Asked by Anonymous
Calculate delta H for the formation of one mole of N2O5 from the elements at 25 degrees C using the following data.
2H2 + O2 - 2H2O delta H = -571.6 kJ
N2O5 + H2O - 2HNO3 deltaH = -73.7 kJ
1/2N2 + 3/2O2 + 1/2H2 - HNO3 delta H = -174.1 kJ
2H2 + O2 - 2H2O delta H = -571.6 kJ
N2O5 + H2O - 2HNO3 deltaH = -73.7 kJ
1/2N2 + 3/2O2 + 1/2H2 - HNO3 delta H = -174.1 kJ
Answers
Answered by
DrBob222
I believe you need to do the following:
reverse equation 1 and change sign of delta H.
Multiply equation 2 x 2 and reverse. Multiply delta H by 2 and change sign.
Multiply equation 3 by 4. Multiply delta H by 4.
Add the equations and add the delta Hs.
When you finish you should have
2N2 + 5O2 ==> 2N2O5 which is just twice what you are looking for so divide the coefficients by 2 throughout and divide total delta H by 2.
reverse equation 1 and change sign of delta H.
Multiply equation 2 x 2 and reverse. Multiply delta H by 2 and change sign.
Multiply equation 3 by 4. Multiply delta H by 4.
Add the equations and add the delta Hs.
When you finish you should have
2N2 + 5O2 ==> 2N2O5 which is just twice what you are looking for so divide the coefficients by 2 throughout and divide total delta H by 2.
Answered by
clarke
I got delta H = 11.3 kJ
Answered by
Veronica
Can I see the calculation
Answered by
Veronica
I want to see the calculation
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.