Spontaneity refers to any process that appears to proceed naturally to a final state without outside intervention. Some processes are spontaneous while others need an energy input in order to take place
Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous
spontaneous nonspontaneous
Entropy, Free Energy and the Direction of Chemical Reactions
Thermodyamics considers the energetics of a reaction but says nothing about the rate of a chemical reaction (kinetics). Potential Energy 1 second 1 week 1 decade Time
Recall, The Internal Energy (E) of a system is a bulk property of matter that describes all energy within a defined system. Internal Energy is the sum of energies that exists in atoms/molecules. Translational kinetic energy. Translational kinetic energy. Rotational energy Molecular rotation. Bond vibrations. Intermolecular attractions. Vibrational energy Electrons. Nuclear Electrochemical Electrostatic energy
Molecules translate, vibrate, rotate in quantized states. The more atoms there are, there are more ways a molecule can vibrate or wiggle and store energy (degrees of freedom).
MICROSCOPICALLY: Internal energy is distributed over all quantized translational, vibronic and rotational energy states in molecules and atoms Rotational modes Vibrational modes Translational modes Energ gy in kj Energ gy in J Energy in10 17 J
RECALL: We cannot measure the absolute internal energy of a system (in terms of all factors). ΔE system = E final E initial ΔE is the measure of a change relative to an initial state: State 1: P1, V1, T1 State 2: P2, V2, T2 This state, is defined by MACROSCOPIC VARIABLES
MACROSCOPICALLY: The internal energy of the system can be changed only by exchanging heat (q) or work (w) with its surroundings. +q = heat added +w =work done on the system Surroundings System with internal energy E Surroundings ΔE system =q+w -q = heat lost from the system -w = work done by the system
The first law of thermodynamics explains which processes are possibleand whichones are impossible. Energy can be converted from one form to another but energy cannot be created or destroyed d The energy of the universe is constant ΔE universe = 0 ΔE universe = ΔE system + Δ E surrounding ΔE system = Δ E surrounding (q + w) system = (q + w) surrounding
However, in the lab, its difficult to measure ΔE, so instead, chemists use another state function:enthalpy, H. This is the heat gained or lost by the system under conditions of constant pressure ΔE system = q P -PΔV ΔH system = q P = ΔE system + PΔV
In the 1870 s, it was thought that ΔH determined the spontaneity of a chemical reaction. This notion was proved wrong as both exothermic and endothermic reactions are spontaneous (no outside intervention to make it happen). Spontaneous reactions CH 0 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) ΔH = 890.4 kj H + (aq) + OH (aq) H 2 O (l) ΔH 0 = 56.2 kj H 2 O (s) NH 4 NO 3 (s) N 2 O 5 (s) H 2 O (l) ΔH 0 = 6.01 kj H 2 O NH 4+ (aq) + NO 3 (aq) ΔH 0 = 25 kj 2NO 2 (g) + ½O 2 (g) ΔH 0 = 109.5 kj
The first law of thermodynamics does not explain why some processes are spontaneouswhile others are not. H 2 O (s) H 2 O (l) H 2 O (l) H 2 O (s)
Suppose a chemical reaction from A to B or B to A. A B Can we develop a parameter or function that can help us predict whether A ==>B will occur spontaneously? The answer is yes and the details are explained in thermodynamic theory.
Boltzmann found that statistics was key to understanding spontaneity of a chemical reactions. He related statistics to a state function called entropy, S, that is a measure of the dispersion or spread of energy that occurs in all spontaneous reactions. Boltzmann s Tomb In Vienna, Austria
Freedom of motion and extent of energy distribution is related to a thermodynamic factor called ENTROPY (S). Entropy S = k ln W Boltzman s constant 1.38 10 23 J/K. = R/N A # of microstates Change in Entropy ΔS = k ln Wf k ln Wi When W f > W i then ΔS > 0 When W f < W i then ΔS < 0
BE CAREFUL: entropy is often illustrated incorrectly as a positionally messy room or disordered cards and described as disorder. This is not correct in the chemical sense! WRONG: Neither case represents thermodynamic entropy because rooms and cards do not exchange heat with its surroundings! This is a fundamental requirement.
A system can be described by bulk or macroscopic that we can observe and measure, and the microscopic or molecular that we don t see but can model statistically. We must see both! Consider 4 labeled molecules A,B,C,D,, Microstate is particular distrubution that corresponds to some macrostate. Multiplicity is the number of microstates that give a specific macrostate. Macrostate is the observed state of the system that represents on a molecular level that microstate with the highest probability or number. 5 observable macrostates
Microstate Left Right Side Side Multiplicity Probability Macrostate Bulk Property A,B,C,D - 1 1/16 A,B,C A,B,D A,C,D B,C,D AB A,B A,C A,D BC B,C B,D C,D D C B A CD C,D B,D B,C AD A,D A,C A,B 4 4/16 6 6/16 - A,B,C,D 1 1/16 16
The statistical interpretation of entropy is that it is a measure of the dispersal or spread of energy among atoms or molecules l in quantized microstates t Example: two block of different temperatures are brought together. Heat flows from a body to a cold body spontaneously and never the opposite direction (unless work or energy is added). Thermal energy flows from the higher h occupied energy levels in the warmer object into the unoccupied levels of the cooler one untilequal numbers of states are occupied. 100 C 10 C 100 C 10 C HOT COLD COMBINED
Matter being dispersed into a larger number of statistical microstates disperses energy into a larger number of thermal energy levels increasing entropy. Gas expands spontaneously into larger volume 1.0 atm evacuated 0.5 atm Quantum mechancis dictates closer energy level spacing as V increases. Energy level Same amount of energy is dispersed or spread among more energy levels. Energy level
The number of microstates scales as 2 N in a two state t system. 1 atom 2 atom 2 1 = 2 2 2 = 4 1 mol of Neon (10 L at 298 K) 3 atom expands to (20 L at 298K) NA 2 3 = 8 2 NA
ΔS = = N sys S 2 S 1 k ln 2 k ln 1 ΔS sys = k ln 2 N sys ΔS sys = R/N * N ln 2 ΔS sys = R ln 2 = 5.76 J/mol*K
But how is the microscopic level calculation of entropy equate to the macroscopic bulk property? How do we observe entropy in the macroscopic world? Reversible heat ΔS sys = q rev T ΔS sys = 1717 J/298 K = 5.76 576J/mol K This is important because here, we see that entropy is not related to position (or disorder) but how heat is utilized/dispersed in the system at a particular temperature
(second law of thermodynamics) All real processes occur spontaneously in the direction that increases the entropy of the universe (system + surroundings) ΔS universe = ΔS system + ΔS surrounding >0
Unlike Enthalpy (H) and Internal Energy (E) where we cannot findan absolute baseline, We can assume that baseline for entropy At 0 K, a perfect crystalline solid has zero entropy. (third law of thermodynamics) Standard Molar Entropies (S o ) Standard States: 1 atm for gases; 1 M for solutions; pure substance in its most stable form for solids and liquids. units: J/mol/K at 25 o C
Parameters and events that give higher number of microstates are of higher entropy (the obsolete imprecise way to say this is thata a system hasmore disorder.) 1. Temperature increases 2. Solid==>Liquid==>Gas phase changes 3. Dissolution of a solid or liquid 4. Dissolution of a gas 5. Atomic size or molecular complexity
Increase in temperature increases ΔS, since there is more energy that needs to be distributed LESS freedom of motion (at low T) MORE freedom of motion (at high T) E E Small amount of ENERGY Large amount of ENERGY
Solids and liquids have less freedom of motion = less microstates to disperseenergy. energy. LESS freedom of motion MORE freedom of motion solid liquid gas LOCALIZED energy of motion DISPERSED energy of motion
Solids and liquids have less freedom of motion = less microstates to disperseenergy. energy.
A dissolution process increases the freedom of motion of an ionic solute leading to a larger number of microstates. LESS freedom of motion MORE freedom of motion Solid solute Liquid solvent Mixed solution LOCALIZED energy of motion DISPERSED energy of motion
Mixing two miscible liquids also result into dispersal of energy among more states.
Mixing two miscible liquids also result into dispersal of energy among more states. ethanol water solution of ethanol and water F d f t i ti ll h d Freedom of movement remains essentially unchanged; S increase is due solely to random mixing effects.
Entropy decreases drastically as a gas is dissolved in a liquid or when a gas condenses to a liquid. O 2 gas O 2 dissolved
An increase in volume also leads to a lot more microstates (as proven by quantummechanics) mechanics) Gas expands spontaneously into larger volume 1.0 atm evacuated 0.5 atm Energy level Energy level
Increased molecular complexity (or even just the number of atoms) increases freedom of motion
Parameters and events that give higher number of microstates are of higher entropy (the obsolete imprecise way to say this is that a system has more disorder.) 1. Temperature increases ΔS increases as temperature rises as more energy states are filled 2. Solid==>Liquid==>Gas phase changes ΔS increases as a less dispersed phase changes to a more dispersed phase. 3. Dissolution of a solid or liquid S o of a dissolved solid or liquid is usually greater than S o of the pure solute. However, the extent depends on the nature of the solute and solvent. 4. Dissolution of a gas A gas becomes less dispersed when it dissolves in a liquid or solid. 5. Atomic size or molecular complexity Insimilar substances, increases inmass relate directly to entropy. In allotropic substances, increases in complexity (e.g., bond flexibility) relate directly to entropy.
How does the entropy of a system change for each of the following processes? (a) Condensing water vapor to a liquid dispersal of energy decreases Entropy decreases (ΔS < 0) (b) Forming sucrose crystals tl from a supersaturated t solution dispersal of energy decreases Entropy decreases (ΔS < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C dispersal of energy increases Entropy increases (ΔS > 0) (d) Subliming dry ice dispersal of energy increases Entropy increases (ΔS > 0)
Choose the member with the higher entropy in each of the following pairs, and justify your choice. Assume constant temperature, except in part (e). (a) 1 mol of SO 2 (g) or 1 mol of SO 3 (g) (b) 1 mol of CO 2 (s) or1molofco of 2 (g) (c) 3 mol of oxygen gas (O 2 ) or 2 mol of ozone gas (O 3 ) (d) 1 mol of KBr(s) or1molofkbr(aq) of (e) seawater in mid-winter at 2 o C or in mid-summer at 23 o C (f) 1 mol of CF 4 (g) or 1 mol of CCl 4 (g) SOLUTION: (d) 1 mol of KBr(aq) - solution > solid (a) 1 mol of SO 3 (g) - more atoms (b) 1 mol of CO 2 (g) - gas > solid (c) 3 mol of O 2 (g) - larger # mols (e) 23 o C - higher temperature (f) CCl 4 - larger mass
Standard Molar Entropy, S : The standard molar entropy is the entropy change that occurs when reactants and products are under thermodynamic standard dconditions: i 1 atm pressure, 1M and at 25 C (298K). aa+ bb cc+ dd ΔS 0 rxn =? ΔS 0 rxn = [ cs (C)+ ds (D) ] [ as (A) + bs (B)] Ssys = ni Si (products) nisi (reactants) Looks like Hess s Law! S for each component is found in a table. Also note that S = 0 for elements.
These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass.
Calculating the standard entropy of reaction, ΔS o rxn What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = 197.9 J/K mol S 0 (O 2 ) = 205.0 0 J/K mol 2CO (g) + O 2 (g) S 0 (CO 2 ) = 213.6 J/K mol 2CO 2 (g) ΔS 0 rxn = 2 x S0 (CO 2 ) [2 x S 0 (CO) + S 0 (O 2 )] rxn ΔS 0 rxn = 427.2 [395.8 + 205.0] = 173.6 J/K mol
When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, ΔS 0 > 0. If the total number of gas molecules diminishes, ΔS 0 < 0. If there is no net change in the total number of gas molecules, then ΔS 0 may be positive or negative BUT ΔS 0 will be a small number. What tis the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s)
But what happens to the ΔS surroundings? For spontaneous reactions in which a decrease in the entropy of the system occurs: must be outweighed by a concomitant increase in the entropy of the surroundings (ΔS univ > 0) For an exothermic process: q sys < 0, q surr > 0, ΔS surr > 0 For an endothermic process: q sys > 0, q surr < 0, ΔS surr < 0 The initial temperature of the surroundings affects the magnitude of ΔS surr. ΔS surr α qq ss sys ΔS surr α 1/T
But what happens to the ΔS surroundings? ΔS surr = ΔH sys / T Exothermic Process ΔS surr > 0 Endothermic Process ΔS surr < 0
So, going back to the second law of thermodynamics ΔS univ = ΔS sys + ΔS surr > 0 univ sys surr Heat that flows into or out of the system also changes the entropy of the surroundings. Standard Entropy of ΔS surr = Reaction (ΔS 0 qsys = ΔH sys from Tables) T T
Components of ΔS o univ for spontaneous reactions exothermic endothermic exothermic
So, going back to the second law of thermodynamics ΔS universe = ΔS system + ΔS surrounding >0 universe system surrounding ΔS universe = ΔS system ΔH system >0 TΔS universe = TΔS system + ΔH system < 0 T This form of the second law of thermodynamics is better because we need to only focus on the system!
Determining reaction spontaneity At 298 K, the formation ato of ammonia ahas a negative ΔS o sys. N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔS o sys = 197 J/K 2 2 3 sys Calculate ΔS o univ and state whether the reaction occurs spontaneously at this temperature. PLAN: ΔS o univ must be > 0 in order for the reaction to be spontaneous, so ΔS o surr must be > +197 J/K. To find ΔS o surr, first find ΔH sys. ΔH o sys = ΔH o rxn which can be calculated using ΔH o f values from data tables. Then apply ΔS o univ = ΔS o surr + ΔS o sys.
SOLUTION: N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔS o sys = 197 J/K PLAN: ΔS o univ must be > 0 in order for the reaction to be spontaneous, so ΔS o o surr must be > +197 J/K. To find ΔS surr,, first find ΔH sys. ΔH o sys = ΔH o rxn which can be calculated using ΔH o f values from data tables. Then apply ΔS o univ = ΔS o surr + ΔS o sys. SOLUTION: ΔH o rxn = [(2 mol)(δh o f NH 3 )] [(1 mol)(δh o f N 2 ) + (3 mol)(δh o f H 2 )] ΔH o rxn = 91.8 kj ΔS o surr = ΔH o sys/t = ( 91.8 x 10 3 J/298 K ) = +308 J/K (>+197 J/K) ΔS o o o univ = ΔS surr + ΔS sys = 308 J/K + ( 197 J/K) = +111 J/K ΔS o univ > 0: the reaction is spontaneous!
What happens to a system at equilibrium? When a system reaches equilibrium, neither the forward nor the reverse reaction is spontaneous; neither proceeds further because there is no driving force: ΔS o univ = 0. ΔS o univ = ΔS o surr + ΔS o sys = 0
To complete our question: TΔS universe = TΔS system + ΔH system < 0 ΔG system = TΔS system + ΔH system < 0 We call ΔG, the GIBB S FREE ENERGY
The Gibb s Free Energy is the criteria chemist use to determine if a reaction will occur spontaneously. ΔS univ > 0 or ΔG sys < 0 for a spontaneous process ΔS univ < 0 or ΔG sys > 0 for a non spontaneous process ΔS univ = 0 or ΔG sys = 0 for a process at equilibrium For a constant temperature process: Gibbs free energy (G) ΔG = ΔH sys TΔS sys ΔG < 0 The reaction is spontaneous in the forward direction. ΔG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. ΔG = 0 The reaction is at equilibrium.
What is Free Energy? ΔG is the maximum useful work that can be produced by a chemical reaction. ΔG = work max For a nonspontaneous process, ΔG is the minimum work that must be done on the system to make the process take place. ΔG o rxn = ΔH o rxn TΔS o rxn Increases in entropy suck energy from a process (it is dispersed energy that can not be snatched from the process). Total Energy available in a reaction ΔH o rxn = ΔG o rxn + TΔS o rxn Free energy! Unusable energy dispersed through the reaction as entropy
What is Free Energy? A reversible process: one that can be changed in either direction by an infinitesimal change in a variable. The maximum work from a spontaneous process is obtained only if the work is carried out reversibly. In most real processes, work is performed irreversibly; thus, maximum work is not obtained; some free energy is lost to the surroundings as heat and is thus unavailable to do work. A reaction at equilibrium (ΔG = 0) cannot do work.
The standard free energy of reaction (ΔG 0 rxn ) is the free energy change for a reaction when it occurs under standard state conditions. aa+ bb cc+ dd Looks like Hess s Law! ΔG 0 dδg rxn = [ cδg 0 (C) + 0 (D)] [ aδg bδg f f 0 (A) + 0 (B)] f f ΔG 0 nδg = Σ 0 (products) mδg Σ 0 (reactants) rxn f f Standard free energy of formation (ΔG 0 ) is the free energy energy change that f occurs when 1 mole of the compound is formed from its elements in their standard dstates. tt ΔG 0 of any element in its stable form f is zero.
What is the standard free energy change for the following reaction at 25 0 C? 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) ΔG 0 rxn = ΣnΔG 0 (products) ΣmΔG f 0 (reactants) f ΔG 0 rxn = [ 12ΔG 0 (CO 6ΔG 0 (H 0 2 ) + 2 O) ] [ 2ΔG (C f f 6 H 6 )] f ΔG 0 rxn = [ 12x 394.4 + 6x 237.2 ] [ 2x124.5 ] = 6405 kj Is the reaction spontaneous at 25 0 C? ΔG 0 = 6405 kj < 0 spontaneous
There are two methods to calculate ΔG 0 sys ΔS 0 form S 0 tables ΔH 0 from ΔH 0 f tables ΔS 0 sys and ΔH 0 sys ΔG 0 f 0 ΔG 0 sys
Chemists have a defined reference reaction condition called the standard state (below). We denote this standard state symbolically with a superscript degree sign above the thermodynamic functions: ΔH, ΔG, ΔS Standand State Reaction Conditions Standard state ΔH o rxn ΔG o rxn Non-standard ΔHrxn ΔGrxn 1M concentration Think About It: It s great to have thermodynamic data for standard conditions but what about non standard conditions that t exist tin the real world?
The standard free energy (ΔG 0 rxn) is the free energy change for the complete reaction of reactants to products with both reactants and products under standard state conditions. aa + bb cc + dd ΔG 0 values are a special hypothetical case of 100% conversion. We are usually more interested in real world non-standard conditions and equilibrium values for ΔG.
What about non-standard reaction conditions? ΔGrxn under non-standard thermodynamic reaction conditions connects the free energy to the equilibrium reaction quotient, Q. It can be shown that: Non-standard Free Energy Standard Free Energy Reaction Quotient G = G + RT ln Q R is the gas constant (8.314 J/K mol) T is the absolute temperature (K)
The reaction quotient, Q is related to ΔGrxn. Both tell us the direction of a chemical reaction! aa(g) + bb(g) cc(g) + dd(g) Q< K Q= 0 = Kc Q> K ΔG sys < 0 ΔG sys = 0 ΔG sys > 0 Product-Favored Equilibrium Reactant-Favored
Recall our expressions for the equilibrium constants (all of them Kc, Kp, Ka, Ksp) aa(g) + bb(g) cc(g) + dd(g) Concentrations or partial pressures at any time Q = [C]c [D] d [A] a [B] b Q = (P c) c (Pd) d (Pa) a (Pb) b Equilibrium concentrations or partial pressures = [C]c [D] d a (P c) c (Pd) d Kc [A]a [B] b Kp = (P a) a (Pb) b
Free Energy and Equilibrium Gravitational potential energy will seek equilibrium at a lower potential if push over Chemical reaction will seek equilibrium at a lower chemcial potential if pushed over its activation energy.
Free energy diagrams show how the free energy changes move (drive a reaction) towards equilibrium Q = K, ΔG o =0. ΔG o (reactants) To otal free e energ gy G o Q < K Spontaneous Direction Q > K Spontaneous Direction Q = K ΔG o (rxn) ΔG o (products) 100% Reactants t Equilibrium 100% Mixture Products Reaction Progress ===>
Plots of free energy diagrams for ΔG o < 0 (left) and ΔG o > 0 (right). Free energy (G) of the reacting system Q < K Pure reactants Gº(reactants) ΔGº = Gº(products) Gº(reactants) ) ( ) < 0 Q = K Equilibrium position Extent of reaction (a) Gº(products) Q > K Pure products ystem Free energy (G) of the reacting s Gº(products) ΔGº = Gº(products) Gº(reactants) d t ) t t ) > 0 Q > K Gº(reactants) Q < K Q = K Pure reactants Equilibrium position Extent of reaction (b) Pure products
G rxn of a reaction and the equilibrium constant K are related. aa + bb cc + dd Kc = [C]c [D] d [A] a [B] b Grxn = G rxn + RT ln Q R = 8.314 J/K mol T in Kelvin 0 = G + RT ln Kc G rxn = - RT ln Kc Kc = exp (- G rxn/rt) At equilibrium: G = 0 and Q = Kc Solving for G rxn Taking antiloge or exp of both sides
ΔG = RT ln Q/K = RT ln Q RT ln K If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (ΔG < 0) If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (ΔG > 0) If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (ΔG = 0) Under standard conditions (1 M concentrations, 1 atm for gases), Q = 1 and ln Q = 0. Therefore: ΔG o = RT ln K
ΔG o and Kc at 25 o Care inversely related...big K means smaller G and vis versa. G o (kj) K Significance 200 9 x 10-36 Essentially no forward reaction; 100 3 x 10-18 reverse reaction goes to completion 50 2 x 10-9 10 2 x 10-2 1 7 x 10-1 Forward and reverse 0 1 reactions proceed to the same extent -1 1.5-10 5 x 10 1-50 6 x 10 8 17 Forward reaction goes to -100 3 x 10 completion; essentially no -200 1 x 10 35 reverse reaction FO ORWARD REACTIO ON REVERSE E REACTI ION
Summary of Key Points: 1. ΔG o tells us the maximum amount of work a system for complete conversion of reactants to products or the amount of energy needed to drive a reaction. 2. When a system is at equilibrium: Q = Kc, ΔGrxn = 0 and ΔSuniverse = 0. 3. ΔG o rxn gives the position of equilibrium and can be calculated three ways: ΔG o rxn = Σ miδg o fi (products) Σ niδg o fi (reactants) ΔG o rxn = ΔH o rxn TΔS o rxn ΔG o rxn = RT ln K 4. ΔG rxn describes the direction in which a non standard reaction proceeds to reach equilibrium calculated by: ΔGrxn = ΔG o + RT ln Q
What affects Free Energy? Temperature plays an important role in determining whether a reaction is spontaneous. It scales the contribution of entropic : ΔG o sys = ΔH o sys TΔS o sys ΔG can be >0 or <0
At what Temp does ΔG switches sign? Switching or Crossover temperature thetemperature temperature at whicha reaction becomes spontaneous (used only when ΔH and ΔS have the same sign) ΔG o = ΔH o TΔS o = 0 ΔH o = TΔS o T = ΔH o /ΔS o
Cu 2 O(s) + C(s) 2Cu(s) + CO(g) ΔH o = +58.1 kj ΔS o = +165 J/K ΔH o :relatively insensitive to T TΔS o : increases with increasing ceas T
An important reaction in the production of sulfuric acid is the oxidation of SO 2 (g) to SO 3 (g): 2SO 2 (g) + O 2 (g) 2SO 3 (g) At 298 K, ΔG o = 141.6 kj, ΔH o = 198.4 kj, and ΔS o = 187.9 J/K. (a) () Determine the switching Temperature and predict how ΔG o will change with increasing T. (b) Assuming ΔH o and ΔS o are constant with increasing T, is ( ) g g, the reaction spontaneous at 900. o C?
CaCO 3 (s) CaO (s) + CO 2 (g) G f f (kj/mol) -1129-603 -394 H f (kj/mol) -1207-635 -394 S f S f (J/K) mol 91.7 38.2 213 Calculate: a. ΔH 0 rxn b. ΔS 0 rxn c. Show ΔG 0 rxn calculated via the GΔHTΔS equation (at 298 K) is the same as the ΔG 0 rxn calculated from ΔG 0 f? d. Calculate for K of the reaction e. Is this reaction spontaneous or non spontaneous given standard conditions (298K)? f. What happens if we increase Temp? g. What is the switching temperature (if there is any)?
Calculate: a. ΔH 0 rxn = (ΔH 0 f CO 2 +ΔH 0 f CaO ) (ΔH 0 f CaCO 3 ) = ( 394 kj ) + ( 635 kj) ( 1207 kj) [NOTE: the mole is canceled by the stoichiometric coefficients) = + 178 kj b. ΔS 0 rxn = (S 0 CO 2 +S 0 CaO ) (S 0 CaCO 3 ) = (213 J/K ) + (38.2 J/K) ( 91.7 J/K) [NOTE: the mole is canceled ldby the stoichiometric t i coefficients) i = + 159.5 J/K
Calculate: c. COMPARE ΔG 0 rxn = ΔH 0 rxn TΔS 0 rxn = 178 kj (298K)(159.55 J/K)(1kJ/1000J) = 130.5 kj ΔG 0 rxn =(ΔG 0 f CO 2 +ΔG 0 f CaO ) (ΔG 0 f CaCO 3 ) = ( 394 kj) + ( 635) ( 1207) = 132 kj * Relatively Close d. Use either (say use ΔG 0 rxn = 130.5 kj) ΔG 0 rxn = 130.5 kj = RT ln K K = exp ( ΔG 0 rxn/rt) = 1.33 x 10 23 NOTE: Because ΔG 0 is very positive, we expect K to be very small!
Calculate: e. Since ΔG 0 rxn>0 Reaction is non spontaneous at 298 K f. Because ΔH 0 rxn > 0 and ΔS 0 rxn > 0, We expect reaction to be non spontaneous below switching Temp, but become spontaneous as we increase Temp (specifically if we go higher than the switching Temp) T ΔH 0 /ΔS 0 g. T = ΔH 0 rxn/δs 0 rxn T = 1116 K
Temperature and Spontaneity of Chemical Reactions CaCO 3 (s) CaO (s) + CO 2 (g)