[Unknown?]

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.

[shengoc]

(Original post by Unknown?)
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.

Hess law works because enthalpy change is independent of the route(path) which the reaction takes to happen.

Given standard enthalpy of formation, enthalpy of reaction = total enthalpy of formation of (products - reactants)

Remember to multiply out by no.of mol as shown by your balanced equation.

In your case, all mole ratios are 1, so it is fine, it would be,
-394 - (-219 -111) = ?

[Unknown?]

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?

[boromir9111]

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

[Unknown?]

(Original post by boromir9111)
isn't it just -64 KJ mol^-1?

Thats what I got.

[boromir9111]

(Original post by Unknown?)
Thats what I got.

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

[Unknown?]

(Original post by boromir9111)
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?

[boromir9111]

(Original post by Unknown?)
Thanks but what about my second question. Have I got the idea for that right?

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.

[shengoc]

(Original post by Unknown?)
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?

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.

[shengoc]

(Original post by boromir9111)
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.

You would need a balanced eqn indeed for hess cycle to work.

[Unknown?]

(Original post by shengoc)
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.

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?

[Unknown?]

(Original post by boromir9111)
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?

[boromir9111]

(Original post by Unknown?)
How did you get the -1014 then?

I pretty sure i am wrong lol but my way was just basically knowing that it's combustion reaction, therefore CH3CH2OH has to react with oxygen to form carbon dioxide and water. Write that out, balance it out which i don't think i've done right tbh. Use the values you are given but in my case since there are 4 carbons, multiply the value by 4 and 5 hydrogens so do the same for that. Then two multiplied by the value for CH3CH2OH and from there just addition i did as shown in my previous post.

[shengoc]

(Original post by Unknown?)
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?

The -2010 kJ/mol is the enthalpy of combustion for complete combustion for 1 mol of propan-1-ol. The unit is important; it indicates this!

As for my answer to this question, I will post in a few minutes, and see if you could get the same as mine.

[shengoc]

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

[boromir9111]

(Original post by shengoc)
I get -316 kJmol-1 as the enthalpy of formation of propan-1-ol.

That seems better, how did you do that?

[shengoc]

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

[boromir9111]

(Original post by shengoc)
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

Ahh yes of course, once i immediately saw combustion i just did that reaction but this definitely makes more sense!!! cheers for that mate.