- Gibbs free energy and spontaneity
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- 2015 AP Chemistry free response 2c
Standard change in free energy and the equilibrium constant
The relationship between standard Gibbs free energy change and the equilibrium constant K. Calculating K when you know the standard free energy of reaction.
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- I'm confused by the word "equilibrium". From the previous video it was said that at (delta)G = 0, the reaction is in equilibrium, But in this video, eventhought it question is asking for equilibrium constance, (delta)G = -33 kJ.
Well this is what I understand,
The "equilibrium" indicated by (delta)G = 0 is the equilibrium of spontaneity. It means by the energy and entropy of that environment, the reaction rate will be constant both forward and backward.
The "equilibrium" indicated by equilibrium constant K however, is the equilibrium of the concentration. It determines whether the reaction goes forward or backward depending on the qoutient (concentration of the products and reactants)
ex1. the reaction has (delta)G = 0, the reaction still goes forward, if Q<K.
ex2. if Q < Kp (the reaction favours products) but (delta)G > 0, the reaction still doesn't occurs, since the reaction is not spontaneous. (unless, energy is added into the system)
ex3 if Q> Kp (the reaction favours reactants, but (delta)g<0 , the reaction occurs (spontaneous), but there are more reactants than products,
Am i understanding it right? please correct me if i'm wrong.(8 votes)
- Equilibrium means that there is no observed change in the concentrations of reactants or products. Think about it as when the reaction has completed, even though it isn't exactly like that. For Example, for the equation:
N2(g) + 3H2(g) --> 2NH3(g)
at Equilibrium, the Reaction might not stop occurring, but the concentrations of N2, H2, and NH3 stay relatively the same.
The K value tells how much the concentrations of products and reactants there are.
If K >1, there are more Products than Reactants in Equilibrium.
If K=1, there are the same amount of Products as Reactants in Equilibrium.
If K <1, there are more Reactants than Products in Equilibrium.
Q is basically what the value of the the ratio of Products to Reactants before Equilibrium is established. so if
Q<K, the Reaction shifts to the Right,
Q>K, the Reaction shifts to the Left,
Q=K, the Reaction is at Equilibrium.
Delta G comes into Play when figuring out if the Reaction is Spontaneous.
when delta G>0, the Reaction is non- Spontaneous, but if
delta G <0, the reaction is spontaneous.
When K<1, the reaction favors the Reactants, so the Reaction is not Spontaneous, making delta G >0. but when K >1, the Reaction favors the Products, so it is Spontaneous, making delta G< 0.
delta G = -RTlnK
Hope that clears up some of your confusion!(22 votes)
- Why the ΔG0 has different values if the temperature is always T=298 K ?(3 votes)
- Thats because, ΔG = ΔH - TΔS
Even if T of environment is constant, but the change in the entropy is dependent on change in disorder. Hence the term TΔS is different for every reaction.(3 votes)
- What is Standard change in free energy?(4 votes)
- I thought dG0 was a constant for the reaction at 25C. To then find the new equilibrium at a new temperature you take e^(-dg0/(RT))... Why does dg0 change in the different examples for the same reaction?(4 votes)
- Is ΔG only applicable to reversible chemical reactions?(2 votes)
- ΔG applies to every reaction, but ΔG = 0 only for a reaction at equilibrium.(3 votes)
- converting from kj to J don't you just divide by 1000 because for every 1 Joule there is 1000 KiloJoules? 106.5kj x (1/1000)= 0.1065. As I see in the video you just made 106.5kj into 106.5x10^3 Joules. I am just confused because when I cancel out units I should be dividing by 1000 not multiplying.(1 vote)
- The best thing to do is use dimensional analysis rather than trying to remember when to mulitply and when to divide. With dimensional analysis, you multiply by the conversion factors in such a way as to take what you started with and get the units to cancel until you get what you wanted. Thus, to convert kJ to J you do the following:
1 kJ × (1000 J / 1 kJ) = 1000 J
Notice that the
kJcancel out and you're left with
Jas you units.
