[{"data":1,"prerenderedAt":109},["ShallowReactive",2],{"cheatsheet-entropy-and-free-energy":3},{"sheet":4},{"sheetSlug":5,"topicSlugs":6,"primaryTopic":7,"title":8,"subtitle":9,"sections":10},"entropy-and-free-energy",[5],25,"Entropy & Free Energy","Entropy trends, ΔS°, ΔG°, spontaneity, the free-energy/K link, and crossover temperature.",[11,23,36,44,52,60,68,76,85,93,101],{"heading":12,"kind":13,"items":14},"Entropy trends","rules",[15],{"ref":16,"mode":19,"provenance":20,"text":21,"source_hash":22},{"type":17,"topic":7,"id":18,"field":17},"rule","TD-R001","transform","owned_workbook","More freedom → more entropy: S(gas) > S(liquid) > S(solid). Melting and vaporizing raise entropy.","541384a5f0fe9a451f9e3d00bd861d83f1a077c59605438c859ff31b0f99953a",{"heading":24,"kind":13,"items":25},"Predicting the sign of ΔS",[26,31],{"ref":27,"mode":19,"provenance":20,"text":29,"source_hash":30},{"type":17,"topic":7,"id":28,"field":17},"TD-R002","Compare moles of gas on each side: more gas moles = higher entropy.","a7a7a843da34931e7ac6d9285ce75a9eb59e6589401ee343964f900bffe2a5f8",{"ref":32,"mode":19,"provenance":20,"text":34,"source_hash":35},{"type":17,"topic":7,"id":33,"field":17},"TD-R003","Dissolving a solid into ions raises entropy; precipitating lowers it.","adc464831edf1e1b01c30d80febbe30e927c408920a514156ccbbb2203237c60",{"heading":37,"kind":13,"items":38},"ΔS° from standard entropies",[39],{"ref":40,"mode":19,"provenance":20,"text":42,"source_hash":43},{"type":17,"topic":7,"id":41,"field":17},"TD-R004","ΔS° = ΣS°(products) − ΣS°(reactants), each S° × its coefficient. (Standard molar entropies are tabulated and nonzero, even for elements.)","5c1fdc0d7d74c3185dfebe0416149a0b407c6508186b3fe1777279e36189b870",{"heading":45,"kind":13,"items":46},"ΔG° from formation values",[47],{"ref":48,"mode":19,"provenance":20,"text":50,"source_hash":51},{"type":17,"topic":7,"id":49,"field":17},"TD-R006","ΔG° = ΣΔG°f(products) − ΣΔG°f(reactants). ΔG°f = 0 for an element in its standard state.","61fd64bbfbc2f585e12dbf77b48380329e8305310fd58759d179b87dd612f368",{"heading":53,"kind":13,"items":54},"ΔG° = ΔH° − TΔS°",[55],{"ref":56,"mode":19,"provenance":20,"text":58,"source_hash":59},{"type":17,"topic":7,"id":57,"field":17},"TD-R007","When ΔG°f isn't available, use ΔG° = ΔH° − TΔS° (T in K). Watch units: ΔH° in kJ, ΔS° in J/K, so divide ΔS° by 1000 first.","a2161ee35552446d8eab547711edf69731810616643be2a659e4d5d9b922fe8f",{"heading":61,"kind":13,"items":62},"Spontaneity",[63],{"ref":64,"mode":19,"provenance":20,"text":66,"source_hash":67},{"type":17,"topic":7,"id":65,"field":17},"TD-R008","ΔG° \u003C 0 spontaneous; ΔG° > 0 nonspontaneous; ΔG° = 0 at equilibrium.","f316cbfcff6f110d1ee3b7417c611de8d8009d03ee126b7221a25c6e49bc19bc",{"heading":69,"kind":13,"items":70},"Free energy & K",[71],{"ref":72,"mode":19,"provenance":20,"text":74,"source_hash":75},{"type":17,"topic":7,"id":73,"field":17},"TD-R010","ΔG° = −RT ln K (T in K, R = 8.314 J/(mol·K)). K > 1 ↔ ΔG° \u003C 0 (product-favored).","eada3567b27433475139c74f64d42654df1bbdadbb793d96a0a453a69ff8d8e0",{"heading":77,"kind":13,"items":78},"Non-standard ΔG",[79],{"ref":80,"mode":19,"provenance":82,"text":83,"source_hash":84},{"type":17,"topic":7,"id":81,"field":17},"TD-R015","original","ΔG = ΔG° + RT ln Q. ΔG \u003C 0 → proceeds forward; ΔG > 0 → reverse; ΔG = 0 → at equilibrium.","e1874d9913d85b554ed24ab26693d51d21a91ad62d6f054dbb1fd24aa9faee88",{"heading":86,"kind":13,"items":87},"Second law",[88],{"ref":89,"mode":19,"provenance":82,"text":91,"source_hash":92},{"type":17,"topic":7,"id":90,"field":17},"TD-R014","ΔS_univ = ΔS_sys + ΔS_surr, with ΔS_surr = −ΔH_sys/T (const T, P). Spontaneous ↔ ΔS_univ > 0.","14cb69eab428b90936fac050c1fe9c0994cd9857d66ef1a5107dd19e4c0d0960",{"heading":94,"kind":13,"items":95},"Crossover temperature",[96],{"ref":97,"mode":19,"provenance":20,"text":99,"source_hash":100},{"type":17,"topic":7,"id":98,"field":17},"TD-R011","When ΔH° and ΔS° are both positive, the reaction turns spontaneous above T = ΔH°/ΔS° (where ΔG° changes sign).","72378f5875e6b74ab0bd69ae4a1ca3ccd9606cee6eff6f8ffd2f42d54309e909",{"heading":102,"kind":13,"items":103},"Phase transitions",[104],{"ref":105,"mode":19,"provenance":20,"text":107,"source_hash":108},{"type":17,"topic":7,"id":106,"field":17},"TD-R012","At a boiling or freezing point the two phases are at equilibrium, so ΔG° = 0 and the transition temperature is T = ΔH°/ΔS° for that phase change.","8251f75eabeff94b3194f22a2c368fd1658f35a816ba900222cc6b47bdc3f2e5",1787246033158]