Hess' Law: Absolutely Desperate

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kmj005
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#1
Report Thread starter 4 months ago
#1
*I'm an IB student in dire need of some help, please*

Question
Calculate the enthalpy change of formation of NO2 (g) from the following data:
2Pb(NO3)2 (s) ---> 4NO2 (g) + 2PbO (s) + O2 (g) ∆H = +602 kJ/mol
∆Hof [Pb(NO3)2 (s)] = -452 kJ/mol
∆Hof [PbO (s)] = -217 kJ/mol
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Abraham_Otaku
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#2
Report 4 months ago
#2
Let's see... A method I know of is one that is pretty tedious.. but it works.
I got my final answer as 33kJmol^-1.. What you need to do is, First simply write out the enthalpy of formation of PbO as well as Pb(NO3)2 and the equation for enthalpy of formation of NO2.. you compare the three equations(from the given information) with the enthalpy of formation of NO2, and with that you add the equations-
2Pb(NO3)2 (s) ---> 4NO2 (g) + 2PbO (s) + O2 (g) ∆H = +602 kJ/mol
Pb(s) +N2(g) +3O2(g)--> Pb(NO3)2 ∆H= -452kJ/mol
Pb(s) + 0.5O2(g) ---> PbO(s) ∆H= -217kJ/mol
and all of these equations should add up(of course, you have to rearrange the equations as well as multiply the equations so that you can cancel out the PbO and Pb(NO3)2 in both sides of the equation, and don't forget to change the signs of the enthalpy change as you flip the equations!)and to finally form the equation( 2N2(g) +4O2(g)---> 4NO2(g) ∆H=132kJ/mol, and now using the definition of enthalpy of formation- the enthalpy change when ONE mole of a substance is formed from its elements in their standard states.. you multiply the entire equation which you found by- 1/4 which gives you 0.5N2(g) +O2(g)--> NO2(g) ∆H=33kJ/mol..
hope that helped! let me know if there's something that needs clarification(or if my final answer is wrong as I may have made a stupid mistake in the maths somewhere lol)
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kmj005
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#3
Report Thread starter 4 months ago
#3
Thanks so freakin much bro. You're a lifesaver
Last edited by kmj005; 4 months ago
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