you're on the right track ... n is the total number of moles
use the percentages to find the moles of each gas
... they have different heat contents
A particular natural gas consists, in mole percents, of 83.0% CH4, 11.2% C2H6, and 5.8% C3H8. A 385L sample of this gas is measured at 25C and 729mmHg, and is burned in an excess of oxygen. How much heat in kJ is involved in this reaction.
I assume you use n=PV/RT to get moles but i don't know how the mole percents come into play. After using the percents in some way do you then add the heat of each component?
6 answers
When you find moles total do you then multiply by each percent by n to get the moles of each gas?
This seems like a multi-step problem.
You will use the percentages to find the number of moles for each hydrocarbon. But first you should write out the combustion reaction for each hydrocarbon. For example, for C3H8:
C3H8 + O2 --------> CO2 + H2O
C3H8 + 5O2 --------> 3CO2 + 4H2O
After you calculate how many moles total using PV=nRT, you will use the percentages to find the number of moles for C3H8:
PV=nRT
n=PV/RT
Let % moles of C3H8 =y
0.058*n=y
Since the above reaction shows that 1 mole of C3H8=3 moles of CO2
y*3= moles of CO2 produced from C3H8
Use Standard Enthalpies of Formation table in your text to find the enthalpy for CO2
Multiply that by the answer above to get the amount of heat involved from this particular hydrocarbon in kJ. Repeat for each hydrocarbon in the vessel.
You will use the percentages to find the number of moles for each hydrocarbon. But first you should write out the combustion reaction for each hydrocarbon. For example, for C3H8:
C3H8 + O2 --------> CO2 + H2O
C3H8 + 5O2 --------> 3CO2 + 4H2O
After you calculate how many moles total using PV=nRT, you will use the percentages to find the number of moles for C3H8:
PV=nRT
n=PV/RT
Let % moles of C3H8 =y
0.058*n=y
Since the above reaction shows that 1 mole of C3H8=3 moles of CO2
y*3= moles of CO2 produced from C3H8
Use Standard Enthalpies of Formation table in your text to find the enthalpy for CO2
Multiply that by the answer above to get the amount of heat involved from this particular hydrocarbon in kJ. Repeat for each hydrocarbon in the vessel.
Mole percents are also mole fractions. That is, 83%CH4 is 0.84mole fraction of the gas mixture. The partial pressure contribution of methane can be calculated as 0.84(729-mmHg) = 605.07. Ethane and Propane the same way give 81.65-mmHg C2H6 and 42.28-mmHg C3H8.
From this use the ideal gas law PV = nRT and calculate moles of each compound.
P = Partial pressure of each gas (calculated above)
V = Total Volume given = 385-L
T = Temperature given = 25C = 298K
R = Gas Constant = 0.08206 L-Atm/mol-K
Once you have moles of each, search for heats of combustion (Kj/mole) for each compound. ( I found CH4 => -890.3 Kj/mole, C2H6 => -2637.45 Kj/mole & C3H8 => *1943.91 Kj/mole)
Calculate heat contribution for each = moles x heat of combustion, then add the results.
I got 15,740-Kj => 3 Sig figs => 1.57E4-Kj
From this use the ideal gas law PV = nRT and calculate moles of each compound.
P = Partial pressure of each gas (calculated above)
V = Total Volume given = 385-L
T = Temperature given = 25C = 298K
R = Gas Constant = 0.08206 L-Atm/mol-K
Once you have moles of each, search for heats of combustion (Kj/mole) for each compound. ( I found CH4 => -890.3 Kj/mole, C2H6 => -2637.45 Kj/mole & C3H8 => *1943.91 Kj/mole)
Calculate heat contribution for each = moles x heat of combustion, then add the results.
I got 15,740-Kj => 3 Sig figs => 1.57E4-Kj
The last entry above is fm Doc48
I have been out of practice using these concept-- it has been years since I have used them. So, I apologize. I said this seems like a multistep problem in my original post because I wasn't 100% sure. After looking at the post by Doc 48, I see that I made an error. Look for *** for the additions/corrections to the problem. There is a little bit more work to do.
You will use the percentages to find the number of moles for each hydrocarbon. But first you should write out the combustion reaction for each hydrocarbon. For example, for C3H8:
C3H8 + O2 --------> CO2 + H2O
C3H8 + 5O2 --------> 3CO2 + 4H2O
After you calculate how many moles total using PV=nRT, you will use the percentages to find the number of moles for C3H8:
PV=nRT
n=PV/RT
Let % moles of C3H8 =y
0.058*n=y
*** Use the Standard Enthalpies of Formation table in your text to find the enthalpy for EACH COMPONENT OF THE REACTION and use the following relationship to calculate the heat of combustion:
***ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]
***ΔH(C3H8) = 2*ΔHf(CO2) + 4*ΔHf(H2O) - ΔHf(C3H8) – 5*ΔHf(O2)
*** Note: the numbers in the equation represent the coefficients for the balanced reaction for the hydrocarbon in question.
***Multiply ΔHc(C3H8) by our 'y' answer to get the amount of heat released from this particular hydrocarbon in kJ. Repeat for each hydrocarbon in the vessel.
***I do not believe that the answer that Doc48 submitted for the problem will return the correct answer. However, if he did not submit an answer for this particular question, I would not have caught my error.
You will use the percentages to find the number of moles for each hydrocarbon. But first you should write out the combustion reaction for each hydrocarbon. For example, for C3H8:
C3H8 + O2 --------> CO2 + H2O
C3H8 + 5O2 --------> 3CO2 + 4H2O
After you calculate how many moles total using PV=nRT, you will use the percentages to find the number of moles for C3H8:
PV=nRT
n=PV/RT
Let % moles of C3H8 =y
0.058*n=y
*** Use the Standard Enthalpies of Formation table in your text to find the enthalpy for EACH COMPONENT OF THE REACTION and use the following relationship to calculate the heat of combustion:
***ΔHr = Σ ΔHf[products] - Σ ΔHf[reactants]
***ΔH(C3H8) = 2*ΔHf(CO2) + 4*ΔHf(H2O) - ΔHf(C3H8) – 5*ΔHf(O2)
*** Note: the numbers in the equation represent the coefficients for the balanced reaction for the hydrocarbon in question.
***Multiply ΔHc(C3H8) by our 'y' answer to get the amount of heat released from this particular hydrocarbon in kJ. Repeat for each hydrocarbon in the vessel.
***I do not believe that the answer that Doc48 submitted for the problem will return the correct answer. However, if he did not submit an answer for this particular question, I would not have caught my error.