00:01
Okay, so we have this reaction occurring at 298 kelvin.
00:04
We need to show that it is not spontaneous under standard conditions by calculating the delta g of the reaction.
00:11
So you need to look up the delta g values from the appendix, which for n2o is 103 .7.
00:21
For n02, it is 51 .3.
00:26
And for n0, it's 87 .6.
00:31
And of course these are in kilojoules.
00:34
So once you have looked up those values from your appendix, you're going to do the products minus reactants.
00:42
So i want to do three times the 87 .6 products minus our reactants, which is the, there's only one each of the 103 .7 and the 51 .3 .3.
00:59
So make sure to use coefficients when you're calculating with that.
01:05
And once you solve, you get that the delta g overall is positive 107 .8 kilojoules.
01:13
So because it is positive, it is a non -spontaneous process.
01:21
Okay, for part b, if the reaction mixture contains only n2 and n2 at partial pressures 1 atm each, reaction will be spontaneous until some n0 forms in the mixture.
01:31
What maximum partial pressure of no builds up before the reaction ceases to be spontaneous.
01:38
So this is a kind of an application of equilibrium with thermodynamics.
01:44
So the first thing we want to do is use the delta g equals negative rtl and k.
01:50
So we have our delta g value from part a.
01:52
And instead of kilojoules, i want to go ahead and put that in joules.
01:56
So 107, 800 joules for that one.
02:00
And r is 8 .314.
02:06
Our temperature at standard conditions is 298 kelvin, and we'll solve for k, and we get that k is equal to 1 .27 times 10 to the negative 19.
02:21
Now that we have our value for the equilibrium constant, let's go ahead and write our equilibrium expression.
02:27
It's the pressure of the no cubed over the pressure of the n2o times the pressure.
02:35
Of the n02 and from there i can plug in the 1 .27 times 10 to the negative 19 is equal to the pressure of the n0 cubed and we're going to plug in one for each of these now i know these are initial initial values for pressure but because the value of k is so small very very very little of it is only to actually convert into the no so i'm just going to assume that according to significant digits it's going to stay one atm for each of those and and again it's because this value is so small so if you solve for the pressure of no you get 5 .0 times 10 to the negative 7th atm which is exceptionally small compared to the one of the other two, so our assumption was valid.
03:38
Okay, now for part c, we want to know, can the reaction be made more spontaneous by an increase or decrease in temperature? and if so, what temperature is required to make the reaction spontaneous? for this, we need to calculate delta h and delta s using values from the appendix again.
03:55
So let's do delta h first...