Question
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
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.
I got delta H = 11.3 kJ
Can I see the calculation
I want to see the calculation