00:01
This question wants us to calculate the equilibrium constants at 655 for some reactions.
00:14
And the first reaction is a reaction between n -o -2.
00:23
So we have 2 n02.
00:27
We have 2 n02 gas.
00:30
Gas going to n204 gas and we are to estimate kp at 655 p so the first thing to do is to calculate the delta g of reaction and the way to do that is to subtract the delta g formation of the products from a supply the delta g of a formation of reactants from the products so the delta g of formation of n204 so that's one times we have one mole of n204 1 mole of n204 multiplied by the delta g of formation of n204 which is running the data table so n204 gas is 99 .8 so we have 99 .8 moles 99 .8 kilojouge from mole minus 2 more no2.
01:54
N02 is 51 .3 kilojouge per more.
02:07
So this will give us 99 .8 minus 2 times 51 .3.
02:28
This is minus 2 .8 kilojouge.
02:32
Then we can find the kp since we know dltg is equals to minus rtl in kp so kp will be equal to exponential minus dlt standard over rt and that's kp will be equals to exponential minus minus 2 .8 kilojoules divided divided by haryl is 8 .314 but since our doughty g is in kilojouze i'm going to convert that to kilojouge by multiplying by 10x power minus 3 kilojougeal for kelvin times temperature is 298 kelvin so this will give us 3 .8 about by 314 times minus 3 times 398 so this would give us exponent here one one one three and this would give us three point zero nine six so kp is equal to three point zero nine six now we now we have to evaluate k at six twenty twenty 5 65 kelvin so this is the k at 298k now we can call this k1 then we need to find k at 655k so we can call that k2 at 655k okay so to find k2 as 655k okay so to find k2 so the expression collaborative clauscious equation k k2 over k1 is equal to negative delta h of reaction over half into 1 over t 2 minus 1 over t 1 1 so um k2 is the polygon constant at 655k and uh t2 is the 655 k k1 is the 655 k and t1 is at temperature at 298 k so we need to find what delta h of reaction is so that's a h standard of reaction would be equals to product minus reactant as usual.
05:43
So we have to find the delta h of formation of n2 .04.
05:48
So n2 .o4 if we check the table, the delta h of formation of n204 is 9 .16.
05:56
So we have one mole of it.
05:59
So this is 1 mole times 9 .16 kilojoules per mole.
06:06
Minus the delta h of formation of n02 multiply by two moles so two mole multiply that's h of formation of n02 and n02 is 33 .2 so that's 33 .2 kilo two kilojouge per mole so what would this give us so you have 9 .16 minus 2 times 33 .2 so this is minus 57 .24 kilojouge so we know that now so link k2 over k1 we calculated to be 3 .096 3 .096 what they start together in bracket is opposed to negative into negative into negative 57 .2 kilojoules divided by how is the gas constant and since this is in kilojou we should have it in kilojol as well so this is 8 .3 .14 times 10 is by minus 3 kilojoules per kelvin into 1 over t2 t2 is 655k minus 1 over t1 which is 298k so 1 about 655 minus 1 divided by 298 times 7 .24 divided by 8 .34 times times by minus 3.
07:59
So link k to logarithm minus lin 3 .96 is equals to minus 12 .592.
08:17
So lean of 3 .096, then 3 .096.
08:27
So link k2 will be equals to minus 12 .592 plus 12 .592 plus 1 .1301.
08:44
So link, so k2 we then be equals to exponential.
08:50
So the exponential minus 12 .592 plus.
09:02
So this would be esponation minus 11 .462.
09:52
So this would be 1 .345.
09:55
This should be 1 .05 times 10x minus 5.
10:06
Okay, let's look at it again to make sure everything is fine.
10:12
So we have theirs, 3 .8 divided by 8 .34, 8 .34 times 10x by minus 3 .98.
10:44
It's 1 .130.
10:49
So, the explanation of that is 3 .96...