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 (- Η) 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 surroundings the system
The Surroundings The Universe The System
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
How does the system impacts the S surr? H < 0 S surr increases! heat
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, less ordered, low S S > 0 high S
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
Benzene Toluene low entropy Very unlikely! high entropy More likely!
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 = +80.3 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
G S vaporization Entropy S fusion S L Temperature
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 = Σn p S o products - Σn r S o reactants stoichiometric coefficient standard entropy in J/K i.e. (@ SATP)
for example, C 8 H 18(g) + 12.5 O 2(g) 8 CO 2(g) + 9 H 2 O (g) 13.5 moles 17 moles (expect S > 0) S rxn = Σn S o products - Σn S o reactants
for example, = [8(213.6) + 9(188.6)] [463.2 + 12.5(204.8)] = +383.0 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
now multiply through by -T -T S univ = -T S sys + H sys or, -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? the energy used up creating disorder the free energy left over G = H-T S the energy transferred as heat
When is G < 0? H o S o G o Spontaneous? - + - always + - + never - - + or - at lower T + + + or - at higher T
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)
4 KClO 3(s) 3 KClO 4(s) + KCl (s) Η rxn = Σn p H o products - Σn 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) = -144.3 kj mol -1
4 KClO 3(s) 3 KClO 4(s) + KCl (s) S rxn = Σn p S o products - Σn 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) = - 36.8 J K -1 mol -1
4 KClO 3(s) 3 KClO 4(s) + KCl (s) G = H - T S -144.3 kj mol -1-298 K (-0.0368 kj K - 1 mol -1 ) = -133.3 kj mol -1 G < 0, thus reaction proceeds spontaneously
4 KClO 3(s) 3 KClO 4(s) + KCl (s) G = H - T S -144.3 kj mol -1-298 K (-0.0368 kj K -1 mol -1 ) = -133.3 kj mol -1 25 o C N.B. conversion to kj!
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) Γ rxn = Σn p G o products) - Σn 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) = - 133.6 kj
Homework: p.g. 512: 1-14