What is the LHV of n-pentane at standard condition in KJ/kgmole n-pentane? (n-pentane --gas)
What is the LHV of n-pentane at standard condition in KJ/kgmole n-pentane? (n-pentane --gas)
Homework Answers
The balanced combustion reaction of n-pentane
C5H12 (g) + 8O2(g) = 5CO2 (g) + 6H2O (l)
Enthalpy change for the reaction = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
Hc = 5*Hf(CO2) + 6*Hf(H2O) - 8*Hf(O2) - Hf(C5H12)
= 5*(-393520)+ 6*(-285820) - 8*0 - (-146760)
= - 1967600 - 1714920 + 146760
= - 3535760 kJ/kmol
Higher heating value of gaseous n-pentane = - (-3535760)
HHV = 3535760 kJ/kmol
LHV = HHV - moles of H2O * enthalpy of vaporization of H2O
= (3535760 kJ/kmol) - (6 kmol H2O / kmol C5H12) * (44010 kJ/kmol H2O)
= 3271700 kJ/kmol n-pentane
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What is the LHV of n-pentane at standard condition in KJ/kgmole
n-pentane? (n-pentane --gas)
n-Pentane is burned with excess air in a continuous combustion chamber. Below is a skeleton flowchart....
n-Pentane is burned with excess air in a continuous combustion chamber. Below is a skeleton flowchart. n (mol n-pentane) Combustion Chamber 100 mol Dry Product Gas (DPG) Xp (mol n-pentane / mol DPG) Xaq (mol O2/mol DPG) Xco, (mol CO2/mol DPG) XNC (mol N2 mol DPG) n (mol H2O) Excess Air n2 (mol O2) na (mol N) Air Composition Correct. 169.25 If n2 were 45.0 moles, what would be the value of n4? mol the tolerance is +/-2% Your answer...
The following information is given for n-pentane at 1 atm: boiling point-36.20 °C S melting point...
The following information is given for n-pentane at 1 atm: boiling point-36.20 °C S melting point -129.7 °C specific heat gas - 1.650 J/g°C specific heat liquid = 2.280 J/g C AH,p(36.20 °C) = 357.6Jg AH :(-129.7 °C) = 116.7J/g kJ of energy are A 40.50 g sample of liquid n-pentane is initially at -68.30 °C. If the sample is heated at constant pressure (P-1 atm), needed to raise the temperature of the sample to 48.10 °C.
1. Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane...
1. Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane isomerization reactions: n-pentane(g) i-pentane(g) + neopentane(g) Please note that there are only TWO independent reactions in this case and choose any two reactions with one initial amount, two unknown extents of Species AG%20% (kJ/mol) reaction, and the standard-state Gibbs free -pentane 40.17 energies of the species at 400 K to determine the i-pentane 34.31 equilibrium composition of the reactive mixture. neopentane 37.61
Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane isomerization...
Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane isomerization reactions among n-C5, 1-C5, and neo-Cs. To study the equilibrium composition, we can consider the following two reactions. We have also found a source that provides the standard-state Gibbs free energy of formation, AGYI, for the three isomers whose values in kJ/mol at 400 K are listed below. (1) n-C,(9) Hi-C (g) Species AG,400K (kJ/mol) (II) i-C, (g) neo-C (g) n-pentane 40.17 i-pentane 34.31...
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Assuming complete combustion, 500.g of pentane (C5H12) reacts with 1.00 kg of oxygen gas. 1. What...
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Assuming complete combustion, 500.g of pentane (C5H12) reacts with 1.00 kg of oxygen gas. 1. What...
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For a mixture of n-butane (1) + n-pentane(2) at 298.15 K, what would be the predicted...
For a mixture of n-butane (1) + n-pentane(2) at 298.15 K, what would be the predicted bubble-point pressure using Raoult's Law if your mixture was 20% n-butane by mole Would you expect Raoult's law to be a good model for this system Why or why not
n-Pentane is burned with excess air in a continuous combustion chamber. Below is a skeleton flowchart. n (mol n-pentane) Combustion Chamber 100 mol Dry Product Gas (DPG) Xp (mol n-pentane / mol DPG) Xaq (mol O2/mol DPG) Xco, (mol CO2/mol DPG) XNC (mol N2 mol DPG) n (mol H2O) Excess Air n2 (mol O2) na (mol N) Air Composition Correct. 169.25 If n2 were 45.0 moles, what would be the value of n4? mol the tolerance is +/-2% Your answer...
The following information is given for n-pentane at 1 atm: boiling point-36.20 °C S melting point -129.7 °C specific heat gas - 1.650 J/g°C specific heat liquid = 2.280 J/g C AH,p(36.20 °C) = 357.6Jg AH :(-129.7 °C) = 116.7J/g kJ of energy are A 40.50 g sample of liquid n-pentane is initially at -68.30 °C. If the sample is heated at constant pressure (P-1 atm), needed to raise the temperature of the sample to 48.10 °C.
1. Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane isomerization reactions: n-pentane(g) i-pentane(g) + neopentane(g) Please note that there are only TWO independent reactions in this case and choose any two reactions with one initial amount, two unknown extents of Species AG%20% (kJ/mol) reaction, and the standard-state Gibbs free -pentane 40.17 energies of the species at 400 K to determine the i-pentane 34.31 equilibrium composition of the reactive mixture. neopentane 37.61
Inside a catalytic reactor at 400 K and 1 bar, equilibrium is established for pentane isomerization reactions among n-C5, 1-C5, and neo-Cs. To study the equilibrium composition, we can consider the following two reactions. We have also found a source that provides the standard-state Gibbs free energy of formation, AGYI, for the three isomers whose values in kJ/mol at 400 K are listed below. (1) n-C,(9) Hi-C (g) Species AG,400K (kJ/mol) (II) i-C, (g) neo-C (g) n-pentane 40.17 i-pentane 34.31...
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For a mixture of n-butane (1) + n-pentane(2) at 298.15 K, what would be the predicted bubble-point pressure using Raoult's Law if your mixture was 20% n-butane by mole Would you expect Raoult's law to be a good model for this system Why or why not
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