[mahi123]

we did a lab in which we dissolved 8-10 g of some unknown salt in water in styrofoam cups( calorimeter) and calculated the enthalpy of solution in kj/g and then compared to values of substances in a table to find what your substance was. I have one question in lab which asks what would have happened had some salt spilled( would enthalpy be high or low). I know we calculate the enthalpy using q=mct and then divide by 8-10 to get it in kj/g. Our reaction was endothermic so temp of water went down. Now I know if there was less mass of salt, the there would be a smaller temperature change so we would get a smaller enthalpy or enthalpy of solution. So I'm saying it would be low.
Also the formula h=nhx, we know if n is smaller, h is smaller also.
         But if we rearrange nhx=mct for hx, we get hx=mct/n,  so if n is smaller, we get bigger hx, so from equation h=nhx, if hx is bigger we get bigger h?  This is so confusing cuz I don't know which method is right. Im thinking it's the first one as it makes more sense, but why with the equation nhx=mct the enthalpy would be high for a smaller n value and lower using other methods. If someone can take the time to answer me, that would help a lot.

[mahi123]

seems like either other people are confused as well like me or they don't know what im trying to say, cmon i need helppp i have to hand this in tomorrow plssss

[macman104]

what is h=nhx?  Also, yes, I agree.  If you spilled some of your salt, you would have calculated a lower enthalpy, because while you might have thought you put in 10 g or whatever of the salt, you might have only put in 9.5g, which would cause your calculation to be low.  I don't know what h=nhx is, so I can't speak to your confusion regarding how that equation works out.

For future reference, please post your questions earlier.  People don't always get a chance to respond the day of, so plan ahead if you need help.  Also, posting a second time about how your assignment is due really soon and you really need help, is generally not helpful.

[mahi123]

hey thanks for replying.

 h=nhx, h is the enthalpy, n is the number of moles, hx is molar enthalpy so in place of x you would write the type of reaction it is hsol or hcomb or hvap etc, and this equation combined with q=mct gives nhx=mct .

Im gonna hand in my lab report tomorrow, and i wrote down enthalpy would be lower  cuz it makes more sense and you can see it from the formula q=mct if mass is less t would be smaller, so smaller q value, but i just don't know why with the other formula nhx=mct if you rearranged variables to see how each would effect the other, you get hx to be bigger if n is smaller so then when you sub hx in h=nhx, you would get higher h and h is samething as q, atleast for my purpose. ill ask my teacher tomorrow

[macman104]

Ah, ok.  For future reference, there are special tags on these forums.  You can use the

Code: [Select]

[sub][/sub]tags to get subscript.  So hx could be h_x.  As far as nh_x=cmt...
You are forgetting that if n decreases, in the equation cmt/n, then m would also decrease (by some ratio, since, m and n are directly related by molar mass).  So, h_x will also go down.  Thus, h would still go down, because n went down so did h_x.  So either way you would look at it, h still decreases, make sense?

As an example (I'm making these numbers up by the way)...

c=4.184
deltaT = 20
m = 10g (this is what we should have measured) - for our purposes, we're also going to say that 10g = 1 mol

h = c*m*deltaT = 836.8

We rearrange for h_x and get (4.184*10*20)/1 = 836.8
Then h=n*h_x = 1*836.8 = 836.8

So we also get 836.8 for the other method

Now, say that accidentally, we spill and only put in 5g

m = 5g (this is now equal to only 0.5 moles)
deltaT = 10

h = c*m*deltaT = 209.2

we rearrange n*h_x=c*m*deltaT for h_x, and get

(4.184*5*10)/0.5 = 418.4

Then h = n*h_x = 0.5*418.4 = 209.2

So we still get 209.2

So you see that no matter which way we calculate it, we still get the same result of h decreasing if we spill some.  I think the main thing you forgot is that by decreasing the number of moles, we also had to decrease m, and thus, we still decrease h_x

[mahi123]

Well, i don't know, the way i was thinking is this. Our reaction was endothermic, so the heat was transferred from the water to the salt. now if there was less salt, it's obvious there would have been less heat transfer, therefore smaller delta t change

so from q=mct, we can conclude if t is smaller so is q i think we both agree with this part


now nh_x=mct, well let's rearrange this for h_x, so we get

h_x=mct/n, now from this we know if mass was spilled, we would have smaller n. So whenever you have a smaller number in a denominator, you would get a bigger quantity, like 1/1=1 but 1/.5 is 2. So if n is smaller delta h_x is big so from the other equation h=nh_x we can see if h_x is big, h is also big. But i know the correct answer is 1st way, but this i just had it in my mind for some reason lol anyway

[macman104]

Well, i don't know, the way i was thinking is this. Our reaction was endothermic, so the heat was transferred from the water to the salt. now if there was less salt, it's obvious there would have been less heat transfer, therefore smaller delta t change

Correct.

so from q=mct, we can conclude if t is smaller so is q i think we both agree with this part

Yes, we can agree on this.

now nh_x=mct, well let's rearrange this for h_x, so we get

h_x=mct/n, now from this we know if mass was spilled, we would have smaller n. So whenever you have a smaller number in a denominator, you would get a bigger quantity, like 1/1=1 but 1/.5 is 2. So if n is smaller delta h_x is big

The part you are leaving out is that, while n is small, which would make h_x large, you also have to make m and t smaller along with n.

Either way, glad it made sense in the end.

[Padfoot]

The units for molar enthalpy of solution are kJmol^-1.

It is enough to look at these units to see that molar enthalpy is independant of of number of moles/mass.

Unless I'm missing something 

[mahi123]

The units for molar enthalpy of solution are kJmol^-1.

It is enough to look at these units to see that molar enthalpy is independant of of number of moles/mass.

Unless I'm missing something 

you're missing something. I think it's just me, i forgot to state the question properly, it asks if some salt was spilled, would your calculated enthalpy of salt be low or high and we used 10 g.

So assuming we didn't drop any salt,
we get let's say 10/10=1kj/g
but if we did spill some salt, it's 10/5 = 2kj/g so if we dropped it's higher, so our calculated enthalpy be low

I know you're saying since it's /g it doesn't depend on quantity, but i think what the question is trying to ask is if we did drop it( we wouldnt know), so we would still divide it by 10, so when we do that our enthalpy comes out to be lower, i hope it clears up but i think my question wasn't clear earllier but this is so confusing tho, i can do all the problems right in the textbook, it's just this question lol

[Padfoot]

Oh, ok, I was wondering what on earth was going on in this thread.  So the spillage was not realised.  But this seems to at odds with a few statements made earlier in this thread:

But if we rearrange nhx=mct for hx, we get hx=mct/n,  so if n is smaller, we get bigger hx, so from equation h=nhx, if hx is bigger we get bigger h?

The part you are leaving out is that, while n is small, which would make h_x large, you also have to make m and t smaller along with n.

n(hx)=mCt
If a spillage occurs (and we don't know about it), in our calculations only delta t will change (it will decrease) and hence the apparent value of (hx) goes down.