ib chemistry on entropy, gibbs free energy, second law of thermodynamics
DESCRIPTION
IB Chemistry on Entropy, Gibbs Free Energy, Second Law Thermodynamics and Spontaneity of reactionTRANSCRIPT
Entropy, Free Energy and Spontaneity
• How will a reaction go ?
• Why gas MIXES and NOT UNMIX ?
• Why concentrated solute DIFFUSE and NOT UNDIFFUSE?
Entropy
• Measures the degree of DISORDER for a system
• Measures the probability or chance for a system, both in distribution of particles in space
and distribution of energy
Entropy and Spontaneity
• Unit for entropy – JK-1mol-1
• Formula for entropy, S = k ln W k = Boltzmann constant
W = Number ways a particle can arrange in space
• Absolute entropy can be measured!
• Entropy of a perfectly order solid crystal at OK is ZERO
• Entropy of an element under standard condition is NOT ZERO
ΔSθ (H2) gas = 130.6JK-1mol-1
• Standard entropy change, ΔSθ = Entropy change per mole of a substance
heated from OK to standard Temp of 298K
Important concepts for Entropy
Entropy and Spontaneity
ENTROPY CHANGE, ΔSθ
• Change in the disorder of a system
• Higher ↑ disorder – Higher entropy ↑
• Change in state from SOLID → LIQUID → GAS → Higher entropy ↑
• Greater ↑ number of particles formed in products → Higher entropy ↑
• Complex molecules (more atoms bonded) → Higher entropy ↑
• Higher temperature ↑ → Particles vibrate faster → More random → Higher entropy ↑
Standard entropy, ΔSθ /298K Sθ / JK-1mol-1
Second Law of Thermodynamics • For all spontaneous reactions, TOTAL entropy of the universe ΔSθ
uni always INCREASES
• Formula for ΔSθsys = ΔSθ (products) - ΔSθ (reactants)
• Reaction will likely to happen if -
• Formula for ΔSθ(surrounding)
• Conclusion : For spontaneous reaction, ΔSθuni must > O (positive)
Gibbs Free Energy, ΔG
To predict if a reaction will likely to happen
• ΔG is used or preferred because it involves the system while ΔS involves system and surrounding
• Easier to determine ΔH and ΔS for a system
ΔG = ΔH -TΔS
• ΔSuni = ΔSsys + ΔSsurr > 0, (+ve) positive
• ΔGsys =ΔHsys –TΔSsys < 0, (-ve) negative
•ΔHsys = -ve (exothermic)
• ΔSsys = +ve (entropy increses ↑)
• Δssurr = +ve (entropy increases ↑ )
•ΔSuni = +ve (entropy increases ↑ )
Gibbs Free Energy, ΔG is used
Unit for ΔGsys = kJmol-1 ΔHsys = kJmol-1 ΔSsys = JK-1mol-1
Combination ΔHsys , ΔSsys and ΔGsys to predict if reaction is spontaneous
ΔHsys +ve ΔSsys +ve ↑
Spontaneous if Temp High ↑
ΔHsys -ve ΔSsys –ve ↓
Spontaneous if Temp low ↓
ΔHsys -ve ΔSsys +ve Always Spontaneous
ΔHsys +ve ΔSsys –ve Non Spontaneous
ΔSθsys ΔSθ
surr
ΔSuni = ΔSsys + ΔSsurr
= (-118.7) + 148.0 = +29.3 JK-1mol-1
Conclusion :
• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding
INCREASES ↑ the entropy of surrounding, ΔSsurr
• Spontaneous - ΔGθ = -ve
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (70.0) – (188.7) = -118.7 JK-1mol-1
ΔSsurr = - ΔHsys /T
= -(-44100)/298 = +148.0JK-1mol-1
ΔGθ = ΔH -TΔS
= (-44.1) – (298 x -118.7/1000) = - 8.72kJmol-1
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
+
ΔSθsys ΔSθ
surr
Conclusion :
• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding
