enthalpy changes
need HELP with these reactions! will give out rep
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
Original post by Deyn_08
need HELP with these reactions! will give out rep
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
here's an example that might help:
Enthalpy of ethanol
Bond Average Bond Enthalpy (Kj/mol)
C-C 348
C-H 412
C-O 360
O-H 463
C=O 743
Bond breaking:
(+348 x 1) + (+412 x 5) + (+360 x 1) + (+463 x 1) = 3231 kJ
Bond making:
(-463 x 6) + (-743 x 4) = -5750 kJ
total sum=
+3231 + - 5750 = -2519Kj mol
you need to read the table to understand the bonds.
hope that helps....
hey do know the chemical balanced equation for butanol and its enthalpy as im quite stuck on it.
(edited 12 years ago)
These are cycle questions i believe, not bond enthalpy as above you need a cycle with a top equation then elements underneath and drawing appropriate arrows from left and right of the top equation down (or indeed up) to the elements.... Now the arrows correspond with the signs of the kj mol value and should be switched when following a path to the other side of the reaction and also when going from elements to you need to multiply accordingly due to the multiple amounts of nitrogen for example 
hope this helps ill try get a pic of a worked out one up in a bit.....
-David
hope this helps ill try get a pic of a worked out one up in a bit.....-David
right here goes on the visual aid ..... and honestly i no of no explanation thats concrete i see them and am able to do them instinctually ... :/
http://photobucket.com/diddy2480
images too large to stick on here, first photo is the structure i use, and then in order of the questions you asked
+694 ammonia
-85 CH4
may have cocked up the third one but i cba to rectify hope this helps
-David
http://photobucket.com/diddy2480
images too large to stick on here, first photo is the structure i use, and then in order of the questions you asked
+694 ammonia
-85 CH4
may have cocked up the third one but i cba to rectify hope this helps
-David
Original post by diddy2480
right here goes on the visual aid ..... and honestly i no of no explanation thats concrete i see them and am able to do them instinctually ... :/
http://photobucket.com/diddy2480
images too large to stick on here, first photo is the structure i use, and then in order of the questions you asked
+694 ammonia
-85 CH4
may have cocked up the third one but i cba to rectify hope this helps
-David
http://photobucket.com/diddy2480
images too large to stick on here, first photo is the structure i use, and then in order of the questions you asked
+694 ammonia
-85 CH4
may have cocked up the third one but i cba to rectify hope this helps
-David
thanks, made things more clear
Original post by Deyn_08
need HELP with these reactions! will give out rep
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
3. Given the data:
4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(l), ?H = -1530kJmol-1
H2(g) + 1/2O2(g) ? H2O(l), ?H = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
8. Given the following data:
CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H = -890 kJmol-1
CO(g) + 1/2O2(g) ? CO2(g) ?H = -284 kJmol-1
C(s) + O2(g) ? CO2(g) ?H = -393 kJmol-1
H2(g) + 1/2O2(g) ? H2O(l) ?H = -286 kJmol-1
Calculate:
a) The enthalpy of formation of methane
b) The enthalpy of formation of carbon monoxide
c) The enthalpy change when methane is burned in limited oxygen to form carbon monoxide and water.
question 8
you can think about it as about simulteneous equations:
1.rearrange the first equation
CO2+2H2O --> CH4 + 2O2, H = 890(reverse reaction)(1)
then we multiply the last equation by 2:
2H2 + O2 --> 2H20, H = -572(2)
ADD (1) AND (2):
CO2+2H2O+2H2+O2 --> CH4+2O2+2H20, cancel 2H20 and O2:
CO2+2H2 --> CH4 + O2
ADD THIRD EQUATION(C+O2 --> CO2)
CO2+2H2+C+O2 --> CH4+CO2+O2
CANCEL OUT
C+2H2 --> CH4
NOW WE DO THE SAME WITH VALUES= 890+2*(-286)-393=-75 is the answer
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