What is that 4556.2 J? I see on the web delta Hc for methaol is 726 kJ/mol
If I calculate using your data I get
mass methanol used = 80.6 - 75.9 = 4.7 g and moles CH3OH = 4.7/32 = 0.147
q H2O = mass H2O x specific heat H2O x (Tfinal - Tinitial)
q = 100 g x 4.184 J/g*c x (32.4 - 21.5) = 4560 J for 0.147 moles CH3OH so for 1 mol we have 4560 x (1 mol/0.147) = 31024 J/mol or 31.024 kJ/mol.
A student investigated the enthalpy of combustion (deltaHc) of methanol under standard conditions using the apparatus shown in the diagram. The measurements the student recorded are shown in the table. Use this information to answer the questions below.
Alcohol mass before burning: 80.6g
Alcohol mass after burning: 75.9g
Water heated: 100g
Methanol Mr: 32
Initial temperature of water: 21.5 C
Final temperature of water: 32.4 C
Enthalpy of combustion of 1 mole of methanol: Q = 4556.2 J
The student wanted to know if the value obtained in part 1 is similar to that calculated using average bond enthalpy data.
a) Using the balanced equation and the data in the table below, calculate the theoretical enthalpy of combustion.
Note: you will need to include the enthalpy of vaporisation for the liquid components which are also given.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Average Bond Enthalpies KJmol-1
C-H 412
C-C 348
C-O 358
O=O 496
C=O 743
O-H 463
Enthalpy of vaporisation KJmol-1
Methanol 35
Water 41
b) Suggest some reasons as to why this value is different to the one obtained in the practical.
21 answers
4556.2 J is from my working out using the information and answering another question. I used Q=mcdeltaT. Part b is asking why this value I got using this equation is different to to the one obtained from the practical in part a. Part a has to be worked out first.
a) I think this is the way to find the average bond enthalpy:
3 x 412 = 1236 kJ
1.5 x 496 = 744 kJ
1 x 463 = 463 kJ
358 kJ
1486 kJ
Formed 2 x 743 = 1486 kJ
4 x 463 = 1852 kJ
I did not want to do anything else before checking this was right.
The note says to include the enthalpy of vaporisation for the liquid components which are also given. I did not. I do not know what to do.
3 x 412 = 1236 kJ
1.5 x 496 = 744 kJ
1 x 463 = 463 kJ
358 kJ
1486 kJ
Formed 2 x 743 = 1486 kJ
4 x 463 = 1852 kJ
I did not want to do anything else before checking this was right.
The note says to include the enthalpy of vaporisation for the liquid components which are also given. I did not. I do not know what to do.
a) I'm told to calculate theoretical enthalpy of combustion.
It says to include the enthalpy of vaporisation for the liquid components.
Why does it say enthalpy of combustion and then to include enthalpy of vaporisation?
It says to include the enthalpy of vaporisation for the liquid components.
Why does it say enthalpy of combustion and then to include enthalpy of vaporisation?
How I got the enthalpy of combustion of 1 mole of methanol:
80.6 – 75.9 = 4.7g
CH3OH: 12 + (3 x 1) + 16 + 1 = 32
75.9/32 = 2.371875 = 2.4 mole (1 d.p.)
Q = m c deltaT
Q = 100g x 4.18 J/g °C x (32.4°C – 21.5°C ) = 100g x 4.18 J/g °C x 10.9°C = 4556.2 J
80.6 – 75.9 = 4.7g
CH3OH: 12 + (3 x 1) + 16 + 1 = 32
75.9/32 = 2.371875 = 2.4 mole (1 d.p.)
Q = m c deltaT
Q = 100g x 4.18 J/g °C x (32.4°C – 21.5°C ) = 100g x 4.18 J/g °C x 10.9°C = 4556.2 J
I did the enthalpy of combustion of 1 mole of methanol calculation again doing it your way and got the same answer.
Why did you do this part:
1 mol we have 4560 x (1 mol/0.147) ?
Why did you do this to get 1 mol?
Why did you do this part:
1 mol we have 4560 x (1 mol/0.147) ?
Why did you do this to get 1 mol?
a) Can you help with what I did
b) Suggest some reasons as to why this value is different to the one obtained in the practical.
31024 J/mol or 31.024 kJ/mol
I do not know reasons for this. Can you help me with all the reasons.
b) Suggest some reasons as to why this value is different to the one obtained in the practical.
31024 J/mol or 31.024 kJ/mol
I do not know reasons for this. Can you help me with all the reasons.
I know I won't cover everything you asked but I can make a start.
1. The calculation for the dH combustion is not correct. See my calculation. Here is what I wrote above.
"If I calculate using your data I get
mass methanol used = 80.6 - 75.9 = 4.7 g and moles CH3OH = 4.7/32 = 0.147
q H2O = mass H2O x specific heat H2O x (Tfinal - Tinitial)
q = 100 g x 4.184 J/g*c x (32.4 - 21.5) = 4560 J for 0.147 moles CH3OH so for 1 mol we have 4560 x (1 mol/0.147) = 31024 J/mol or 31.024 kJ/mol."
I see you don't have the moles correct and you also incorrectly called the q value heat combustion/mol. Let me know it the calculation I show is confusing.
For the calculation of bond energies I agree partly with what you have done but not completely.First, here is the equation.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Yes, 4 x (C-H) = ?
Yes, 1 x OH = ?
Yes, 1.5 x O=O = ?
Sum of those three gives you dH reactants. Now for the products:
2 x C=O for CO2 = ?
4 x OH for 2H2O
Bond enthalpy calculations are good ONLY for gases. Since the H2O in the problem is specified as (liquid) you correct for that by enthalpy of vaporization by -41 kJ/mol or 2 x -41 = -82 kJ so you add those three together for dH products.
Then dHrxn = dHreactants - dH products = ?
(Why that negative sign for 2 x -41 kJ/mol? You know you must ADD energy to vaporize a liquid to a gas so you get energy back (i.e., -) when a gas condenses)
Finally, I wasn't present when you did the calorimeter work for the CH3OH so you need to think back through it to see what may have gone wrong. One thing I can think of is that CH3OH is rather volatile so it will evaporate while you're weighing it or just letting it stand around.
Let me know if I need to explain anything more. If, you may want to keep questions to 1 at a time. Also it helps, since this is getting long, to copy and paste that part where you havae questions and explain with that what your question is. Good luck.
1. The calculation for the dH combustion is not correct. See my calculation. Here is what I wrote above.
"If I calculate using your data I get
mass methanol used = 80.6 - 75.9 = 4.7 g and moles CH3OH = 4.7/32 = 0.147
q H2O = mass H2O x specific heat H2O x (Tfinal - Tinitial)
q = 100 g x 4.184 J/g*c x (32.4 - 21.5) = 4560 J for 0.147 moles CH3OH so for 1 mol we have 4560 x (1 mol/0.147) = 31024 J/mol or 31.024 kJ/mol."
I see you don't have the moles correct and you also incorrectly called the q value heat combustion/mol. Let me know it the calculation I show is confusing.
For the calculation of bond energies I agree partly with what you have done but not completely.First, here is the equation.
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Yes, 4 x (C-H) = ?
Yes, 1 x OH = ?
Yes, 1.5 x O=O = ?
Sum of those three gives you dH reactants. Now for the products:
2 x C=O for CO2 = ?
4 x OH for 2H2O
Bond enthalpy calculations are good ONLY for gases. Since the H2O in the problem is specified as (liquid) you correct for that by enthalpy of vaporization by -41 kJ/mol or 2 x -41 = -82 kJ so you add those three together for dH products.
Then dHrxn = dHreactants - dH products = ?
(Why that negative sign for 2 x -41 kJ/mol? You know you must ADD energy to vaporize a liquid to a gas so you get energy back (i.e., -) when a gas condenses)
Finally, I wasn't present when you did the calorimeter work for the CH3OH so you need to think back through it to see what may have gone wrong. One thing I can think of is that CH3OH is rather volatile so it will evaporate while you're weighing it or just letting it stand around.
Let me know if I need to explain anything more. If, you may want to keep questions to 1 at a time. Also it helps, since this is getting long, to copy and paste that part where you havae questions and explain with that what your question is. Good luck.
oops. Forgot. You also need to ADD the dHvap for methanol on the left for the reactants since that extra 35 kJ must be added before the CH3OH is vaporized.
3 x C-H = 3 x 412 = 1236 kJ
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
2443 kJ
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256 kJ
2443 - 3256 = -813
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
2443 kJ
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256 kJ
2443 - 3256 = -813
3 x C-H = 3 x 412 = 1236 kJ
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
35 kJ
2478 kJ
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256 kJ
2478 - 3256 = -778
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
35 kJ
2478 kJ
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256 kJ
2478 - 3256 = -778
Is the second post right?
(35 + 412 x 3 + 358 + 463 + 496 x 1.5) − (41 x 2 + 463 x 4 + 743 x 2) = -584 kJ/mol
I looked online and found that someone said this for this question for the enthalpy of combustion.
I looked online and found that someone said this for this question for the enthalpy of combustion.
_Clamp stand
_
_ _I__Thermometer
_ I I Beaker
_ I-----I
_ I___I Water
_ I Wick
_ -----
_ I I
_ ----- Alcohol
_ I __I
______________
_
_ _I__Thermometer
_ I I Beaker
_ I-----I
_ I___I Water
_ I Wick
_ -----
_ I I
_ ----- Alcohol
_ I __I
______________
I forgot to post this diagram. I did not do calorimeter work.
b) Suggest some reasons as to why this value is different to the one obtained in the practical.
b) Suggest some reasons as to why this value is different to the one obtained in the practical.
What are the reasons that the the theoretical and actual enthalpy changes are different (part b)
Not quite and it may be because my instructions were not clear. Here is what you asked about if it was correct. Before I post that let me try to write a good set of instructions. BE = bond energy.. rxn = reaction
CH3OH(l) + 1.5 O2(g) ==> CO2(g) + 2H2O(l)
dHrxn = (BEreactants) - BE products)
So adding up the bond energies of the reactants tells you how much energy it costs to break all those bonds. Adding up the bond energies of the products tells how much energy we get back when those bonds are formed. If dH is + it means the reaction is endothermic. If dH is - it means the reaction is exothermic. Now to your post.
"3 x C-H = 3 x 412 = 1236 kJ
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
35 kJ
2478 kJ
The reactants look OK to me but I didn't do the math.
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256
I would change that -41 to +41 and add all of those + numbers together. Each of those values tell you how much energy is released, including that 41 for the condensation of the H2Ovapor to liquid). Then all of them become negative when you subtract from the BEreactants.. That way you're keeping that dHrxn = BEreactants - BE products correct.
Next, you post something from the web and ask about it. It is not correct. Here are some reasone.
1. That 358 for C-O doesn't belong there. There are no C-O bonds in the equation. The rest of it looks ok. You will notice that has the +41 x 2 inside the parentheses(for BEproducts) with the negative sign just preceding the parentheses so the +82 eventually becomes a negative. .
2478 - 3256 = -778
Why is"
CH3OH(l) + 1.5 O2(g) ==> CO2(g) + 2H2O(l)
dHrxn = (BEreactants) - BE products)
So adding up the bond energies of the reactants tells you how much energy it costs to break all those bonds. Adding up the bond energies of the products tells how much energy we get back when those bonds are formed. If dH is + it means the reaction is endothermic. If dH is - it means the reaction is exothermic. Now to your post.
"3 x C-H = 3 x 412 = 1236 kJ
1.5 x O=O = 1.5 x 496 = 744 kJ
1 x O-H = 1 x 463 = 463 kJ
35 kJ
2478 kJ
The reactants look OK to me but I didn't do the math.
2 x C=O = 2 x 743 = 1486 kJ
4 x O-H = 4 x 463 = 1852 kJ
2 x -41 = -82 kJ
3256
I would change that -41 to +41 and add all of those + numbers together. Each of those values tell you how much energy is released, including that 41 for the condensation of the H2Ovapor to liquid). Then all of them become negative when you subtract from the BEreactants.. That way you're keeping that dHrxn = BEreactants - BE products correct.
Next, you post something from the web and ask about it. It is not correct. Here are some reasone.
1. That 358 for C-O doesn't belong there. There are no C-O bonds in the equation. The rest of it looks ok. You will notice that has the +41 x 2 inside the parentheses(for BEproducts) with the negative sign just preceding the parentheses so the +82 eventually becomes a negative. .
2478 - 3256 = -778
Why is"
Here is at least ONE BIG reason why these bond energy calculations won't add up to other calculations. The main reason is that these bond energies are not in stone; i.e., they are averages. That is the -OH bond energy in CH3OH is not the exact same as that of -OH in C2H5OH or C3H7OH etc etc. Same for the C-H bond energy in CH4, C2H6, C3H8 etc etc. or the C=C bond energy in various C=C compounds. So when you calculate dHrxn from Bond Energies you get an average of all those numbers. However, when you get dHrxn values from dH values either calculated or measured from calorimeter data you're getting individual values that have been measure and not averages. If I wanted a dHrxn I would want to do it with the good data. It amazes me that so many instructures use this Bond Energy thing when it is only good for a good guess. And frankly, at least in my opinion, I think using the tabulated values are easier to calculate AND you end up with values that are more likely to be correct.
H
I
H- C - O - H
I
H
I thought there is a C-O bond? Is that not used?
If it is ignored because we used the full equation average bond enthalpy then why in the products side when the water equation average bond enthalpy is used do we also use the O-H avergae bond enthalpy information?
I
H- C - O - H
I
H
I thought there is a C-O bond? Is that not used?
If it is ignored because we used the full equation average bond enthalpy then why in the products side when the water equation average bond enthalpy is used do we also use the O-H avergae bond enthalpy information?
The equation posted wrong.
How does the diagram come into the question? Is there anything I can say about it for part b (why the value is different to the one obtained in the practical)?
How does the diagram come into the question? Is there anything I can say about it for part b (why the value is different to the one obtained in the practical)?
"If I wanted a dHrxn I would want to do it with the good data. It amazes me that so many instructures use this Bond Energy thing when it is only good for a good guess. And frankly, at least in my opinion, I think using the tabulated values are easier to calculate AND you end up with values that are more likely to be correct."
Is the good data and the tabulated values that are easier to calculate the ones from the practical?
Is the good data and the tabulated values that are easier to calculate the ones from the practical?