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Current time:0:00Total duration:9:57

have this reaction here where if I had a mole of methane and I react that with two moles of oxygen I'll produce the mole of carbon dioxide and two moles of water but we want to answer in this video is whether this reaction is spontaneous and we learned in the last video that to answer that question we have to turn to Gibbs free energy or the change in Gibbs free energy and the change in Gibbs free energy is equal to the enthalpy change for the reaction minus the temperature which it is occurring times the change in entropy and if this is less than zero then it's a spontaneous reaction spontaneous so I got us a I gave us a little bit of a head start I just calculated the change in enthalpy for this reaction and that's right here and we know how to do that we've done that several videos ago it you could just look up the heats of formation of each of these products it for water you'll multiply it by two since you have two moles of it and so you have the heats of formation of all the products and then you subtract out the heats of formation of all the reactants and of course the heat of formation of o2 is 0 so this won't even show up in it and you'll get minus 890 point 3 kilojoules what's tells us that this is an exothermic reaction that this side of the equation has less energy in it you can kind of think of it that way is that side so some energy must have been released we could even put here you know plus e for energy let me write plus some energy is going to be released so that's why it's exothermic but our question is is this spontaneous so to figure out if it's spontaneous we also have to figure out our Delta s our Delta s and to help figure out the Delta s I head of time looked up the standard molar entropy s for each of these molecules so for example the standard I'll write it here in a different color the standard and this is just the notation the standard no there's no Delta there let me erase I could just write it here the standard standard and standard you put a little knot symbol there the standard molar entropy so when we say standard it's at 298 degrees Kelvin actually I shouldn't say degrees Kelvin is at 290 Kelvin you don't use the word degrees necessarily when you talk about Kelvin so set 298 Kelvin which is 25 degrees Celsius so at room temperature so that's why it's considered standard temperature so the standard entropy of methane at room temperature is equal to this number right here 186 joules per Kelvin molar not molar per Kelvin mole so if I have one mole of methane I have a 186 joules per Kelvin of entropy if I have 2 moles I multiply that by 2 if I have 3 moles I've multiplied with that by 3 so our total entropy is the entropy or the total change in entropy of this reaction is the total standard entropies of the products minus the total standard enterpise of the reactions just like what we did with enthalpy so that's going to be equal to 213 0.6 plus I have 2 moles of water here so it's plus 2 times 2 times let's just write 70 there 69 point nine almost 70 plus two times 70 and then I want to subtract out the entropy of the reaction or this side of the the reactants or the side of the reaction so the entropy of one mole of ch4 is 186 186 plus 2 times 205 2 times 205 so just eyeballing already this number is close to this number but this number is much larger than this number liquid water has a much lower this is liquid waters entropy it has a much lower entropy than oxygen gas and that makes sense because liquid form there's a lot fewer states it kind of takes the shape of it it you know it all falls to the bottom of the container as opposed to kind of taking the shape of the room and expanding so a gas is naturally going to have much higher entropy than a liquid so just eyeballing it we can we can already see that our products are going to have a lower entropy than our reactants so this is probably to be a negative number but let's let's confirm that let's confirm that so I have 200 let me see I have 200 to 13.6 plus well plus 140 right two times seventy plus 140 is equal to 350 three point six so this is three fifty three point six and then from that I'm going to subtract out so 186 plus two times now let me see plus 2 times 205 is equal to five 96 so minus 596 and what is that equal to so let me put the minus five ninety six and then plus the 350 350 3.6 and we have minus two hundred forty two point four so this is equal to minus two forty-two point four joules per Kelvin is our Delta s minus two so we lose that much entropy those units might not make sense to you right now and actually these are a bit of arbitrary units but you can just say hey this is getting more ordered and it makes sense because we have a ton of gas we have three and we have three separate molecules one here and two molecules of water of oxygen and then we go to three molecules again but the water is now liquid so it makes sense to me that we lose entropy there's there are fewer states that the liquid especially can take on but let's figure out whether this reaction is spontaneous so our Delta G Delta G is equal to our Delta H we're releasing energy so it's minus 890 I'll just get rid of the decimals we don't have to be that precise minus our temperature where we're assuming that we're at room temperature 298 degrees Kelvin that's 28 I should just say 298 Kelvin I should get in the habit of not saying degrees when I say Kelvin which is 25 degrees Celsius times times our change in entropy now this is going to be minus now you might say okay - 240 - you might want to put that there we have to be very very very careful this right here is in kilojoules this right here is in joules so we want to write if we want to write everything in kilojoules since we already wrote that down let's write this in kilojoules so it's point two for two kilojoules kilojoules per Kelvin so I can put a point two for two let me get rid of this right there kilojoules per Kelvin and so now our Gibbs free energy right here is going to be minus 890 kilojoules minus 209 so the minus and the minus you get a plus and that makes sense that this the entropy term is going to make our Gibbs free energy more positive which as we know since we want to get a we want to get this thing below zero this is going to fight the spontaneity but let's see if it can overwhelm if it can overwhelm the actual enthalpy the exothermic nature of it and it seems like it will because you multiplied a fraction times this it's going to be a smaller number than that but let's just figure it out so divided by one two three that's our change in entropy times 298 that's our temperature is minus 72 so this so this term becomes and then we put a minus there so it's plus 72 point two so this is the entropy term at standard temperature it turns into that and this is our enthalpy term so we can already see that the enthalpy is a much more negative number than our positive term from our temperature times our change in entropy so this term is going to win out this term is going to win out even though we lose entropy in this reaction it releases so much energy that's going to be spontaneous this is definitely less than zero so this is going to be a spontaneous reaction as you can see these Gibbs free energy problems are really not too difficult you just really need to find these values and to find these values you it'll either be given the Delta H but we know how to solve for the Delta H you just look up the heats of formations of all the products subtract out the reactants and of course you wait by the number by the coefficients and then to figure out the change in entropy you do the same thing you have to look up the standard molar entropy of the products weight by the coefficients subtract out the reactants and then just substitute in here and then you essentially have the Gibbs free energy and case it was negative now you could imagine a situation where with a much higher temperature let's wear it like we're at the surface of the Sun or something where all of a sudden instead of a 298 here if you had like a I don't know you had a two thousand or a four thousand there then all of a sudden things become interesting if you know I mean if you could imagine if you had a a forty thousand Kelvin temperature here then all of a sudden the entropy term the loss of entropy is going to matter a lot more and so this term this positive term is going to outweigh this and maybe it wouldn't be spontaneous at a very very very very high temperature another way to think about it a reaction that generates heat that lets out heat doesn't matter that the heat being released doesn't matter so much when there's already a lot of heat or kinetic energy in the environments if the temperature was high enough this reaction might would not be spontaneous because maybe then the entropy term would win out but anyway I just wanted to do this calculation for you to show you that you know there's nothing to abstract here you can look up everything on the web and then figure out if something is going to be spontaneous