(1) Benzene (C6H6) as a gas burns in air at 200% theoretical air. In answering t

Question

(1) Benzene (C6H6) as a gas burns in air at 200% theoretical air. In answering the following parts, you may assume the products of combustion are CO2, H O, O2, and N2. You may also assume the following (Total of 35 points): CPCO 0.0578 kJ/molK CpH2o= 0.0462 kJ/mol. CP,N2=0.0345 kJ/mol K Cp02= 0.0363 kJ/molK Cp,02 Table 2.2 Heat of formation A in kJ/mole (at 25 C and 1 atm)" (abstracted from Reference [2D Formula State hi(kJ/mole) Oxygen Nitrogen 0 0 Diamond 1.88 Carbon dioxide 393.5 -110.5 0 Carbon monoxide CO Hydrogen water- H2O 285.9 0. -92.3 +135.1 74.9 103.8 124.7 187.8 +82.9 115.9 -201.2 -238.6 -277.7 ater Clz Hydrogen chloride HCI Hydrogen cyanide HCN CH4 C3Hs n-Butane 10 n-Heptane Benzene CH4O CH40 Ethylene C,H Values for gaseous substances not in equilibrum at the standard state a) For products and reactants at 25°C, calculate the heat of combustion of benzene in b) For the reactants at 25°C, compute the adiabatic flame temperature in Kelvin of c) For the reactants at 25°C, compute the adiabatic flame temperature in Kelvin of have been determined from the liquid ard the heat of vaporization. kJ/mol (10 points). benzene burning in air at 200% theoretical air (10 points). benzene burning in pure oxygen. Assume 200% of the required stoichiometric oxygen is supplied (molar basis). Is this a realistic temperature? Explain your answer (10 points) d) Why is there a difference between b) and c) above? Give a physical argument (5 points)

Explanation / Answer

-6436.8The equation for combustion of Benzene can be written as 2C6H6(l)+15O2 (g)--->12CO2 (g)+6H2O (l)

Heat of combustion = enthalpy of products -enthalpy of reactants =12*-393.5+6*-285.9-{2*40.9+0)=-6519.2 Kj

Heat of combustion per mole of Benzene = -6519.2/2 =-3259.6 Kj/mol

b) Oxygen required for complete combustion from the reaction=15 moles , Air required= 15/0,21 ( Since air contains 79% Nitrogen and 21% Oxygen)=71.43 molea of air, Nitrogen =71.43-15= 56.43 moles

Actual air suppled = 200% of that theoretically required =2*71.43= 142.86,

Nitrogen = 142.86*0.79=112.86 moles, Oxygen =142.86-112.86=30 moles

Now the reaction becomes

2C6H6+ 30 O2 + 112.86 N2 ----> 12CO2+6H2O+15O2+112.86N2

Enthalpy of reactants = [-3259.6-(6*-393.5+3*-285.9)=-40.9 Kj/mol

Enthalpy of products

CO2 = 12*[-393.5+cp*(T-298)= 12*[-393.5+0.0578*(T-298)

T is adiabatic flame temperature of products and this occurs when the enthalpy of products= Enthalpy of reactants

H2O= 6*-[-241.8+0.0462*(T-298)

N2= 112.86*0.0345*(T-298)

O2= 15*0.0343*(T-298)

Equating the enthalpies of prodcuts= reactants

Standard heat of formation of Benzene= -40.9*2

+[T-298] {12*0.0578+6*0.0462+112.86*0.0340+15*0.0343}-6172.8= -2*40.9=-81.8

5.32*[T-298]= -81.8+6172.8 =6091

T-298 = 6091/5.32=1144.92 T= 1144.92+298=1442.9K

c) When combusted with 200% Oxygen, Oxygen supplied = 15*2=30 moles

The reaction becomes 2C6H6+3OO2-----> 12CO2+6H2O+15O2

Enthalpy of reactants= 2*-40.9=- 81.8

Enthalpy of products 12*-393.5+6*-241.8 +(T-298)[12*0.0578+6*0.0462+15*0.0343)= -6172.8+(T-298)*1.4853=

-81.8 = -6172.8+T-298)*1.4853

6091= (T-298)*1.4853

T-298= 6091/1.4853=4100.9

T=4100.9+298 =4398.9K

d) Nitrogen acts as an inert blancket resulting in reduction in adiabatic flame temperature,

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