How to solve Enthalpy Problems

Could use some help on the following question. Last chemistry class we went quickly over Hess's law but we didn't really try any questions so I could use an example.

Caculate he standard enthalpy change in the reaction

PbO + CO = Pb + CO2

Given the standard enthalpies of formation of lead oxde, carbon monoxide and carbon dioxide are -219, -111 and -394 KJmol^-1 respectively.

isn't it just -64 KJ mol^-1?

Thats what I got.

Yeah that is right cause it is an exothermic reaction. You got it right

Thanks but what about my second question. Have I got the idea for that right?

Okay thankyou I see how that works your working out but what about this question.

Caculate the standard enthalpy of formation of propan-1-onal, CH3CH2CH20H given the standard enthalpies of combustion in KJmol^-1 are -2010 propan-1-ol, carbon -394 and hydrogen -286.

I imagine I would write the reaction so that one mole of propan-1-ol would be fomed. Would I then subtract the the number of carbon and hydrogens from -2010? Also what does the -2010 represent. its not combusting so it isn't realeasing any energy. Is that just the amont of energy it would give out and the energy level its at?

This question to me is weird. Because i think the reaction goes like this

CO2 + H20 -----> CH3CH2OH + O2.... but this isn't balanced, so i think we need to balance it first.

Caution : Balanced chemical equation with combustion will most often involve stoichiometric ratios higher than 1, so it is very difficult to just do the calculation without mistake(not that I am saying it is impossible)

Best approach(my opinion) : Draw up a Hess cycle for this. Write an eqn for formation of propan-1-ol, then down from the equation, come up with the balanced eqn for combustion. It shouldn't matter if you have 1 or n mol of propan-1-ol, but you just need to remember to divide your enthalpy change of combustion by this number of mol of propan-1-ol you have in your eqn, ie n mol; but if it is 1, then obviously it isn't going to make a difference.

I wouldn't do the calculation here for you, but you can try it your self, it will be very clear once you get the cycle right. One clue, enthalpy of combustion will always be exothermic, ie negative sign.

This question to me is weird. Because i think the reaction goes like this

CO2 + H20 -----> CH3CH2OH + O2.... but this isn't balanced, so i think we need to balance it first.

4CO2 + 5H20 -----> 2CH3CH2OH + 5.5O2

4*(-394) + 5*(-286) ---->-2010*2

i believe the answer may be -1014 Kj mol^-1....but i may be totally wrong.

How did you get the -1014 then?

Now I'm kind of confused. Just to help clear thins up what does the 2010 for the propan-1-ol represent? Its own energy level?

I get -316 kJmol-1 as the enthalpy of formation of propan-1-ol.

3C + 4H2 + 1/2 O2 ----> C3H7OH

3CO2 + 4H20

Imagine an arrow coming down from both LHS and RHS.

The LHS would be the combustion of C and H, remember to multiply by no of mol,

LHS = 3(-394) + 4(-286) = -2326

RHS = -2010(1) = -2010

Therefore enthalpy H(with both arrows coming down)
= -2326 - ( -2010)
= -316 kJ/mol