: 2. Consider the combustion reaction of the normal butane gas: : C4H10(g) + 13/2 O2(g) →4 CO2(g) + 5 H2O(g) : With the help of the related thermodynamic data as listed below, aswer the : following questions: : (a) At 1.00 atm, calculate △H of this reaction at 298K and 1000K, : respectively. (5%) (i) Molar heat capacities at constant pressure (J/mol): O2(g)=29.5 N2(g)=29.1 H2O(l)=75.2 H2O(g)=24.8 CO2(g)=37.3 C4H10(g)=97.5 (ii) Standard enthalpies of formation (KJ/mol) at 298K: CO2(g)=-394 C4H10(g)=-125 H2O(l)=-286 H2O(g)=-242 C4H10(g) + 13/2 O2(g) →4 CO2(g) + 5 H2O(g) Hess Law 298k , △H = 5*(-242) + 4*(-394) - [ 13/2*(0) + 1*(-125) ] = -2661 KJ/mole 1000k , △H(1000K) = △H(298K) + Cp dT △H of C4H10(g) = -125 + 37.3(1000-298) △H of ....................................... 寫出四個 最後帶入 Hess Law 即可求得1000K的燃燒熱 : (b) Calculate the theoretical maximum temperature that could be reached by : igniting a mixture of butane and the air (O2/N2 mole ratio 1/4) under 1 atm : constant pressure (i.e. the operation condition of a butane stove used in : camping or in a picnic). The initial system temperature is assumed to be : 298K. (5%) 題目我看不太出來系統的量耶 不會算 : (c) Instead of the air, if pure oxygen was used in (b), what would be the : theoretical maximum temperature one could reach in the butane stove? (5%) : Related thermodynamic data: : (i) Molar heat capacities at constant pressure (J/mol): : O2(g)=29.5 N2(g)=29.1 H2O(l)=75.2 : H2O(g)=24.8 CO2(g)=37.3 C4H10(g)=97.5 : (ii) Standard enthalpies of formation (KJ/mol) at 298K: : CO2(g)=-394 C4H10(g)=-125 H2O(l)=-286 H2O(g)=-242 同上 --

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