You can use dimensional analysis for many of the processes in Chemistry.(3 votes)
- I’m a bit confused about Kp and Kc, how can they be equal since isn’t the formula the Kp is Kp=Kc(RT)^delta N(N is the coefficient of the elements) and N is clearly not 0 here(2 votes)
- Hi, how can I link equilibrium to the effectiveness of hot and cold packs?(2 votes)
- Interesting question! The central idea here is that systems will always tend toward equilibrium. So if there is a difference in temperature between two entities that are in contact, heat will flow from the hotter entity to the colder one; this decreases the temperature of the former and increases that of the latter until both are at the same temperature. For instance, if you are cold and apply a hot pack to your skin, some heat from the hot pack will flow to your skin, and thus you will get warmer and the hot pack colder. Of course, this process will only occur until your skin and the hot pack are at the same temperature and then there will be an equilibrium, which means not net change in temperature between the two systems. Heat will still flow between the hot pack and your skin, but now at the same rate in both directions, so there is no overall change at equilibrium.
The inverse of the above process occurs with the cold pack, expect in this case your body is the heat donor and the cold pack the acceptor. :)(1 vote)
- So when delta G = 0, what does that tell us about the free energies of the reactants and products? They have the same magnitude but opposite charges? If so, how's the reaction still going in both directions? Is the reaction spontaneous in both directions?(2 votes)
- At5:39he said "since we're dealing with gases, if you wanted to put in a Kp here you could." But because Kp and Kc aren't interchangeable (related by a factor of (RT)^∆n), how do you know if the value you got is Kp or Kc? I know the example given is with gases, but if I had denoted the K as Kc instead, it would supposedly be wrong, right? How is the K in the larger expression -RTlnK defined?(2 votes)
- [Voiceover] In the previous video, we looked at the relationship between the change in free energy, delta-G, and the reaction quotient, Q. And we plugged in different values for Q and we saw how that affected our answer for delta-G. The sign of delta-G told us if a reaction was spontaneous or not. We also said that at equilibrium, Q, the reaction quotient is equal to the equilibrium constant, K. And we plugged K into the equation and solved for delta-G. Delta-G was equal to zero. So, we know, at equilibrium, the change in free energy is equal to zero. So, there's no difference in free energy between the reactants and the products. Let's plug in delta-G is equal to zero into our top equation here, so, we have zero is, zero is equal to delta-G zero, the standard change in free energy, plus R times T, and since we're at equilibrium, delta-G is equal to zero, this would be the natural log of the equilibrium constant, K. So, we solve for delta-G zero. Delta-G zero is equal to negative RT, natural log of K. So, we have another very important equation to think about. Delta-G zero is the standard change in free energy, or the change in free energy under standard conditions. R is the gas constant, T is the temperature in Kelvin, and K is our equilibrium constant. So, if you're using this equation, you're at equilibrium, delta-G is equal to zero. And we know at equilibrium, our equilibrium constant tells us something about the equilibrium mixture. Alright, do we have more products or do we have more reactants at equilibrium. And this equation relates the equilibrium constant K to delta-G zero, the standard change in free energy. So, delta-G zero becomes a guide to the ratio of the amount of products to reactants at equilibrium, because it's related to the equilibrium constant K in this equation. If you're trying to find the spontaneity of a reaction, you have to use this equation up here, and look at the sign for delta-G. So, if you're trying to find if a reaction is spontaneous or not, use this equation. If you're trying to find or think about the ratio of the amount of products to reactants at equilibrium, then you wanna use this equation down here, and that ratio is related to the standard change in free energy, delta-G zero. Now we're ready to find some equilibrium constants. Remember, for a specific temperature, you have one equilibrium constant. So, we're going to find the equilibrium constant for this reaction at 298 K. So, we're trying to synthesize ammonia here, and at 298 Kelvin, or 25 degrees C, the standard change in free energy, delta-G zero, is equal to negative 33.0 kilojoules for this balanced equation. So, let's write down our equation that relates delta-G zero to K. Delta-G zero is equal to negative RT, natural log of K. Delta-G zero is negative 33.0 kilojoules, so, let's write in here, negative 33.0, and let's turn that into joules, so times ten to the third, joules. This is equal