chemistry

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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?

• chemistry -

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.

• chemistry -

I saw, and it was SO helpful! Thank you!!
Okay, here is how I did (b):
(4)x(-285.8) - (1)(11.1)=-1150

• chemistry -

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.

• chemistry -

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!

• chemistry -

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?

• chemistry -

Ah, yes it does!
Here are the three complete:
(a)
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?)

(b)
N2O4(g) + 4 H2(g) -> N2(g) + 4 H2O(g)
[4(-285.8)] - [1(11.1)]= -1154.3

(c)
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!

• chemistry -

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.

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