c) 1 kmol of liquid pentanol is burned completely in pure oxygen under stoichiometric conditions, with temperature and p

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C 1 Kmol Of Liquid Pentanol Is Burned Completely In Pure Oxygen Under Stoichiometric Conditions With Temperature And P 1 (92.12 KiB) Viewed 18 times

c) 1 kmol of liquid pentanol is burned completely in pure oxygen under stoichiometric conditions, with temperature and pressure maintained at 298 K and 105 Pa respectively. Calculate the quantity of energy available for external heating, assuming that the water condenses to liquid. Enthalpies of formation are available in Table Q2-1 and you may take the chemical equation for the combustion of pentanol to be: 2 C5H₁1OH +15 0₂ 10 CO₂ + 12 H₂O Enthalpy of Formation [kJ-kmol ¹] Substance Carbon Dioxide Water (liquid) -393,522 -285,838 0 Oxygen Pentanol (liquid) -351,900 Table Q2-1:Enthalpy of formation of relevant reactants and products [4 marks] d) If the pentanol was now burned in air under adiabatic conditions at 105Pa, show that the flame temperature lies between 2300 and 2400K. The following data in table Q2-2 shows the total enthalpy of the reaction products at these temperatures. Alternatively, a complete table can be found on page 8. Ideal gas Enthalpies h+hr/298 at 105 Pa (kJ kmol-¹) Temperature K CO CO₂ H₂ H₂O N₂ 2300 -42,853 63,371 -153,532 67,007 -283,851 -277,734 2400 -39,183 66,915 -148,223 70,651 Table Q2-2: The total enthalpy of gases at various temperatures. [6 marks]