The direction that spontaneous change is the direction in i beg your pardon total entropy increases. Total entropy change, also called the entropy adjust of the universe, is the sum of the entropy readjust of a system and the its surroundings:
DSuniv = DSsys + DSsurr
According to the second law that thermodynamics, the entropy of the universe, Suniv must always increase for a spontaneous process, the is, DSuniv>0.
You are watching: Delta s is positive for the reaction
Free energy and complimentary Energy Change—the Gibbs complimentary energy, G, is offered to describe the spontaneity of a process.
G = H  TDS
The free energy change, DG is equal to TDSuniv and also it applies just come a device itself, without regard because that the surroundings. That is defined by the Gibbs equation:
DG = DH  TDS
For a spontaneous procedure at constant temperature and also pressure, DG must be negative. In many cases, we have the right to predict the sign of indigenous the indications of DH and DS.Standard complimentary Energy Change, DGo —the standard complimentary energy change, DGo deserve to be calculated (1) through substituting conventional enthalpies and entropies the reaction and a Kelvin temperature into the Gibbs equation or (2) by combine standard free energies of development through the expression
Summary the Gibbs costfree energy
Enthalpy change  Entropy change  Gibbs complimentary energy  Spontaneity 
positive  positive  depends ~ above T, may be + or   yes, if the temperature is high sufficient 
negative  positive  always negative  always spontaneous 
negative  negative  depends on T, might be + or   yes, if the temperature is low enough 
positive  negative  always positive  never voluntary 
Problems:
1) In the Haber process for the produce of ammonia
Equation  Enthalpy change  Entropy change  
N2 + 3H2 2NH3  ΔH = 93Kj / mol  ΔS = 198 j / mol K  
At what temperature will certainly the reaction above become spontaneous?
The fact that both terms are negative way that the Gibbs costfree energy equation is balanced and temperature dependent:
ΔG = ΔH  TΔS
ΔG = 93000  (T x 198) note that the enthalpy is provided in kilojoules 

if ΔG = 0 then the system is at the limit of reaction spontaneity 

When ΔG = 0 then (T x 198) = 93000 

and T = 93000/198 Kelvin 

therefore the reaction i do not care spontaneous when T = 469 K (196 ºC) 

