Thermochemistry Synthlet Select Reaction Acetaldehyde reacting with water to give acetic acid and ethanolAcetylene decomposing to its elementsCarbon monnoxide plus water gives hydrogen plus carbon dioxide equilibriumCarbon monoxide + hydrogen gives methane + water equilibriumCarbon plus carbon dioxide gives carbon monoxide equilibriumCarbon plus water shift reactionCombustion of acetoneCombustion of butane to carbon dioxide and steamCombustion of methane to carbon dioxide and steamCracking butane to ethylene and hydrogenDecomposing butane to its elementsDecomposing steam into its elementsDecomposition of calcium carbonate to CaO and CO2Decomposition of ozone to oxygenDimerisation of NO2 to N2O4Dimerising methane to acetylene and hydrogenDioxygen (O2) ozone (O3) equilibriumFormation of methane from its elementsHaber ammonia synthesisHydrogen + iodine in equilibrium with hydrogen iodideHydrogen radical plus oxygen gives hydrogen peroxide radicalHydrogenation of 1-buteneHydrogenation of benzeneHydrogenation to ethylene to ethaneHydrolysis of carbon disulfideOxidation of ammonia to nitric oxideOxidation of ammonia to nitric oxide with hydrogen peroxideOxidation of ammonia to nitrogen and steamOxidation of ammonia to nitrogen with hydrogen peroxideOxidation of nitrogen to nitric oxidePartial combustion of carbon to carbon monoxidePartial oxidation of methane to acetylene, H2 and CORadical coupling of methane radicals to ethaneReduction of iron(III) oxide with hydrogenThe Contact processWater-steam equilibrium      0K 200K 298K 500K 1000K 3000K 0°C 25°C 100°C 500°C 1000°C 2500°C    or    Kelvin

Haber ammonia synthesis            +       3 2

Nitrogen + Hydrogen   Reactants   Ammonia   Products   Products – Reactants ΔfH   ΔfH   Σf-reactH   ΔfH       Σf-prodH   Σf-prodH - Σf-reactH = ΔrH (0.0 x 1) + (0.0 x 3) = 0.00   (-46.1 x 2) = -92.2   -92.2 - 0.0 = -92.2 kJmol-1                               S   S   ΣreactS   S       ΣprodS   ΣprodS - ΣreactS = ΔrS (0.192 x 1) + (0.131 x 3) = 0.584   (0.192 x 2) = 0.385   0.385 - 0.584 = -0.199 kJK-1mol-1

Gibbs free energy at 298K (standard conditions and standard temperature):   ΔrG  = ΔrH – (T298 x ΔrS)   =   -92.2 – (298 x -0.199)  = -33.0 kJmol-1

ΔrH = -92.2 kJmol-1 Negative value: the reaction is exothermic ΔrS = -0.199 kJK-1mol-1 Negative value: the system becomes less dispersed ΔrG 298K = -33.0 kJmol-1 Negative value: the reaction is feasible (spontaneous) Keq 298K = 6.07 x 105 Value is greater than 1: the reaction proceeds to the right ΔG° = 0.0, and the system is at equilibrium, at a temperature of 464K (191°C)

© Mark R. Leach 1999-

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