Delta and ksp relationship problems

Standard change in free energy and the equilibrium constant (video) | Khan Academy

delta and ksp relationship problems

This is called confirmation bias and is actually a problem in Physics/Space/ KSP is only a hobby of mine. The exact relation is Δv = Isp * ln (total mass / dry mass) assuming your specific impulse is measured in m.s Go, for a chemical process at some temperature T, we can calculate the equilibrium constant for the process at that temperature using the relationship between. We can use this relationship to derive an equation to convert directly between K c K_\text .. To solve this problem, we can use the relationship between the two.

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.

Calculating the equilibrium constant from the standard cell potential (video) | Khan Academy

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 K.

So, we're trying to synthesize ammonia here, and at Kelvin, or 25 degrees C, the standard change in free energy, delta-G zero, is equal to negative So, let's write down our equation that relates delta-G zero to K. Delta-G zero is negative This is equal to the negative, the gas constant is 8. 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 So, we say kilojoules, or joules, over moles of reaction just to make our units work out, here.

  • Standard change in free energy and the equilibrium constant
  • Calculating equilibrium constant Kp using partial pressures
  • Calculating the equilibrium constant from the standard cell potential

Temperature is in Kelvin, so we have K, so, we write 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 So, we're going to divide that by negative 8.

And so, we get So, now we have So, how do we solve for K here? Well, we would take E to both sides. So, if we take E to the So, let's take E to the 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 Kelvin, so we're still trying to make ammonia here, and our goals is to find the equilibrium constant at this temperature. At Kelvin, the standard change in free energy, delta-G zero, is equal to zero.

Cell Potential & Gibbs Free Energy, Standard Reduction Potentials, Electrochemistry Problems

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 Kelvin.

delta and ksp relationship problems

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. Then over here, right, for our oxidation half reaction, we have three electrons. We need to have the same number of electrons.

So we need to have six electrons for both half reactions, because remember the electrons that are lost are the same electrons that are gained. So we need to multiply our first half reaction by three.

delta and ksp relationship problems

All right, if you multiply our first half reaction by three, we'll end up with six electrons. And our second half reaction, we would need to multiply the oxidation half reaction by two, in order to end up with six electrons. So let's rewrite our half reactions. So, first we'll do the reduction half reaction, so we have, let me change colors again here.

And let's do this color. So we have three times, since we have three I two now, We have three I two, and three times two gives us six electrons. So three I two plus six electrons, and then three times two, all right, gives us, three times two gives us six I minus. All right, so we multiplied our half reaction by three, but remember, we don't multiply the voltage by three, 'cause voltage is an intense of property.

So the standard reduction potential is still positive. So we have positive. Next, we need to multiply our oxidation half reaction by two. So we have two Al, so this is our oxidation half reaction, so two Al, so two aluminum, and then we have two Al three plus, so two Al three plus, and then two times three gives us six electrons.

So now we have our six electrons and once again, we do not multiply our standard oxidation potential by two, so we leave that, so the standard oxidation potential is still positive 1. Next we add our two half reactions together. And if we did everything right, we should get back our overall equation.

So our overall equation here.

Chemical Forums

We have six electrons on the reactant side, six electrons on the product side, so the electrons cancel out. And so we have for our reactants three I two, so we have three I two, plus two Al, and for our products right here, we have six I minus, so six I minus, plus two Al three plus, plus two Al three plus.

So this should be our overall reaction, all right, this should be the overall reaction that we were given in our problem. Let's double check that real fast. So three I two plus two Al, all right, so right up here, so three I two plus two Al, should give us six I minus plus two Al three plus.

So six I minus plus two Al three plus. So we got back our original reaction. Remember our goal was, our goal was to find the standard cell potential E zero, because from E zero we can calculate the equilibrium constant K. So we know how to do that, again, from an earlier video.

To find the standard cell potential, all right, so to find the standard cell potential, all we have to do is add our standard reduction potential and our standard oxidation potential.

So if we add our standard reduction potential and our standard oxidation potential, we'll get the standard cell potential. So that would be positive. So the standard potential for the cell, so E zero cell is equal to. All right, now that we've found the standard cell potential, we can calculate the equilibrium constant. So we can use one of the equations we talked about in the last video that relates the standard cell potential to the equilibrium constant.

So I'm gonna choose, I'm gonna choose one of those forms, so E zero is equal to 0. So again, this is from the previous video. So, the standard cell potential is 2. Remember what n is, n is the number of moles transferred in our redox reaction. So we go back up here and we look at our half reactions and how many moles of electrons were transferred?

Well, six electrons were lost, right, and then six electrons were gained. So n is equal to six. So we plug in n is equal to six into our equation.