Question

For each of the following hydrocarbons, find i) the balanced chemical reaction equation when burned with oxygen, ii) the amount of energy released, in MJ, per kg of fuel, iii) the mass of CO2 released, in kg, per kg of fuel, and iv) the mass of CO2 produced per MJ of energy relased.

(a) Methane (CH4)

You will need the following enthalpies of formation: H◦ f,CO2 = −393.5 kJ/mol; H◦ f,H2O = −241.8 kJ/mol; H◦ f,CH4 = −74.9 kJ/mol; H◦ f,C3H8 = −103.9 kJ/mol; H◦ f,C16H34 = −456.0 kJ/mol. Note, water value assumes gaseous product.

I was looking for help on how to do parts iii and iv so that I may do the other fuels on my own. I already have parts i and ii. Thanks.

Answer 1

combustion of methane CH4+ 2O2--------->CO2+ 2H2O

Basis : 1 mole of methane, heat of reaction = sum of enthalpy of products-sum of enthalpy of reactants

=1* enthalpy of formation of CO2+2* enthalpy of formation of H2O- { 1* enthalpy of formation of CH4+2* enthalpy of formation of O2), 1,2, 1 and 2 are stoichiometric coefficients of CO2, H2O, CH4 and O2 respectively.

since heat of formation of oxygen =0

=-393.5+2*(-241.8)-{1*(-74.9)} = -802.2 KJ/mole

1 mole corresponds to 16 gm of CH4 ( molar mass of CH4= 16)

16 gm of fuel produces 802.2 KJ

1000 gm (1Kg produced 802.2*1000/16 Kj =50138 Kj =50138/10⁶ MJ=0.050138 MJ/kg

moles of CO2 produced per mole of CH4= 1, molar mass of CO2= 44, mass of CO2= 44 gm

16 gm of CH4 prodcues 44 gm of CO2. hence (44/16) kg of fuel is produced per kg of fuel =2.75 kg of CO2/ kg of fuel

1000 gm of CH4 produces 0.050138 MJ of energy and 1000*44/16 kg of CO2=2750 gm of CO2= 2.75 Kg of CO2

CO2 per /MJ= 2.75/0.050138=54.85 Kg/MJ

2. C3H8+ 5O2-------->3CO2+ 4H2O   Basis :1 mole of C3H8= 44 gm of C3H8

heat of reaction =3* heat of formation of CO2+4* heat of formation of H2O-{ 1* heat of formation of C3H8}

= 3*(-393.5)+4*(-241.8)-(-1*103.9)=-2044 Kj/ mole

44 gm of C3H8 gives 2044 Kj

1000 gm of C3H8 gives 1000*2044/44 =46455 Kj/ Kg= 0.465 MJ/kg

44 gm of C3H8 produces 3*44=132 gm of CO2

Kg of CO2/ kg of fuel= 132/44=3

44 gm of C3H8 produces 132 gm of CO2

1000 gm of C3H8 produces 132*1000/44= 3000 gm of CO2= 3 Kg of CO2

CO2 produced per MJ= 3/0.465 =6.45 MJ of CO2/ kg of fuel

3. C₁₆H₃₄+ 24.5O₂------>16CO₂+ 17H2O

Basis : 1 mole of C16H34,    Enthalpy change= 16* heat of formation of CO2+17* heat of formation of H2O- heat of formation of C16H34 = 16*(-393.5)+17*(-241.8) -(-456) = -5840 Kj/mole

mole of C16H34 = 16*12+34= 226 gm

226 gm of C16H34 gives -5840 Kj

1000 gm of C16H34 gives -5840*1000/226=25841 KJ =25841MJ/10⁶= 0.0258 MJ/kg

226 gm of C16H34 gives 16*44=704 gm of CO2, kg of CO2/ kg of C16H34= 704/226=3.115 kg CO2/ kg fuel

1000 gm of C16H34 gives 1000*704/226= 3115 gm of CO2= 3.115 kg of CO2

Kg of CO2/ MJ= 3.115//0.0258= 120.74 Kg/MJ