INCREASES ↑ the entropy of surrounding, ΔSsurr
• Spontaneous - ΔGθ = -ve
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (213.6 + 2x171) – (186.0 + 2x205) = -40.4 JK-1mol-1
ΔSsurr = -ΔHsys /T
= -(-890000)/298 = +2986.5JK-1mol-1
+
ΔSuni = ΔSsys + ΔSsurr
= -40.4 + 2986.5 = +2946 JK-1mol-1
ΔGθ = ΔH -TΔS
= (-890) – (298 x -40.4/1000) = - 878 kJmol-1
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
+ ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (2 x 115) – (130.6) = +99.4 JK-1mol-1
ΔSsurr = -ΔHsys /T
= -(+436000)/298 = -1463 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= +99.4 + (-1463) = -1363.6 JK-1mol-1
ΔGθ = ΔH -TΔS
= (436) – (298 x 99.4/1000) = +406 kJmol-1
Conclusion : • Non Spontaneous - Entropy of system ΔSθ
sys INCREASES ↑ BUT heat absorbed from surrounding
DECREASES ↓ the entropy of surrounding, ΔS surr
• NON spontaneous - ΔGθ = +ve
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
+
Is it possible to freeze water at 298K ( 25oC)
At 25oC
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (48) – (70) = -22 JK-1mol-1
ΔSsurr = -ΔHsys /T
= -(-6010)/298 = +2016 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= -22 + (+2016) = -1.84 JK-1mol-1
ΔGθ = ΔH -TΔS
= (-6.01) – (298 x -22/1000) = +0.54 kJmol-1
Conclusion :
• Non Spontaneous - Entropy of system ΔSθsys DECREASES ↓ more than > the
INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding
• NON spontaneous - ΔGθ = +ve
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 263K
AND
+
Is it possible to freeze water at 263K ( -10oC)
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (48) – (70) = -22 JK-1mol-1
At -10oC
ΔSsurr = -ΔHsys /T
= -(-6010)/263 = +22.85 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= -22 + (+22.85) = +0.85 JK-1mol-1
ΔGθ = ΔH -TΔS
= (-6.01) – (263 x -22/1000) = -0.23 kJmol-1
Conclusion :
• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to surrounding
INCREASES ↑ the entropy of surrounding, ΔSsurr
• Spontaneous - ΔGθ = -ve
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
+
Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g)
At 25oC
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (39.7 + 213.6) – (92.9) = + 160.4 JK-1mol-1
ΔSsurr = -ΔHsys /T
= -(+178300)/298 = - 598.3 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= +160.4 + (- 598.3) = - 438 JK-1mol-1
ΔGθ = ΔH -TΔS
= (+ 178.3) – (298 x +160.4/1000) = + 130.5 kJmol-1
Conclusion :
• Non Spontaneous - Entropy of system ΔSθsys INCREASES ↑ BUT heat absorbed from surrounding
DECREASES ↓ the entropy of surrounding, ΔSsurr
• NON spontaneous- ΔGθ = +ve
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 1500K
AND
+
Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g)
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (39.7 + 213.6) – (92.9) = + 160.4 JK-1mol-1
At 1227oC
ΔSsurr = -ΔHsys /T
= -(+178300)/1500 = - 118.8 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= +160.4 + (- 118.8) = + 41.6 JK-1mol-1
ΔGθ = ΔH -TΔS
= (+ 178.3) – (1500 x +160.4/1000) = - 62.3 kJmol-1
Conclusion :
• Spontaneous - Entropy of system ΔSθsys INCREASES ↑ more than >
