The Student Room Group

Hess' Law: Really Stuck!

Hi!

I understood this topic in class, now i just don't know how to start the questions!

Please could somebody explain to me how to do the following question:

Calculate the Standard Enthalpy Change for the decomposition of Lea Nitrate:

2Pb(NO3)2 ----> 2PbO + 4NO2 + O2

Given:

Pb(NO3)2: Std Enthalpy of Formation (-452 kJ/mole)

PbO: Std Enthalpy of Formation (-217 kJ/mole)

NO2: Std Enthalpy of Formation (33kJ/mole)


I know to construct equations for everything, which i've done: and i'm familiar in using the triangle rule/idea, but how do i start?!

Thanks very much for your help, rep points are available!

Cheers.
Reply 1
Remember Hess's Law states that the enthalpy of a reaction is independent of route.

Now write the equation of the decomposition.

Below, draw a box of elements: 2Pb(s) + N2(g) + 3O2(g). This links to both the LHS and RHS of the decomposition to show the formation of compounds.

Now to the LHS: -452 * 2 = -904 ...(1)
RHS: -217 * 2 + 33*4 = -302 ...(2)

Now since you want the enthalpy of the decomposition reaction (to the RHS), then

Enthalpy of decomposition = - (1) + (2) = 904 - 302 = 602 kJ/mol
Reply 2
^^To add to that, a positive value of enthalpy suggests that lead nitrate is thermodynamically stable wrt its decomposition products.
Reply 3
Thanks so much for your reply.

I'm still a little confused, however!

I think i've got the LHS sorted; but how does 2Pb(s) + N2(g) + 3O2(g) link in with the products of the decomposition of 2Pb(NO3)2?

Thank You!
Reply 4
Ah!!

Can anyone help me out further pleeeease if possible!!

I'm really stuck; i thought i had it, but i just don't understand it!

Thank you very much!
Reply 5
Well 2Pb(s) + N2(g) + 3O2(g) can be used to form the decomposition products too.
Reply 6
Right. I'm with you.

Please can i ask why you've multipyed each side of the triangle by what you have?

Cheers.
Reply 7
LHS: you are forming 2 moles of lead nitrate.

RHS: you are forming 2 moles of lead oxide and 4 moles of nitrogen dioxide; oxygen is just redundant, because it is already in its standard state.