Enthaly Change and Hess's Law

Theres a question that I am completely stuck on so would love some help :smile:

1/2 N2 + O2 --> NO2
Enthalpy of formation: +34kJmol-1

C + O2 --> CO2
Enthalpy of formation: -394kJmol-1

H2 + 1/2 O2 --> H2O
Enthalpy of formation: -242kJmol-1

C3H5N3O9 + 11/4 O2 --> 3 CO2 + 5/2 H2O + 3 NO2
Enthalpy of Combustion: -1540kJmol-1

a. Use the standard enthalpy changes given above to calculate the standard enthalpy of formation of nitroglycerine

Is it: 3(-394) + 5/2(-242) + 3(34) = -1685
-1685 - -1540 = -145kJmol-1 ??

b. Calculate the enthalpy change for the following decomposition of nitroglycerine.
C3H5N3O9 --> 3 CO2 + 5/2 H2O + 3/2 N2 1/4 O2

I have no idea how to do this...

c. Suggest one reason why the reaction in part (b) occurs rather than combustion when a bomb containing nitroglycerine explodes on impact

No idea about this one either...

Sorry, its a bit long... it just that this whole question has slightly confused me :tongue:

Edit: And just realised I missed off the p in enthalpy on the title... and eerr... dont know how to change it... ah well... :rolleyes:

(edited 13 years ago)

Original post by spoinkytheduck

Theres a question that I am completely stuck on so would love some help :smile:

1/2 N2 + O2 --> NO2
Enthalpy of formation: +34kJmol-1

C + O2 --> CO2
Enthalpy of formation: -394kJmol-1

H2 + 1/2 O2 --> H2O
Enthalpy of formation: -242kJmol-1

C3H5N3O9 + 11/4 O2 --> 3 CO2 + 5/2 H2O + 3 NO2
Enthalpy of Combustion: -1540kJmol-1

a. Use the standard enthalpy changes given above to calculate the standard enthalpy of formation of nitroglycerine

Is it: 3(-394) + 5/2(-242) + 3(34) = -1685
-1685 - -1540 = -145kJmol-1 ??

The enthalpy of formation:

3C + 5/2H₂ + 3/2N₂ + 9/2O₂ --> C₃H₅N₃O₉

If you combust all of the elements on the LHS you get the products of burning the RHS

But to go from the products of buring TO the RHS it's the REVERSE of combustion of the products (so you change the sign).

So, your formula is:

Combustion enthalpy(reactants) - combustion enthalpy (products) = enthalpy of reaction

3(-394) + 5/2(-242) + 3/2(34) + (+1540) = enthalpy of reaction

ΔHf(nitroglycerine) = -196 kJ

Original post by spoinkytheduck

b. Calculate the enthalpy change for the following decomposition of nitroglycerine.
C3H5N3O9 --> 3 CO2 + 5/2 H2O + 3/2 N2 1/4 O2

I have no idea how to do this...

You now have the enthalpy of formation of nitroglycerine (part a)

So, use Hess's law:

nitroglycerine ----> elements in their standard states ------> products

Original post by spoinkytheduck

c. Suggest one reason why the reaction in part (b) occurs rather than combustion when a bomb containing nitroglycerine explodes on impact

No idea about this one either...

Look at the relative energy values ...

Original post by spoinkytheduck

Sorry, its a bit long... it just that this whole question has slightly confused me :tongue:

Edit: And just realised I missed off the p in enthalpy on the title... and eerr... dont know how to change it... ah well... :rolleyes:

Check it out and have a go...

Reply 2

spoinkytheduck

OP

12

Original post by charco

The enthalpy of formation:

3C + 5/2H₂ + 3/2N₂ + 9/2O₂ --> C₃H₅N₃O₉

If you combust all of the elements on the LHS you get the products of burning the RHS

But to go from the products of buring TO the RHS it's the REVERSE of combustion of the products (so you change the sign).

So, your formula is:

Combustion enthalpy(reactants) - combustion enthalpy (products) = enthalpy of reaction

3(-394) + 5/2(-242) + 3/2(34) + (+1540) = enthalpy of reaction

?Hf(nitroglycerine) = -196 kJ

I'm with you right up until the plus sign... why is it not a minus?

Original post by charco

You now have the enthalpy of formation of nitroglycerine (part a)

So, use Hess's law:

nitroglycerine ----> elements in their standard states ------> products

So it would be:
Hf products - Hf reactants
3(-394) + 5/2 (-242) - (-196)
-1591kJmol-1 ??

Original post by charco

Look at the relative energy values...

I'm going to guess that there is less of an enthalpy change in the decompostition...??? which would make my answer to (c) wrong... :rolleyes:

Reply 3

Ari Ben Canaan

16

Original post by charco

So, your formula is:

Combustion enthalpy(reactants) - combustion enthalpy (products) = enthalpy of reaction

3(-394) + 5/2(-242) + 3/2(34) + (+1540) = enthalpy of reaction

?Hf(nitroglycerine) = -196 kJ

Why 3/2 ? Shouldnt it just be 3*34 ?

Our workings differ only in that sense. I get -145 KJ/mol.

(edited 13 years ago)

Reply 4

shengoc

12

Original post by Ari Ben Canaan

Why 3/2 ? Shouldnt it just be 3*34 ?

