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DraconicAcid
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Quote: Originally posted by KonkreteRocketry  | Ok cool,
I cant find any enthalpy of Cl = O bonds, so then i found Chlorine dioxide on wikipedia which is O=Cl=O says the enthalpy is 104.60, so can i assume
Cl = O is half of 104.60 ? |
No, that's the enthalpy of formation, not a bond enthalpy.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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KonkreteRocketry
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but if i know they got the enthalpy from Cl2 + O2 i can still work out the C=O energy right ?
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DraconicAcid
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You can work it out. Show your calculations, and I'll tell you if you're on the right track.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Ral123
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Don't work it out. Use wikipedia, youtube and people's opinion about propellant performance.
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Dornier 335A
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You don't have to care about the enthalpy of single bonds inside the molecule. The enthalpy of formation is easy to find for many compounds.
Neither is all the steps in the reaction relevant. All you have to know is the reactants' ΔHf and the products' ΔHf.
The enthalpy change of reaction is the sum of ΔH for the products minus the sum of ΔH for the reactant(s).
The decomposition of NH4ClO4
2 NH4ClO4(s) → Cl2(g) + N2(g) + 2 O2(g) + 4 H2O(g)
(1*0 + 1*0 + 2*0 + 4*-241.8)-(2*-295.3)
The ΔH of this reaction is -380.6 kJ/mol so it is exothermic. And yes, it is the water produced in the reaction that liberates this energy.
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