Question 85

Moderators: Chem_Mod, Chem_Admin

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Question 85

Postby Andrea Grigsby 1I » Sun Jan 21, 2018 11:56 pm

specifically part b, why is the final step in the answer book shown as the mole (from after calculating it with PV=nRT) being multiplied by 180.6kJ? I would've thought it would be multiplied by 90.3kJ which is reaction enthalpy that produces 1 mole of NO.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 85

Postby Kelly Kiremidjian 1C » Mon Jan 22, 2018 9:13 am

For part B, you must use the gas law, PV=nRT. In this problem we are given everything except for moles, so you can rearrange the equation to look like n=PV/RT. Then you can plug the given numbers in.
n=(1.00 atm)(5.45 L)/ (0.08206)(273K)
n=0.243 mol

Now that you have moles of Nitrogen, you can multiply the 0.243 moles x ΔH required for 1 mole of N2 to find the desired heat absorbed.
If you do this, you would get 0.243 moles N2 x 180.6 kJ/mol N2= 43.9kJ

Hope this helps!


Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 3 guests