Many power plants produce energy by burning carbon-based fuels, which also produces CO2. CO2 is a greenhouse gas, so over-production can have negative effects on the environment.
Use enthalpy of formation data to calculate the number of moles of CO2 (g) produced per mega joule of heat released from the combustion of each fuel under standard conditions.
a) Coal c (s, graphite)
b) Natural gas, CH4 (g)
c) propane, C3H8 (g)
d) Octane, C8H18 (l) delta H f = -250 kJ Mol -1
a) 2.54
b) 1.12
c) 1.35
d) 1.46
First write out the combustion reaction.
Second calculate the energy released (delta Hf).
Third divide 1MJ = 1000kJ by the energy released in step 2 to obtain the number of complete reactions needed to release 1MJ.
Fourth, multiply by the coefficient of CO2, to get the moles of CO2 produced in total for all the reactions in step 3.
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Batman is correct.
Okay, First write out all of the balanced equations of combustion(A combustion equation is just adding O2 to the reactant, and ending with CO2 in the products. Notice, if H is in the reactants, you'll need H2O to balance the products:
Coal:
C+O2-->CO2
Natural Gas:
CH4+2O2-->CO2+2H2O
Propane:
C3H8+5O2-->3CO2+4H2O
Octane:
2C8H18+25O2-->16CO2+18H2O
Then find the energy released for both the products and the reactants. Since Octane was given to us in the question, I'll use that as an example for the equation:
[2C8H18(-250.1)+25O2(0)]-[16CO2(-393.5)+18H2O(-285.8)] = 10940.2 KJ
(assuming reaction is under standard conditions of 1 atm, and about 25 deg C.)
Now take the coefficient of CO2 that was formed in the product and divide it by the number of J found.
(16/10940.2) = 1.46*10^-3
Then multiply it by 1000 KJ to get it into MJ.
(1.46*10^-3)1000 = 1.46 mol/MJ
Batman is indeed correct.
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In response to lobu6058, the deltaHf value for liquid water is correct IF we're assuming standard conditions for this reaction. Specifically, that it is taking place at 25C. Because water is a liquid at 25C, the deltaHf value should be for liquid water. Not all combustion reactions produce gaseous H2O. If the reaction is taking place at 100+C, where water is a gas, then you would use the deltaHf value for gaseous H2O, if it is taking place between 0-99C, where water is a liquid, you use the deltaHf value for liquid H2O.
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Can someone write out the combustion reaction?
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I disagree with batman. Although his set up is correct the use of the KJ/mol is incorrect. In a combustion reaction all of the products will be in gas, not liquid. The number batman is giving is for liquid H2O, not gaseous H2O. SO if your answer is wrong this is why.
Should be:
a. 2.54 mol/MJ
b. 1.25 mol/MJ
c. 1.47 mol/MJ
d. 1.58 mol/MJ