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Current time:0:00Total duration:10:46

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 right do we have more products or do we have more reactants at equilibrium and this equation relates the equilibrium constant k2 Delta g0 the standard change in free energy so Delta g0 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 we'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 a for reaction to 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 want to use this equation down here and that ratio is related to the standard change in free energy Delta g0 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 0 is equal to negative 33.0 kilojoules for this balanced equation so let's write down our equation that relates Delta G 0 to K Delta g0 is equal to negative RT natural log of K Delta g0 is negative 33.0 kilojoules so let's write in here negative 33.0 and let's turn that into joules so times 10 to the 3rd 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 g0 is equal to negative 33.0 kilojoules 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 we'll start with the value for Delta g0 which is negative 33.0 times 10 to the 3rd so we're going to divide that divide that by negative 8.314 and we'd also need to divide by 298 and so we get 13 point 3 2 so now we have 13 point 3 2 right so the R 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 13.3 2 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 thirteen point three two so let's do that so let's take let's take a to the thirteen point three two and that's equal to this would be six point one six point one times 10 to the one two three four five so six point one six point one times 10 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 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 g0 let's do the same problem again but let's say that 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 goal 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 g0 is equal to negative RT and natural log of the equilibrium constant K and this time for Delta g0 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 0 so we have 0 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 0 is equal to e to the natural log of K each of the natural log of K is just equal to K so k is equal to e to the 0 and each of the zero is equal to one so when when delta g zero is equal to zero so let's write this down here so when your standard change in free energy Delta G zero is equal to zero K is equal to 1 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 1,000 K and our standard change in free energy Delta G 0 is equal to positive one hundred and six point five kilojoules so Delta G zero is equal to negative RT natural log of K this time we're putting in positive 100 6.5 kilojoules which is positive 106 point five 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 we is 1,000 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 point five 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 1,000 and that gives us negative twelve point eight one so we have negative twelve point eight one 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 twelve point eight one so what is that equal to e to the negative twelve point eight one is equal to two point seven times ten to the negative six so K the equilibrium constant is equal to two point seven times ten to the negative six so when 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 right 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