Why do I say at a given temperature? Because we're going to say that temperature directly affects our equilibrium constant k. Increasing the temperature will increase our k value. Decreasing the temperature will decrease our k value. Now, we're going to say it's a ratio of products to reactants, so we're going to say here that \( k = \frac{\text{products}}{\text{reactants}} \). Now, we're going to say that k is important because its magnitude tells us how far to the left or to the right our chemical reaction is at a given temperature. So we're going to say if \( k > 1 \), then products are favored over reactants. Think about it. This makes sense because we said \( k = \frac{\text{products}}{\text{reactants}} \). So let's say our products are 10 and our reactants are 1. \( k = 10 \), definitely greater than 1. We're going to say when products are favored over our reactants, that means we're making more products. How do we make more products? By going in the forward direction. So the forward direction would be favored. Now in the opposite way, if \( k < 1 \), so let's say we have in this case, products are 1 but our reactants are 10. We'd have a \( k < 1 \). So in this case, if \( k < 1 \), then reactants are favored over products, which means our reaction is heading in the reverse direction. So the reverse direction would be favored. Just remember these two differences, when \( k > 1 \) and when \( k < 1 \). Now, what if we say \( k = 1 \)? We know that this is products over reactants, so that means that our products and reactants would have to be equal. Let's say they're both 10. \( \frac{10}{10} = 1 \). So when \( k = 1 \), we're going to say both our reactant and our product amounts are equal to one another. Now we're going to say the equilibrium constant \( k \) takes into account all the states of matter except 2. It doesn't look at solids and it doesn't look at liquids. It ignores those 2 states of matter.
The Equilibrium Constant - Video Tutorials & Practice Problems
Equilibrium Constant K is associated with any reaction at equilibrium. Its numerical value determines if reactants or products are more greatly favored within a reaction.
The Equilibrium Constant K
The Equilibrium Constant Concept 1
Video transcript
The Equilibrium Constant K is a ratio of products to reactants. It only deals with gaseous or aqueous compounds.
The Equilibrium Constant Example 1
Video transcript
So it looks at, it only cares about really gases and also aqueous. Aqueous just means that the solvent is water, so a compound is dissolved in water. It would be aqueous. Now, knowing this, let's take a look at the following example. It says, Write the equilibrium expression for the following reaction. Here we have N2O5 (aqueous) gives us 4NO2 (aqueous) plus O2 (aqueous). It's important to look at the phases because remember if it's a solid or a liquid, we ignore it. Now we're going to say here k= products over reactants. So we're going to have our products as NO2 and remember, you have to use the coefficients in these calculations. The number in front of NO2 is a 4, so that 4 will become the power. So it's going to be NO2 to the 4 times O2. O2 just has a coefficient of 1 in front of it, which we don't have to show. Divided by N2O5. Again, the coefficient in front of N2O5 is a 2, so that becomes the power. So we would say that this represents our equilibrium expression or our equilibrium equation. Same thing. Equilibrium expression, equilibrium equation. Now that we've seen this one, let's look at B.
For B, we're going to say here, we're going to ignore this compound because it's a solid and we're going to ignore this compound because it's a solid. Here we're going to say k= products over reactants like before. But here's the thing, we're going to say we don't have any products available. But you have to put something on top. We're going to say it's equal to 1. Solids and liquids are ignored and in place of them, we're going to put 1. We're going to say this because things such as pressure don't really affect solids and liquids like they do gases and aqueous compounds. That's why they're equal to 1 because their number is being held constant. So we're going to replace solids and liquids with the number 1. And on the bottom we'd have O2. Now, a different way your professor could give you this is, we know there's a one here, so this could also be O2-1. So be aware of both types of answers. Both are correct. Both are saying the same thing. Inverse 1 just means 1 over whatever it is.
For the next one, again we ignore solids. The only thing that we don't ignore is this gas. Here, k= products over reactants, so just equal Xe3 because of the 3 over 1. But we don't need to put the one because anything over 1 is the same exact thing. So that would be our equilibrium C. We ignore the solids and the liquids.
Now that you've seen this, I want you guys to attempt to answer a question that follows this one, the practice one. Here, I want you to tell me who's greater in amount? Is it products or is it reactants? From that, you have to remember what do we say about K. When it's greater than 1, who's favored? When it's less than 1, who's favored? When it's equal to 1 who's favored. Remembering that will be a great way for you to approach this problem. Good luck, guys.
State which is greater in amount: reactants or products, based on the given equilibrium constant, K.
a) N2 (g) + O2 (g) ⇌ 2 NO (g) K = 1.0 x 1020
b) 2 CO (g) + O2 (g) ⇌ 2 CO2 (g) K = 2.2 x 10-22
c) 2 BrCl (g) ⇌ Br2 (g) + Cl2 (g) K = 1
a) products, b) products, c) reactants
a) reactants, b) products, c) reactants
a) products, b) reactants, c) equal
a) reactants, b) products, c) equal
The decomposition of nitrogen monoxide can be achieved under high temperatures to create the products of nitrogen and oxygen gas.
6 NO(aq) ⇌ 3 N2(aq) + 3 O2(aq)
a) What is the equilibrium equation for the reaction above?
b) What is the equilibrium expression for the reverse reaction.
The equilibrium constant, K, for 2 NO (g) + O2 (g) ⇌ 2 NO2 (g) is 6.9 x 102.
What is the [NO] in an equilibrium mixture of gaseous NO, O2, and NO2 at 500 K that contains 1.5 x 10 –2 M O2 and 4.3 x 10 –3 M NO2?
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