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Enthalpy
Questions (3)
Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion.
Note: you will
1 answer
178 views
I finished my working out for a question (question not needed):
1) 6C + 6O2 → 6CO2 ΔH = -2360.4 2) 3H2 + 3 ½ O2 → 3H2O ΔH
11 answers
180 views
Enthalpy of reaction of nitrogen dioxide (use Hess Law)
N2(g) + 2O2(g) → 2NO2(g) N2 + O2 → 2NO deltaHr= +180 kJmol-1 2NO + O2
7 answers
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Answers (10)
I saw your response again. It answers everything.
How would you get to that same answer from the question given? The question was: Use the information provided in the below equation and Table, to show how the enthalpy of formation for benzene was determined using Hess’ Law. You should include an
I understand the answer. You did do everything.
I wrote that I got the same equation and the same deltaHf as the question. My values did add up to this.
The question was: Use the information provided in the below equation and Table, to show how the enthalpy of formation for benzene was determined using Hess’ Law. You should include an enthalpy cycle diagram and show your workings clearly. 6C(s) + 3H2(g)
Do you not need to multiply the whole equation by the same thing? With 3H2 + 3/2 O2 ==> 3H2O this was not done.
What about: 1) 6C + 6O2 → 6CO2 2) 3H2 + 3 ½ O2 → 3H2O 3) 6CO2 + 3H2O → C6H6 + 7 ½ O2 Looking at my equations (1, 2, 3) why is oxygen not equal on both sides? Hydrogen and carbon are equal. I still left my answer as the starting equation: 6C (s) +
Also, why write 2 ΔH (NO2) instead of ΔH (2NO2)?
In the calculations why should I write 2 ΔH (NO2(g)) instead of ΔH (2NO2(g))? I did this myself after seeing what the teacher had done in class for another calculation in enthalpy.
That is 68. N2(g) + 2O2(g) → 2NO2(g) has 2O2. N2 + O2 → 2NO deltaHr= +180 kJmol-1 does not have 2O2. What about that?