below this temperature the reaction is spontaneous.
2) recognize the Delta G under standard problems using Gibbs free Energies that Formation uncovered in a perfect Thermodynamics table for the complying with reaction:
4HCN(l) + 5O2(g) > 2H2O(g) + 4CO2(g) + 2N2(g)
examine to make sure the equation is well balanced Look increase the Standard complimentary Energy of formation of H2O(g) and multiply by its coefficient(2) in the equation.2 mole ( 237.2 kj/mole) = 474.4 kj = Standard costfree Energy of development for 2 moles H2O(l)
Look for the Standard free Energy of formation of CO2(g) and also multiply through its coefficient (4)4 moles ( 394.4 kj/mole) = 1577.6 kJ = Standard totally free Energy of development for 4 moles CO2
Look increase the Standard free Energy of formation of N2(g) and multiply by its coefficient(2)2 moles(0.00 kj/mole) = 0.00 kJ = Standard free Energy of formation for 2 moles of N2(g)
add the results of steps 2,3, and 4 to acquire the Standard costfree Energy for the products(474.4 kJ) + (1577.6 kJ) + 0.00 = 2052 kJ = Standard complimentary Energy for commodities
Look increase the Standard free Energy of formation for HCN(l) and also multiply through its coefficient(4)4 mole ( 121 kj/mole) = 484 kJ = Standard totally free Energy because that 4 mole HCN(l)
Look increase the Standard free Energy of development of O2(g) and also multiply through its coefficient(5)5 moles (0.00 kJ/mole) = 0.00 kJ = Standard totally free Energy that 5 mole of O2(g)
add the results of measures 7 and 8 to obtain the Standard complimentary Energy the the reactants(484) + (0.00) = 484 kJ = Standard totally free Energy for Reactants
Subtract the an outcome of action 9 from the result of step 5 to get the Standrad complimentary Energy adjust for the ReactionSum of complimentary Energy of assets  amount of totally free Energy of reaction = (2052 kJ)  (484 kJ) = 1568 kJ = Standard free Energy change for the Reaction.
3) for the complying with reaction using the Thermodynamics table:
CoCl2(g) > CO(g) + Cl2(g)
calculation at 127°C the DG calculation the Temperature once the over reaction is in ~ equilibrium (DG = 0) at what temperature will this reaction it is in spontaneous? inspect to view if the equation is balanced identify the Delta H that the Reaction making use of Hess Summation Law and Standard Enthalpies of developmentDelta H = amount of Delta Hf of commodities  sum Delta Hf of reactants
Delta H = < 1(110.5) + 1(0.00)>  < 1(220)>
Delta H = 110.5  (220) = +110.5 kJ
recognize Delta S because that the reaction using conventional Molar Entropies and Hess law of SummationDelta S = sum Standard Molar Entropies of products  amount of traditional Molar Entropies of reactants
Delta S = < 1 mole(197.5 J/moleK) + 1 mole(223) J/moleK>  < 1 mole(283.7 J/moleK)>
Delta S = 420.5 J/K  283.7 J/K = 136.8 J/K
transform Delta S native J/K come kJ/K136.8 J/K X 1 kJ / 1000 J = 136.8 / 1000 = .1368 kJ/K
transform 127 C come KK = C + 273 = 127 + 273 = 400 K
Plug outcomes of step 2 and 4 into Gibbs Helmholtz Equation along with Kelvin Temperature to acquire Delta G the the ReactionDelta G = Delta H  T(Delta S)
Delta G = 110.5 kJ  400 K(.1368 kj/K)
Delta G = 110.5  54.72 kJ = + 55.78 kJ
Because this reaction has actually a confident Delta G it will be nonspontaneous as written.
Name of types Delta Hf(kJ/mole) Delta Gf(kJ/mole) S(J/moleK)
CO2(g) 393.5 394.4 213.7
CH3OH(l) 238.6 166.2 127
COCl2(g) 220 206 283.7
CO(g) 110.5 137.2 197.5
C2H2(g) 227 209 200.9
Cl2(g) 0 0 223
H2O(g) 241.8 228.6 188.7
H2O(l) 285.8 237.2 69.9
HNO3(aq) 206.6 110.5 146
N2(g) 0 0 191.5
NO2(g) 33.2 51 239.9
NO(g) 90.3 86.6 210.7
O2(g) 0 0 205
CS2 151
H2 0 0 130.6
CH4 74.8 50.8 186.3
H2S 20.17 33.01 205.6
SO2 248.1
Free Energy and Equilibrium.
Because DG is a measure up of how favorable a reaction is, it likewise relates come the equilibrium constant.
A reaction through a an unfavorable DG, is really favorable, for this reason it has actually a big K. A reaction with a optimistic DG is not favorable, so it has a small K. A reaction v DG = 0 is in ~ equilibrium. There room several different DG"s. The is vital to distinguish in between them. DGo (a delta G, with a superscript o), is the complimentary energy change for a reaction, with everything in the standard states (gases in ~ 1 bar, and solutions at 1 M concentration), and also at a details temperature (usually 25°C) DG (just delta G). This is the costfree energy readjust for a reaction that is not at the typical state. The DG"s are connected as follows. Recognize THIS EQUATION and understand exactly how to use it:DG =DGo + RT ln Q
Where Q is the exact same Q we offered for calculating equilibrium (K is the special situation for Q when at equilibrium.)
Spontaneity and also Speed the Reactions.See more: Windows Cannot Print Due To A Problem With The Printer Setup
This section just points out that over there is not any direct relationship in between DG and the rate of a reaction (kinetics).
Problem:
Starting through the adjust in totally free energy at constant temperature: DG° = DH°  TDS°, and with the relation in between DG and also equilibrium constant, K: DG° = RT lnK, derive a straight equation the expresses lnK as a function of 1/T (a straight equation is of the kind y = mx + b).
(b) The equilbrium constants for the conversion of 3phosphoglycerate come 2phosphoglycerate at pH 7 is provided: