8.85

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Robert Estrada
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

8.85

Postby Robert Estrada » Thu Feb 01, 2018 8:20 pm

The oxidation of nitrogen in the hot exhaust of jet engines and automobile engines
occurs by the reaction
N2 (g) + O2 (g) ----> 2NO (g) ΔHº =+180.6kJ
a. How much heat is absorbed by the formation of 1.55 mol of NO?
b. How much heat is absorbed by the reaction of 5.45 L of nitrogen measured at
1.00 atm and 273K?
c. when the oxidation of N2 to No was completed in a bomb calorimeter, the heat was measured at 492J. What mass of nitrogen gas was oxidized?

Esin Gumustekin 2J
Posts: 57
Joined: Thu Jul 27, 2017 3:01 am

Re: 8.85

Postby Esin Gumustekin 2J » Thu Feb 01, 2018 9:35 pm

For part a, delta H is 180.6kJ/mol reaction. Since there are 2 moles of NO per mole of reaction, delta H is 180.6KJ/2mol NO. If you multiply this number by 1.55 mol NO you will find the heat absorbed in the reaction. In this case, the answer comes out to be 140kJ

Jasmine Wu 1L
Posts: 55
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time

Re: 8.85

Postby Jasmine Wu 1L » Thu Feb 01, 2018 9:48 pm

To answer this question:

a) Find the correlation between the enthalpy change and the number of moles of NO from thermochemical equation. The heat absorbed to form 2 mol NO is the given 180.6 kJ. Then, just multiply the 1.55 mol NO by the ratio described above: (180.6kJ / 2 mol).

b) Use the ideal gas formula (PV=nRT) to solve this. Plug in the given pressure, volume, temperature, and the R constant (about 0.08286 L.atm.K-1.mol-1 to find the number of moles being reacted. Then, multiply the number of N2 moles being reacted by (180.6kJ / 1 mol N2).

c) Convert the given joules to kilojoules. Then divide that by the given 180.6 kJ/mol to get the amount of moles of the nitrogen that was oxidized. Convert those moles to grams.

Robert Estrada
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

Re: 8.85

Postby Robert Estrada » Fri Feb 02, 2018 1:10 pm

Thank You!


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