Our workings differ only in that sense. I get -145 KJ/mol.

If you look at charco working carefully, notice that the enthalpy of formation equation for NO2 involves only 1/2N2, so now he is using 3/2N2 for the consequent reaction, so he has to multiply by 3/2.

Reply 5

spoinkytheduck

OP

12

Original post by Ari Ben Canaan

Why 3/2 ? Shouldnt it just be 3*34 ?

Our workings differ only in that sense. I get -145 KJ/mol.

Thats what I got first time... so -145 is right?

Reply 6

shengoc

12

Original post by spoinkytheduck

I'm with you right up until the plus sign... why is it not a minus?

So it would be:
Hf products - Hf reactants
3(-394) + 5/2 (-242) - (-196)
-1591kJmol-1 ??

I'm going to guess that there is less of an enthalpy change in the decompostition...??? which would make my answer to (c) wrong... :rolleyes:

If you are trying to make nitroglycerine, your hess cycle would involve adding in the reverse arrow for combustion.

combustion is exo in forward sense, so when it is reversed, you get the endo reaction. you add the endo reaction based on hess cycle, and you'd end up with +(+combustion enthalpy given)

Reply 7

Ari Ben Canaan

16

Original post by shengoc

If you look at charco working carefully, notice that the enthalpy of formation equation for NO2 involves only 1/2N2, so now he is using 3/2N2 for the consequent reaction, so he has to multiply by 3/2.

Ahhhhhh. I see. Thanks :smile:

Reply 8

spoinkytheduck

OP

12

Uuumm.. now very confused...

So for the first one...
(3(-394) + 5/2 (-242) + 3/2 (34)) + 1540 = -196

And thats definately the right answer??

Reply 9

shengoc

12

Original post by spoinkytheduck

Uuumm.. now very confused...

So for the first one...
(3(-394) + 5/2 (-242) + 3/2 (34)) + 1540 = -196

And thats definately the right answer??

i am not doing the calculations, but the important thing to keep in mind in these longer type calculation question is to do the individual steps correctly, such that when you do the final summation, you'd end up with a much closer, if not correct answer.

if your steps are correct, your answer is most probably correct. even if they aren't, you'd be penalised on your final answer only, not the steps which you have done correctly

Original post by Ari Ben Canaan

Why 3/2 ? Shouldnt it just be 3*34 ?

Our workings differ only in that sense. I get -145 KJ/mol.

You are quite correct, my apologies, I didn't read the equation carefully...

1/2N₂ + O₂ --> NO₂

is equivalent to one nitrogen atom only so the equation for formaion of nitroglycerine involves 3 times this...

Hence:

3(-394) + 5/2 (-242) + 3(34) + 1540 = -145 kJ

Reply 11

Stuckonchemistry

Original post by spoinkytheduck

b. Calculate the enthalpy change for the following decomposition of nitroglycerine.
C3H5N3O9 --> 3 CO2 + 5/2 H2O + 3/2 N2 1/4 O2

I have no idea how to do this...

c. Suggest one reason why the reaction in part (b) occurs rather than combustion when a bomb containing nitroglycerine explodes on impact

Sorry but I've been reading your thread and I have to do this question myself for school but I still do not understand part b or c although a is completely fine! Please can someone give me some help? :smile:

Sorry if I'm being slow :colondollar:

Reply 12

Curtyn

3

mate, I got 145KJ for A and -145 for B using the energy cycle formation diagram

Original post by Stuckonchemistry

Sorry but I've been reading your thread and I have to do this question myself for school but I still do not understand part b or c although a is completely fine! Please can someone give me some help? :smile:

Sorry if I'm being slow :colondollar:

Original post by Curtyn

mate, I got 145KJ for A and -145 for B using the energy cycle formation diagram

Your "Mate" left the building 7 years ago...

Reply 14

burnt_t0ast

10

Original post by spoinkytheduck

Theres a question that I am completely stuck on so would love some help :smile:

1/2 N2 + O2 --> NO2
Enthalpy of formation: +34kJmol-1

C + O2 --> CO2
Enthalpy of formation: -394kJmol-1

H2 + 1/2 O2 --> H2O
Enthalpy of formation: -242kJmol-1

C3H5N3O9 + 11/4 O2 --> 3 CO2 + 5/2 H2O + 3 NO2
Enthalpy of Combustion: -1540kJmol-1

a. Use the standard enthalpy changes given above to calculate the standard enthalpy of formation of nitroglycerine

Is it: 3(-394) + 5/2(-242) + 3(34) = -1685
-1685 - -1540 = -145kJmol-1 ??

b. Calculate the enthalpy change for the following decomposition of nitroglycerine.
C3H5N3O9 --> 3 CO2 + 5/2 H2O + 3/2 N2 1/4 O2

I have no idea how to do this...

c. Suggest one reason why the reaction in part (b) occurs rather than combustion when a bomb containing nitroglycerine explodes on impact

No idea about this one either...

Sorry, its a bit long... it just that this whole question has slightly confused me :tongue:

Edit: And just realised I missed off the p in enthalpy on the title... and eerr... dont know how to change it... ah well... :rolleyes:

y'all don't understand how much help this is.
i have this EXACT question to do right now.
wow.
13 years, and still doing the same questions.
i wonder what everyone's doing now

Enthaly Change and Hess's Law

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