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Learn to calculate temperature for spontaneous reactions using ΔG, ΔH, and ΔS values, understanding equilibrium and phase transitions.
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HReaction = ΣnH(products) - ΣnH(reactants) exothermic, ∆H = - points toward spontaneous endothermic, ∆H = + points toward non-spon. SReaction = ΣnS(products) - ΣnS(reactants) more disorder, ∆S = + points toward spontaneous more order, ∆S = - points toward non-spon. GReaction = ΣnG(products) - ΣnG(reactants) ∆G = - reaction IS spontaneous as written ∆G = + reaction IS non-spon. ∆G = 0 reaction is at equilibrium
ΔG = ΔH - TΔS ΔG = ΔH – TΔS = 0 @Equilibrium and.. This allows us to perform phase transition calculations (melting “fusion” and boiling “vaporization”.
Equilibrium R = 8.314 J∙mol-1∙K-1 T = Temperature in Kelvin Keq is the thermodynamic equilibrium constant. We will ignore activities. Gases will use partial pressure values in atmospheres and solutes will have concentrations expressed in molarity.
Assuming ∆S and ∆H do not vary with temperature, at what temperature will the reaction shown below become spontaneous? C(s) + H2O(g) → H2(g) + CO(s)∆S = 133.6 J ; ∆H = 131.3 kJ Use ΔG = ΔH – TΔS to solve for T. Since ΔS Points toward spontaneous (+ value) and ΔH points toward non-spontaneous (+ value), the temperature at which the term TΔS dominates is when the reaction becomes spontaneous.
Coupled Reactions: ∆G = ∆G1 + ∆G2 + ...... 2Fe2O3(s) → 4Fe(s) + 3O2(g) ∆Go = 1487 kJ 6CO(g) + 3O2(g) → 6CO2(g) ∆Go = -1543 kJ 2Fe2O3(s) + 6CO(g) → 4Fe(s) + 6CO2(g) ∆Go = -56 kJ