What is a spontaneous reaction? One, that given the necessary activation energy, proceeds without continuous outside assistance

Why do some reactions occur spontaneously & others do not? Atoms react to achieve greater stability Therefore products are generally more energetically stable than reactants In general, exothermic reactions (-  ) tend to proceed spontaneously

EXCEPTIONS Some endothermic reactions and those that produce less energetically stable products proceed spontaneously EXAMPLES: Ba(OH) 2(aq) + 2 NH 4 NO 3(aq)  Ba(NO 3 ) 2(aq) + 2 NH 4 OH (l) NH 4 NO 3(s)  NH 4 + (aq) + NO 3 - (aq)

Entropy, S - a measure of the disorder of a system or the surroundings

Entropy of “The Universe” the system the surroundings

The Universe The System The Surroundings

1 st law of thermodynamics: The total energy of the universe is constant (The best you can do is break even) 2 nd law of thermodynamics: The entropy of the universe is increasing (You can’t break even)

Low entropy is less probable

 S universe  S system  S surroundings If  S universe  0, reaction is spontaneous If  S universe  0, reaction is nonspontaneous

heat S surr increases!  H < 0 How does the system impacts the  S surr ?

Entropy is a State Function  S = S final - S initial path taken is irrelevant rate of change is irrelevant

 S > 0 for: - melting - vaporizing - making a solution - a reaction that produces an increased number of moles - heating a substance

H 2 O (s)  H 2 O (l)  ordered, low S less ordered, high S  S > 0

 S > 0 for: - melting - vaporizing - making a solution - a reaction that produces an increased number of moles - heating a substance

H 2 O (l)  H 2 O (g) high entropy low entropy

 S > 0 for: - melting - vaporizing - making a solution - a reaction that produces an increased number of moles - heating a substance

low entropy high entropy Very unlikely! More likely! Benzene Toluene

 S > 0 for: - melting - vaporizing - making a solution - a reaction that produces an increased number of moles - heating a substance

Ba(OH) 2  8H 2 O (s) + 2 NH 4 NO 3(s)  Ba(NO 3)2(aq) + 2 NH 3(aq) + 10 H 2 O (l)  H = kJ (unfavorable) 3 moles  13 moles  S > 0 (favorable)

 S > 0 for: - melting - vaporizing - making a solution - a reaction that produces an increased number of moles - heating a substance

Temperature Entropy S L G  S fusion  S vaporization

Entropy tends to increase In general, a system will increase in entropy (  S > 0) if: the volume of a gaseous system increases the temperature of a system increases the physical state of a system changes from solid to liquid to gas the number of moles in a system increases

Calculating  S for a reaction  S rxn =  n p S o products -  n r S o reactants standard entropy in J/K i.e. SATP) stoichiometric coefficient

for example, C 8 H 18(g) O 2(g)   8 CO 2(g) + 9 H 2 O (g) 13.5 moles  17 moles (expect  S > 0)  S rxn  =  n S o products -  n S o reactants

for example, = [8(213.6) + 9(188.6)] – [ (204.8)] = J K -1 mol -1

Temperature and pressure are strongly connected to ideas of enthalpy and entropy. (Remember that -∆H and +∆S are favourable). Consider the following three examples:For each reaction, identify the sign of ∆H and ∆S. Indicate whether the reaction is likely to be spontaneous. 1.Zn (s) + 2 HCl (aq) ↔ ZnCl 2 (aq) + H 2 (g) 2.3 C (s) + 3 H 2 (g)  C 3 H 6 (g) 3.2 Pb(NO 3 ) 2 (s)  2 PbO (s) + 4 NO 2 (g) + O 2 (g) In a case where both ∆H and ∆S are favourable, we consider the reaction to be spontaneous and very likely to occur. What about in cases where only one is favoured?

Gibbs Free Energy  S univ  S sys  S surr and,  S surr = -  H sys T thus,  S univ  S sys  -  H sys T

-T  S univ  = -T  S sys  +  H sys now multiply through by -T -T  S univ  =  H sys  -T  S sys or,  G sys  =  H sys  -T  S sys

 G  =  H  -T  S Gibbs energy change or the “free energy change”

 G and spontaneity recall that  G sys = -T  S univ since  S univ > 0 for a spontaneous change,  G sys < 0 for a spontaneous change

What’s “free” about free energy?  G  =  H  -T  S the energy transferred as heat the energy used up creating “disorder” the “free” energy left over

 H o  S o  G o Spontaneous? - + -always + - +never - -+ or -at lower T + ++ or -at higher T When is  G < 0?

G is a state function  G = G final - G initial path is irrelevant rate of reaction is irrelevant

How do we find  G values? 1. Calculate  H,  S values, then use  G =  H - T  S 2. Look up  G o f values

for example, Will this reaction proceed at 25 o C? 4 KClO 3(s)  3 KClO 4(s) + KCl (s)

 rxn =  n p H o products -  n r H o reactants = 3  H o f (KClO 4(s) ) +  H o f (KCl (s) ) - 4  H o f (KClO 3(s) ) = 3(-432.8) + (-436.7) - 4(-397.7) = kJ mol -1

4 KClO 3(s)  3 KClO 4(s) + KCl (s)  S rxn =  n p S o products -  n r S o reactants = 3 S o (KClO 4(s) ) + S o (KCl (s) ) - 4 S o (KClO 3(s) ) = 3(151.0) + (82.6) - 4(143.1) = J K -1 mol -1

4 KClO 3(s)  3 KClO 4(s) + KCl (s)  G =  H - T  S kJ mol K ( kJ K - 1 mol -1 ) = kJ mol -1  G < 0, thus reaction proceeds spontaneously

4 KClO 3(s)  3 KClO 4(s) + KCl (s)  G =  H - T  S kJ mol K ( kJ K -1 mol -1 ) = kJ mol -1 N.B. conversion to kJ! 25 o C

How do we find  G values? 1. Calculate  H,  S values, then use  G =  H - T  S 2. Look up  G o f values (standard free energies of formation)

4 KClO 3(s)  3 KClO 4(s) + KCl (s)  rxn =  n p G o products ) -  n r G o reactants = 3 G o (KClO 4(s) ) + G o (KCl (s) ) - 4 G o (KClO 3(s) ) = 3(-303.2) + (-409.2) - 4(-296.3) = kJ

Homework: p.g. 512: