Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on January 26, 2013, 04:05:03 PM
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Complete combustion of 25.0 L (30 C, 100.8 kPa) of a methane-ethane mixture (40% ethane by mass) in oxygen produces 1055 kJ of heat. Combustion of an equal amount of methane alone produces only 882 kJ of heat. Estimate the specific heat of combustion of higher alkanes (in kJ/mol).
As I have the answer and I did it similarly I will just copy the answer:
There was 1 mole of the mixture, n(ethane)=0.262 mol and n(methane)=0.738 mol.
From the heat released: 882·0.738 + х·0.262 = 1055, x=1542 kJ/mol (enthalpy of combustion of ethane).
Comparing the structure of ethane and methane we see that the heat of combustion of the CH2 unit is 1542-882=660 kJ/mol. The general formula of alkanes is CnH2n+2 or, equivalently, [CH4 + (n-1)CH2]. Thus, 1 mol of CnH2n+2 will produce 882 + (n-1)*660 = 222 + 660*n kJ of heat...
Until here everything is ok, but now I am confused:
...The molar mass of CnH2n+2 is (14n+2) g/mol. Combustion of 1 kg of an alkane produces (222+660n)/(14n+2) kJ of heat. As n increases indefinitely, the value of the fraction approaches 660n/14n=47. Thus, the specific heat of combustion of higher alkanes is 47 MJ/mol.
Isn't 222 + 660*n kJ/mol the answer? Why doing the extra calculations? Can someone explain why is 1kg used?
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Complete combustion of 25.0 L (30 C, 100.8 kPa) of a methane-ethane mixture (40% ethane by mass) in oxygen produces 1055 kJ of heat. Combustion of an equal amount of methane alone produces only 882 kJ of heat. Estimate the specific heat of combustion of higher alkanes (in kJ/mol).
As I have the answer and I did it similarly I will just copy the answer:
There was 1 mole of the mixture, n(ethane)=0.262 mol and n(methane)=0.738 mol.
From the heat released: 882·0.738 + х·0.262 = 1055, x=1542 kJ/mol (enthalpy of combustion of ethane).
Comparing the structure of ethane and methane we see that the heat of combustion of the CH2 unit is 1542-882=660 kJ/mol. The general formula of alkanes is CnH2n+2 or, equivalently, [CH4 + (n-1)CH2]. Thus, 1 mol of CnH2n+2 will produce 882 + (n-1)*660 = 222 + 660*n kJ of heat...
Until here everything is ok, but now I am confused:
...The molar mass of CnH2n+2 is (14n+2) g/mol. Combustion of 1 kg of an alkane produces (222+660n)/(14n+2) kJ of heat. As n increases indefinitely, the value of the fraction approaches 660n/14n=47. Thus, the specific heat of combustion of higher alkanes is 47 MJ/mol.
Isn't 222 + 660*n kJ/mol the answer? Why doing the extra calculations? Can someone explain why is 1kg used?
Why have you got n(ethane)=0.262 mol and n(methane)=0.738 mol? I agree if you can somehow be sure that 25 dm3 of this mixture is 1 mol but how do you know?
Does "equal amount of methane alone" mean 0.738 mol methane or 1 mol methane?
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"Complete combustion of 25.0 L (30 C, 100.8 kPa)"
From this you get the number of moles of the mixture (pV=nRT), n=1mol, if I mark the number of moles if methane with x and of ethane with y I can make a system of two linear equations:
x+y=1
30y/(30y+16x)=0.4
When solved, it is x=0.738 mol and y=0.262 mol.Does "equal amount of methane alone" mean 0.738 mol methane or 1 mol methane?
1 mol. I would like to return to my question now. Any idea why is 1kg used and why isn't 222 + 660*n kJ/mol the answer?
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"Complete combustion of 25.0 L (30 C, 100.8 kPa)"
From this you get the number of moles of the mixture (pV=nRT), n=1mol, if I mark the number of moles if methane with x and of ethane with y I can make a system of two linear equations:
x+y=1
30y/(30y+16x)=0.4
When solved, it is x=0.738 mol and y=0.262 mol.Does "equal amount of methane alone" mean 0.738 mol methane or 1 mol methane?
1 mol. I would like to return to my question now. Any idea why is 1kg used and why isn't 222 + 660*n kJ/mol the answer?
Ah I see! Sorry for that elementary oversight. It's rare to get P, V and T in my usual problems.
The way I see it 1g should be used - specific enthalpy pertains to the enthalpy change of combustion divided by the Mr. You thus would rightly end up with 47.14 kJmol-1. Using 1 kg goes against the normal definition of specific enthalpy.
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2 terms are included here: Heat of combustion and Specific heat capacity. What it then the Specific heat of combustion?
You said "specific enthalpy pertains to the enthalpy change of combustion divided by the Mr." and I thank you for this info as it sounds logic and correct to me.
Thus, the Specific heat of combustion is the amount of energy released when 1kg of an alkane gets combusted (as 1kg is the SI unit of mass).
1 mol of an alkane releases 222 + 660*n kJ of heat when combusted. The amount of heat released when 1kg gets combusted is: Q=ΔH*n=(222+660n)*1000/(14n+2) kJ/g or (222+660n)/(14n+2) kJ/g=MJ/kg. Then the answer for big n values is 47 MJ/kg and not MJ/mol as they wrote in the answer and asked in the problem, so there is the mistake. The units were wrong, kg≠mol. I hope its okay now.
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No, I am still wrong. The formula I used isn't correct (Q=ΔH*n). What formula should be used?
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I should reformulate my question. The posts before this one don't matter. In the problem I posted:
Complete combustion of 25.0 L (30 C, 100.8 kPa) of a methane-ethane mixture (40% ethane by mass) in oxygen produces 1055 kJ of heat. Combustion of an equal amount of methane alone produces only 882 kJ of heat. Estimate the specific heat of combustion of higher alkanes (in kJ/mol).
the mistake is, that the specific heat of combustion should be expressed in kJ/kg. The enthalpy of combustion of higher alkanes is 222 + 660*n kJ/mol. How to convert this to kJ/kg? If someone can answer, thanks, if not I will have to leave this, at first sight, interesting problem :'(.
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I should reformulate my question. The posts before this one don't matter. In the problem I posted:Complete combustion of 25.0 L (30 C, 100.8 kPa) of a methane-ethane mixture (40% ethane by mass) in oxygen produces 1055 kJ of heat. Combustion of an equal amount of methane alone produces only 882 kJ of heat. Estimate the specific heat of combustion of higher alkanes (in kJ/mol).
the mistake is, that the specific heat of combustion should be expressed in kJ/kg. The enthalpy of combustion of higher alkanes is 222 + 660*n kJ/mol. How to convert this to kJ/kg? If someone can answer, thanks, if not I will have to leave this, at first sight, interesting problem :'(.
Divide by MW and add in a factor of 1000?
[ kJ/gmol ] / [gm /gmol] * [gm /kg] = kJ / kg
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Combustion of 1 kg of an alkane produces (222+660n)/(14n+2) kJ of heat.
Not kJ.
Should be MJ I think.
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Yes, MJ it is. I found that mistake and wrote it few posts before, but I am interested in the formula for calculating the specific heat of combustion of higher alkanes.
"Divide by MW and add in a factor of 1000?"
Why doing this? Why not using 2kg=2000g (the units are still the same)?
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Why doing this? Why not using 2kg=2000g (the units are still the same)?
How does that let you convert a molar quantity to a per kg quantity.
If x is molar heat and M is MW and y is per kg heat what is the relation between x,y and M?
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y=x(J/mol)*1000(g)/M(g/mol), this would be then the formula Q=ΔH*n, but Q is expressed in J only, not in kg ???.
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y=x(J/mol)*1000(g)/M(g/mol), this would be then the formula Q=ΔH*n, but Q is expressed in J only, not in kg ???.