DECREASES ↓ in entropy of surr, ΔSsurr due to heat absorbed from surrounding
• Spontaneous - ΔGθ = -ve
ΔSθsys ΔSθ
surr
Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K
AND
+
Is the reaction possible 2NO (g) + O2 (g) → 2NO2 (g)
ΔSsys = ΔSsys (product) – ΔSsys (reactant)
= (2 x 240) – (2 x 210.7+102.5) = - 43.9 JK-1mol-1
ΔSsurr = -ΔHsys /T
= -(-114000)/298 = +382.5 JK-1mol-1
ΔSuni = ΔSsys + ΔSsurr
= -43.9 + (+382.5) = +339 JK-1mol-1
ΔGθ = ΔH -TΔS
= (- 114) – (298 x -43.9/1000) = -100.9 kJmol-1
Conclusion :
• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ less than <
INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding
• Spontaneous - ΔGθ = -ve
Reaction which is Temperature Dependent whereby ΔHsys (+ve) and ΔSsys (+ve)
Finding the temperature which make reaction spontaneous
ΔGsys = (-ve) Spontaneous ΔGsys = (+ve) Non Spontaneous
ΔGsys = O (Equilibrium)
Equilibrium Temperature whereby reaction becomes spontaneous happen when ΔG = O
ΔG = ΔHsys - TΔSsys
0 = ΔHsys -TΔSsys
TΔS = ΔH
T = ΔH/ΔS
At what temperature will reaction becomes spontaneous
0 = ΔH -TΔS
TΔS = ΔH
T = ΔH/ΔS
T = 178300/160.4 = 1111.6K
ΔHsys = +178.3 kJmol-1
ΔSsys = +160.4 JK-1mol-1
Standard Enthalpy Change of Formation, ΔHf
• Energy released when ONE mole of compound form from its constituent elements in their std states
EX : C(s) + 2H2 → CH4 (g) ΔHf = -75 kJmol-1
• Using standard enthalpy of formation to calculate ΔH reaction
• Free energy released when ONE mole of compound form from its constituent elements in their std states
EX : C(s) + 2H2 → CH4 (g) ΔGf = -51 kJmol-1
• Using standard free energy of formation to calculate ΔG reaction
Standard Free Energy of Formation, ΔGf
• Standard free energy of formation, ΔGf of compound
C(s) + 2H2 → CH4(g) ΔGf = -51 kJmol-1
H2(g) + ½ O2(g) →H20(g) ΔGf = -228 kJmol-1
2C (s) + 3H2(g) → C2H6(g) ΔGf = -33 kJmol-1
• Standard free energy of formation, ΔGf of element is ZERO
C(s) → C(s) ΔGf = 0 kJmol-1
ΔGreaction = (Sum ΔGf (product) – Sum ΔGf(reactant) )
ΔGreaction = (-ve) Reaction is Spontaneous
Using ΔG to predict if reaction is Spontaneous
Ex 1
Ex 2
Ex 3
ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )
= ( -394 + 2 x -237) – (-51 + 0)
= - 817 kJmol-1 (Spontaneous)
ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )
= ( 3 x -303.2 + (-409.2) ) – ( 4 x -296.3)
= -134 kJmol-1 (Spontaneous)
ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) )
= (-305) – (-267.8 + 0 )
= -37.2 kJmol-1 (Spontaneous)
Combination ΔH, ΔS and ΔG to predict if reaction is Spontaneous
Ex 1
Ex 2
ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) )
= ( -986.6) – (-635.5 + (-285.9) )
= - 65.2 kJmol-1
ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) )
= ( 3 x -432.8 -436.7) – ( 4 x -397.7)
= - 144 kJmol-1
ΔSrxn = (Sum S(product) – Sum S(reactant) )
= ( +76.1) – (+39.7 + 70.0)
= - 33.6 JK-1mol-1
ΔSrxn = (Sum S(product) – Sum S(reactant) )
= ( 3 x 151 + 82.6) – ( 4 x 143.1)
= - 36.8 JK-1mol-1
ΔGθ = ΔH -TΔS
= (-65.2) – (298 x -33.6/1000) = -55.2 kJmol-1
ΔGθ = ΔH -TΔS
= (-144) – (298 x -36.8/1000) = -133 kJmol-1
(Spontaneous)
(Spontaneous)