Question

Consider an internal combustion engine like that used in automobiles. The most common fuel used here is gasoline which is a mixture of hydrocarbons, and an octane rating is given for standardization. Therefore, we will approximate gasoline as being composed of the hydrocarbon, octane, C8H18. Another fuel that is common in racing applications is nitromethane, CH3NO2, used in top fuel dragsters. Let’s consider the combustion of each fuel with oxygen in an internal combustion engine. Let the 8-cylinder engine displacement be 500 in3 total. GOAL: Determine and compare the maximum enthalpy of reaction, ∆Ho rxn, for the combustion of both octane and nitromethane in a single cylinder of the engine. This means determine the maximum amount of fuel that can burn in the cylinder filled with 1 atm of air and the amount of heat that reaction would generate. Here are some suggested steps to guide you along. You will have to look up some data in your book or on the internet. 1. Write a balanced chemical equation for the reaction of octane and nitromethane with O2 gas, assuming complete combustion. For octane, the products would be CO2 and H2O. For nitromethane, the products would be CO2, H2O, and N2. 2. Determine ∆Hrxn, at 298 K, for both in kJ mol fuel. Which is greater? 3. Determine the amount of O2 gas available for the reactions in a single cylinder 500 8 in3 . Assume T = 298 K. 4. Determine how much fuel can be reacted with the amount of O2 in the cylinder. 5. Determine the heat released, ∆Ho at 298 K, for burning that much fuel for both octane and nitromethane. Answers: 1. ∆Ho for burning maximum amount of octane. 2. ∆Ho for burning maximum amount of nitromethane. 3. Which fuel gives the greatest heat output in the engine?

Answer 1

Geven that, A comparison of Engine Gardene (octane) performance geor and Top Fuel (nitromethane). Goal :- Determine and compare the man mum entholopy of Deactions. Ah for the combution of both Octase and netromethane in a single eylender of the engine. 1) Balanced chemical equate on for the reaction of Octane ą netsomethane with Oz gas. · combustion equation yol octane (qHq (e) ? 2/ Ce Hig (e) +251202(9) ► 8602(9),60+ 9 H, 0(e)) 269 H, (C) +2502 (9) 16602 (3) +18 #20(e) _0 Balanced equation. combustion equation for Nitromethane Ct, No, ta) 22 C#gwoq (e)+{ 07 (9) ► 200, (9) + 3440 (e) + N2 (9)] 4 CH3 100q (e) + 302 (9) ► 4602(9) + 6H20 (0)+2x02 (9) Balanced equatione

combustion equation correspond to I Hole of the compound the combuskon equation are Cg Eng (e) + 2 sl, O2 (9) > 80, (3) + 4 Hyo (e) x CHgnoos + 02 (9) → CO2 (9) + 3HO (e) +4N2 (9) wow deviding question traghout by H. 2) Determine A Har xn Altran = EAHU( Products) - EHP ( Reactarts) AHI [Coq (9)] = - 394kJ/mol AHF [H20 (e)] = -286 ks/mol. Ahtº (Celing (e)] = -252.1 to - 248.5 kg/mol on an average we take ~ 280 kJ/mol AH tº [Chy NO2 Ce)] = - 11365lmo AHfº = standard enthalaply are taken as zero. combustion of Octane. Cottag(e) + 25 02(g)_3 8002 (9) + 2 H2O (e) -200 kolmol o . 86-894) kJ/mol i (-286 kg/mol). Ausreaction = 8 (-394) Kylmond +9 (-286) Kylmol] - (-25okelmel All reaction =- 5476 kJ/mol

3) Determine the amount of Oz at T = 298K. volume of 1 cylinder in e volume of B Ejenders = 351995, ndeme of . I. cylender = 35 ms. = 43.9 m3 1m3 {imbec inch) = 0,01642 1. 43.9 m3 = 43.9 x 0.0162 = 0.72L i PV=nRT ne PV RT Palatms V= 0.722 R=0.082 (Atmosl kmol) of T=298k 0 - 2 S xo. 22 0.82 [atus & mol X2983 in = 0.0295 mol The amount of Op avallable at T = 298 K is 0.0295 mol combustion of nitramethane CH3 No.2(e). CHy nog (e) + 02 (9) 5 CO2(g) + ? H20 (e) + (9) -39440)nd 3 (-298) K5 Imol. A fiscations f-394 Kylmal + (-286)*9|mol]-[-113 kJ/molj

:. All reaction =-71065/mol in the Entholopy of combustion Egitiple) is (-5476 kolma] is greater than that of CHg No 2(e) [rzioks/mo]. 4) Determine how much fuel can be reached with the amount of Og in the eylender. Cs vale) + 95 oz (3) do, (4) + Hạo (e) I mol 25 mol 0.0295 mol 25 mol O2(g) reacts with it mol Is Hig(e) 1:0.0295 mol O2(g) seacts with Son X 0.0295 mol of Co Hag (e). = 25 20.0295 = 0,004% mol of Cf Hig(e). 25 Eltę No(e) + 3 02 (9) --> 002(9) +? Hbo (e) +1_802 (9) Samole oz (g) reacts with 1 mol CH3 No.2 (C) $X4 X 0.0295 mol of CHg No.2(e) .= 0.0393 mol of CH3 NO2 --

5) Determine the Heat released At at T = 298K For octane and witromethane For Octare : Entholopy of combustion of actare Cg Hig(e) is - 5476K5|nd 1. 1 mol (e) will release 5446 kJ theat energy. 1. 0.0047 nd Co Hig(e) will release (0.047 *5446) kJ heat evergy is 25.43 K5 beat energy. for Nitro methane :- Enthopy of combustion of CH3 NO2 (e) is 70k5|md Ind CH3 NO2 (e) Loill release F10 ks I heat energy :. 0.0393 nel Elt3 NO2 (e) will release (0.0393x410) KJ breat energy. i Heat energy is 27.90 kJ.