posted by DeeDee .
Use standard enthalpies of formation to calculate ΔHrxn° for each reaction. (See the appendix. Enter your answer to the tenth place.)
(a) 2 H2S(g) + 3 O2(g) 2 H2O(l) + 2 SO2(g)
(b) N2O4(g) + 4 H2(g) N2(g) + 4 H2O(g)
(c) SO2(g) + 1/2 O2(g) SO3(g)
These are so easy, but no matter how I do it, the program won't except my answers. Could someone walk me through this?
It would be so much easier to find what you're doing wrong if you posted how you did one of them (with numbers).
dHrxn = (n*dHf products) - (n*dHf reactants)
Watch the number of significant figures.
BTW I posted a response to your NH4NO3 calculation.
I saw, and it was SO helpful! Thank you!!
Okay, here is how I did (b):
(4)x(-285.8) - (1)(11.1)=-1150
My table has 9.16 for N2O4 but I've assume you looked this up correctly in your table.
4(-285.8)-(11.1) = -1154.3 is what I obtained. Also, note the question tells you to enter to the tenths place. You've rounded to 1150 AND didn't enter to the tenths.
I noticed that right after I wrote you, and typed in -1154.3 thinking I had solved it, but the program rejected it. I don't know any other possible answer. I did the equations the same way for all three reactions, and the third one was counted correct while the others are not, no matter how many times or different ways I try to write it. All the enthalpy numbers I got were provided by the program too.
Thank you so much for helping and clarifying!! It was very helpful. I will keep this page bookmarked as to show my prof.
Thanks for sharing!
If you care to type your three solutions, including the one that was counted right, I might pick out the problem. I'm thinking it is a significant figure problem. If we multiply -285.8 x 4 we get -1143.2 but we are allowed only 4 places so this would round to -1143. Then we add in -11.1 for -1154.1 as the answer. Does that make sense to you?
Ah, yes it does!
Here are the three complete:
2 H2S(g) + 3 O2(g) -> 2 H2O(l) + 2 SO2(g)
[2(-285.8)+2(-296.8)] - [2(-20.6)] =
(-571.6 + -593.6) - (-41.2) = -1124 (-1124.0?)
N2O4(g) + 4 H2(g) -> N2(g) + 4 H2O(g)
[4(-285.8)] - [1(11.1)]= -1154.3
SO2(g) + 1/2 O2(g) -> SO3(g)
-395.7 - -296.8 = -98.9
(c) -98.9 was correct, but the others were not, even though I did the exact same way, to a T. Also, I tripled checked all my numbers and they are consistent with the numbers provided on the site for the problem.
This question is due tonight at 11:30 and I have used up all my free tries.
Thank you so much for your continued help!
I think a is 1124. Putting the zero on makes it too many s.f.
I think b is 1154.1 as suggested earlier.
c works out to be -98.90 but the problem says to the tenth; therefore the 0 should not be there. Hope this takes care of the problem.