"Thus, 1 mol of CnH2n+2 will produce 882 + (n-1)*660 = 222 + 660*n kJ of heat..."
222 + 660*n has units of kJ/gmol
(14n+2) has units of gm/gmol
(222+660n)/(14n+2) has units of kJ/gm
(222+660n)/(14n+2)*1000 has units of kJ/kg
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PS. Of all the problems you posted this was a particularly elegant one I thought.
It's interesting that per mol heat does't tend to a an asymptotic constant value but per kg does. In hindsight it is obvious but I'd never analysed it like that.
Wonder if this can be generalized to every extensive quantity.
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I got some really really tough ones, but it all depends on the area people like. As I am preparing for competition I am doing all the areas (organic, inorganic, analytic, physical...). I asked some tough problems that are still unsolved after many months (I have only numerical answers on my papers).
What are gmol and gm? How did you get them? Shouldn't it be 222 + 660*n has units of kJ/mol?
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I got some really really tough ones, but it all depends on the area people like.
Tough ≠ Elegant :)
What are gmol and gm? How did you get them? Shouldn't it be 222 + 660*n has units of kJ/mol?
gmol = mol
gm = grams
kJ/mol is the same as kJ/gmol
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y=x(J/mol)*1000(g)/M(g/mol), this would be then the formula Q=ΔH*n, but Q is expressed in J only, not in kg ???.
"Thus, 1 mol of CnH2n+2 will produce 882 + (n-1)*660 = 222 + 660*n kJ of heat..."
222 + 660*n has units of kJ/gmol
(14n+2) has units of gm/gmol
(222+660n)/(14n+2) has units of kJ/gm
(222+660n)/(14n+2)*1000 has units of kJ/kg
Then it's not the formula Q=ΔH·n. It is Q=ΔH/M. While I don't completely understand how is this formula derived, I believe that it is concluded from the definition of the specific heat of combustion, which I am not sure of how it sounds. Thank you for helping me.
For me, tough=elegant 8).
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PS. Of all the problems you posted this was a particularly elegant one I thought.
It's interesting that per mol heat does't tend to a an asymptotic constant value but per kg does. In hindsight it is obvious but I'd never analysed it like that.
Wonder if this can be generalized to every extensive quantity.
Yes it is a lovely problem, what confuses me is converting from kJg-1 to kJkg-1, because this seems unnecessary and I rarely see answers in MJ anyway! On the whole it is a sweet problem though.
I got some really really tough ones, but it all depends on the area people like. As I am preparing for competition I am doing all the areas (organic, inorganic, analytic, physical...). I asked some tough problems that are still unsolved after many months (I have only numerical answers on my papers).
What are gmol and gm? How did you get them? Shouldn't it be 222 + 660*n has units of kJ/mol?
Which competition? And if you have lots of hard questions I would love to try them. Any type (provided they are high school-level, as my knowledge of analytical at undergrad level is shaky), as many as you can - I love these types of problems and would really appreciate it. :) If any are unanswered I would do my best to help you with them as well.
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Just go through my earlier posts, there are many problems. Unsolved, that I could find, are:
http://www.chemicalforums.com/index.php?topic=59059.0 which I don't think that can be solved.
http://www.chemicalforums.com/index.php?topic=55769.msg201714#msg201714 the first one.
http://www.chemicalforums.com/index.php?topic=65293.0 this one I solved actually.
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Just go through my earlier posts, there are many problems. Unsolved, that I could find, are:
http://www.chemicalforums.com/index.php?topic=59059.0 which I don't think that can be solved.
http://www.chemicalforums.com/index.php?topic=55769.msg201714#msg201714 the first one.
http://www.chemicalforums.com/index.php?topic=65293.0 this one I solved actually.
The first seems impossible (perhaps you can work towards a ratio but not an exact mass), the second I don't understand the wording of and the third seems fairly standard.
Where do you get these challenging questions, though?
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They are mostly from my country olympiad and these can't be found in English. I also posted few from other olympiads (IChO, Bellarus, Poland, USA...)