to the negative, the gas constant is 8.314 joules over moles times K. So, we need to write over here, joules over moles of reaction. So, for this balanced equation, for this reaction, delta-G zero is equal to negative 33.0 kilojoules. So, we say kilojoules, or joules, over moles of reaction just to make our units work out, here. Temperature is in Kelvin, so we have 298 K, so, we write 298 K in here, Kelvin would cancel out, and then we have the natural log of K, our equilibrium constant, which is what we are trying to find. So, let's get out the calculator and we'll start with the value for delta-G zero which is negative 33.0 times 10 to the third. So, we're going to divide that by negative 8.314, and we'd also need to divide by 298. And so, we get 13.32. So, now we have 13.32, right, so our units cancel out here, and this is equal to the natural log of the equilibrium constant, K. So, how do we solve for K here? Well, we would take E to both sides. So, if we take E to the 13.32 on the left, and E to the natural log of K on the right, this would cancel out and K would be equal to E to the 13.32, so let's do that. So, let's take E to the 13.32, and that's equal to, this would be 6.1, 6.1 times ten to the one, two, three, four, five. So, 6.1, 6.1 times ten to the fifth. And since we're dealing with gases, if you wanted to put in a KP here, you could. So, now we have an equilibrium constant, K, which is much greater than one. And we got this value from a negative value for delta-G zero. So, let's go back up to here, and we see that delta-G zero, right, is negative. So, when delta-G zero is less than zero, so when delta-G zero is negative, what do we get for our equilibrium constant? We get that our equilibrium constant, K, is much greater than one. So, what does this tell us about our equilibrium mixture? This tells us that at equilibrium, the products are favored over the reactants, so the equilibrium mixture contains more products than reactants. And we figured that out by using our value for delta-G zero. Let's do the same problem again, but let's say our reaction is at a different temperature. So now, our reaction is at 464 Kelvin, so we're still trying to make ammonia here, and our goals is to find the equilibrium constant at this temperature. At 464 Kelvin, the standard change in free energy, delta-G zero, is equal to zero. So, we write down our equation, delta-G zero is equal to negative RT, a natural log of the equilibrium constant, K. And this time, for delta-G zero, we're plugging in zero. So, zero is equal to, we know that R is the gas constant, and we know that the temperature here would be 464 Kelvin. So, for everything on the right to be equal to zero, the natural log of K must be equal to zero. So, we have zero is equal to the natural log of K. And now, we're solving for K, we're finding the equilibrium constant. So, we take E to both sides. So, E to the zero is equal to E to the natural log of K. E to the natural log of K is just equal to K. So, K is equal to E to the zero, and E to the zero is equal to one. So, when delta-G zero is equal to zero, so let's write this down on here, so, when your standard change in free energy, delta-G zero, is equal to zero, K is equal to one. And that means that at equilibrium, your products and your reactants are equally favored. Let's do one more example. So, let's find the equilibrium constant again at another temperature. So, now we're at 1000 K, and our standard change in free energy, delta-G zero, is equal to positive 106.5 kilojoules. So, delta-G zero is equal to negative RT, natural log of K. This time, we're putting in positive 106.5 kilojoules, which is positive 106.5 times ten to the third joules is equal to negative, R is our gas constant, 8.314, I'll leave units out of this just to make it a little bit clearer, times the temperature, which is 1000 K, so this would be 1000 Kelvin times the natural log of the equilibrium constant, K. So, let's do the math there. We'll start with our delta-G zero, which is 106.5 times ten to the third. So, we're going to take that value and divide it by negative 8.314, and then, we need to divide by 1000, and that gives us negative 12.81. So, we have negative 12.81 is equal to the natural log of the equilibrium constant. So, to solve for the equilibrium constant, we take E to both sides, and we get that K is equal to E to the negative 12.81. So, what is that equal to? E to the negative 12.81 is equal to 2.7 times 10 to the negative six. So, K, the equilibrium constant, is equal to 2.7 times ten to the negative six. So, when delta-G zero is positive, when the standard change in free energy is positive, let's write this one down. So, when delta-G zero is greater than zero, so, when it's positive, your equilibrium constant, K, is less than one. Alright, so K is less than one. And we know what that means at equilibrium. The reactants are favored at equilibrium. So, your equilibrium mixture contains more reactants than products.