2H2S (g) + SO2 (g) ?? 3S (s) + 2H2O (g) ?H Solution 2 H2S(g) + 3 O2(g) ---> 2 SO

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Explanation / Answer

2 H2S(g) + 3 O2(g) ---> 2 SO2(g) + 2 H2O(g) ?H°f H2O(g) = -241.83 kJ/mole ?H°f H2S(g) = -20.63 kJ/mole ?H°f SO2(g) = -296.84 kJ/mole ?H°f O2(g) = 0 kJ/mole that... ?Hf means heat of formation. The ° means at standard conditions. and anything in it's elemental state, like O2(g) and H2(g) etc.. is assumed to have a ?H°f = 0... ok? that's where that last line came from... now... (rxn is shorthand for reaction) ?H°rxn = ?H°f products - ?H°f reactants. ?H°rxn = [2x(-241.43) + 2x(-296.84)] - [2x(-20.63) + 3x(0)] kJ = -1035 kJ since ?H°rxn is negative, by convention, the reaction is EXOthermic. *************** now we find ourselves in a stoichiometry problem.. ie.. how much of this or that reacts with this or that to produce so much of this or that OR yield or % yield, etc. When you get one of these problems follow these 6 steps... MEMORIZE THEM 1) write a balanced equation 2) convert everything given to moles 3) determine limiting reagent 4) convert moles limiting reagent to moles other species 5) convert moles back to mass. this is theoretical mass.. aka theoretical yield 6) % yield = actual measured mass / theoretical mass x 100% ok? the idea being the coefficients of the balanced equation are in MOLE ratios. Once we convert everything to moles, we can use those coefficients to convert between different chemical species. *** 1 *** 2 H2S(g) + 3 O2(g) ---> 2 SO2(g) + 2 H2O(g) *** 2 *** moles H2S = 225.0g H2S x (1 mole / 34.082g) = 6.6017 moles moles O2 = 400.0g O2 x (1 mole / 31.999g) = 12.500 moles *** 3 *** The easiest way to do this is to pick a reactant. Either one will work. Take the moles we calculated in step 2, and determine how many moles of the OTHER reactant we actually need in accordance with the balanced equation. Then we compare that to the moles calculated in step 2... here. we go...I'll Pick O2 from step 2, we have 12.500 moles O2 from the balanced equation, 3 O2 reacts with 2 H2S... so we need this much H2S to completely consume all the O2... 12.500 moles O2 x (2 moles H2S / 3 moles O2) = 8.3333 moles H2S BUT.. we only have 6.6017 moles H2S available. So we don't have enough H2S to consume all the O2. So H2S is going to limit the reaction and we'll have XS (shorthand for eXcesS) O2. now I could have done it the other way to.. start with H2S.. like this.. 6.6017 moles H2S x (3 moles O2 / 2 moles H2S) = 9.9026 moles O2 since we have 12.500 moles O2 available, O2 is in XS, H2S is limiting. btw, at the end of the reaction we have 12.500 - 9.9026 = 2.5974 moles O2 remaining. *** 4 *** from the balanced equation, 2 H2S --> 2 SO2.. so that.. 6.6017 moles H2S x (2 moles SO2 / 2 moles H2S) = 6.6017 moles SO2 and 2 H2S --> 2 H2O.. so that.. 6.6017 moles H2S x (2 moles H2O / 2 moles H2S) = 6.6017 moles H2O *** 5 *** 6.6017 moles SO2 x (64.07 g / mole) = 423.0g SO2 *** 6 *** % yield = actual / theoretical x 100% = 350.0 g / 423.0g x 100% = 82.7% ******************* e is a bit complicated....but really.. not too bad... let's try it this way... 1) start with P1V1 / (n1T1) = P2V2 / (n2T2) 2) identify and cancel anything constant 3) rearrange and solve for you unknown.. only V is constant... P, T and n change... so... P1V1 / (n1T1) = P2V2 / (n2T2) becomes P1 / (n1T1) = P2 / (n2T2) rearranging... P2 = P1 x (T2 / T1) x (n2 / n1) where... P1 = 1 atm T1 = 273.15K n1 = 6.6017 + 12.500 = 19.1017 n2 = 2.5974 + 6.6017 + 6.6017 = 15.8008 P2 = ? T2 = ? now.. you notice that we don't know P2 nor T2.. BUT we do know how much heat was evolved and we do know T1...AND we know the process was constant volume. So here's what we'll do.. We'll look up Cv (constant volume heat capacities) for the 3 gases remaining after the reaction and calculate a J/K for our system.. then via J, convert that to T2... from here http://www.engineeringtoolbox.com/spesif… Cv O2 = 0.659 kJ/kgK x (1kg / 1000g) x (31.999g / mole) = 0.02109 kJ/moleK Cv SO2 = 0.51 kJ/kgK x (1kg / 1000g) x (64.07g / mole) = 0.03268 kJ/moleK Cv H2O = 1.46 kJ/kgK x (1kg / 1000g) x (18.02g / mole) = 0.02631 kJ/moleK heat cap of the products... 2.5975 moles x (0.02109 kJ/moleK) + 6.6017moles x (0.03268 kJ/moleK) + 6.6017moles x (0.02631 kJ/moleK) = 0.444 kJ/K now.. that -1035 kJ released is based on 2 moles H2S...since we started with 6.6017 moles H2S (which was our limiting reagent).. we should expect this much heat to be liberated. 6.6017 moles H2S x (-1035 kJ / 2 moles H2S) = 3416 kJ and then... ?T = 3416 kJ x (1 K / 0.444 kJ) = 7700°C T2 = 0°C + 7700°C = 7700°C P2 = 1 atm x (7973 / 273) x (15.80 / 19.10) = 24.2 atm ******** got all that? a) H2S b) 423.0g SO2 c) 82.7% d) exothermic e) T2 = 7700°C, P2 = 24.2 atm *********** why on earth "Roger the Chemist" would make these statements is beyond me. 1) Einstein would have trouble with this problem 2) an engineer would have trouble with this problem 3) chemistry is the worst taught subject. 4) this is an entire book of chemistry in 1 problem 5) burn characteristics of fuel in a piston/cylinder 6) engineers use supercomputers to solve this. Because parts a-d are simple. Part e is a bit more complicated but it isn't anything we didn't cover in general chemistry and thermodynamics in college. Roger should have seen problems like this before. Einstein wouldn't waste his time solving this. I'm an engineer and didn't find it too difficult. I was a TA at my school and am a bit offended when someone says chemistry isn't taught well. The students that paid attention and did the homework and kept regular office hours with me knew the subject matter. The idea in chemistry is you build a foundation and keep adding to it. So yes. this problem covered moles. balancing equations, stoichiometry. gas laws. heat capacity, etc. This problem didn't have anything to due with burn characteristics. That's a completely different ball of wax. Supercomputers? I used an excel spreadsheet and the internet. Could have done it with a CRC or Perry's and a calculator.

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