CAS CH 171 Lecture Notes - Lecture 3: Reaction Rate, Endergonic Reaction, Enthalpy
30 Apr 2018
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• Rate of Reaction - (change in concentration)/(change in time)
• Reaction Mechanism
• 1. Reactant particles MUST collide (interact)
• 2. Reactant particles must collide with a specific orientation
• ex: AB + CD —> AC + BD
• A C
• B D
• collide in this orientation
• 3. Reactant particles must collide with enough energy
Influencing reaction rate
• 1. increase the temperature —> increase the reaction rate
• 2. changing the nature of the reactant (ex: state, solution, etc.)
• 3. increase the concentration —> increase the reaction rate
• 4. increase the surface areas —> increase the reaction rate
• 5. adding a catalyst — increases the reaction rate
• catalyst - something that increases the reaction rate but is not used up in the reaction
• ^should be unchanged at the end of the reaction
Chemical Reactions - require energy in order to start
• Free Energy (G) - measures spontaneity
• EXERGONIC - energy released
• reactants are higher than the products
• ∆G < 0 = exergonic (spontaneous)
• ENDERGONIC - energy required
• reactants are lower than the products
• ∆G > 0 = endergonic (non-spontaneous)
• EQUILIBRIUM - rate forward = rate reversed
• no net change in concentration
• ∆G = 0
• ex: H2O (l) <—> H2O (g)
• ^establishes vapor pressure
• Enthalpy (∆H) - heat
• EXO THERMIC - heat released
• ∆H < 0 = exothermic
• ENDO THERMIC - heat absorbed
• ∆H > 0 = endothermic
• Entropy (S) - measure of disorder
• ∆S > 0 = more disorder (favor spontaneity)
• ∆S < 0 = less disorder (disfavor spontaneity)
• *increase temperature = increase molecular motion = increase entropy
• ex: melting
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• H2O (s) —> H2O (l)
• ^increase in entropy, molecules are free to move around
• Trends:
• If ∆S is positive, it is favoring the reaction
• If ∆G or ∆H are positive, they are not favoring the reaction
• If ∆S is negative, it is not favoring the reaction
• If ∆G or ∆H are negative, they are favoring the reaction
1. ex: which process below represents a decrease in entropy?
N2O4 (g) —> 2NO2 (g)
C6H12O6 (s) + 6O2 (g) —> 6CO2 (g) + 6H2O (g)
Sublimation of dry ice: CO2 (s) —> CO2 (g)
N2 (g) + 3H2 (g) —> 2NH3 (g)
∆G = ∆H - T(∆S)
direct relationship between ∆G and ∆H
∆S is inverse of ∆G and ∆H
T = in kelvin
The effect of ∆H and ∆S on ∆G
Sign of ∆H / Sign of ∆S / Sign of ∆G
- (exothermic) / - (increase in entropy) / - (exergonic) = SPONTANEOUS
+ (endothermic) / - (decrease in entropy) / + (endergonic) = NONSPONTANEOUS
+ (endothermic) / + (increase in entropy) / More info needed = More info needed
- (exothermic) / - (decrease in entropy) / More info needed = More info needed
EXAMPLES:
1. ex: C6H12O6 (s) + 6O2 (g) —> 6CO2 (g) + 6H2O (g)
∆G > 0, ∆H < 0 (favored), ∆S > 0 (favored)
2. ex: NH4NO3 (s) —> NH4+ (aq) + NO3- (aq)
∆G < 0, ∆H > 0 (disfavored), ∆S > 0 (favored)
*if ∆G < 0, ∆S —> dominant term
3. ex: H2O (l) —> H2O (s)
∆G < 0, ∆H < 0 (favored), ∆S < 0 (disfavored)
*decrease in entropy, molecules are consolidated from liquid form to solid form
*if ∆G < 0, ∆H < 0 —> dominant term
4. ex: at 20 degrees celsius, the melting of ice is spontaneous and why?
H2O (s) —> H2O (l)
Spontaneous? Yes (∆G < 0)
Enthalpy? Endothermic (∆H < 0; disfavors spontaneity)
Entropy? More Disorder (∆S > 0 favors spontaneity)
5. ex: aA + bB <—> cC + dD
*[X] = concentration of ‘X’ in M (molarity)
K = (products)/(reactants) = [D]d [C]c / [A]a [B]b
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6. ex: predict ∆G for the following reaction:
H2 (g) + I2 (g) <—> 2HI (g) + 10kJ/mol
answer: ∆G < 0 (exergonic)
Equilibrium Constant
• Keq = (products)/(reactants)
• Measured in ‘M’ (molarity, concentration)
• Unitless
• solids/liquids don’t appear in Keq
• unit of 1
• gases will USUALLY be expressed as concentrations
• however, can be converted from pressure (using PV=nRT)
• if products > reactants, keq > 1
• “lies to the right” - concentration of products is favored over the concentration of reactants
• if products < reactants, keq < 1
• “lies to the left” - concentration of reactants is favored over the concentration of products
1. ex: H2 (g) + I2 (g) <—> 2HI (g) + 10kJ/mol
answer: Keq = (HI)2/(H2)(I2)
2. ex: 2H2 (g) + O2 (g) <—> 2H2O (g)
answer: Keq = (H2O)2/(H2)2(O2)
3. ex: 2H2 (g) + O2 (g) <—> 2H2O (l)
answer: Keq = 1/(H2)2(O2)
4. ex: predict Keq for the following reaction: H2 (g) + I2 (g) <—> 2HI (g) + 10kJ/mol
answer: Keq> 1 because the concentration of 2HI > the concentration of H2 + I2
5. ex: H2 and O2 react in a sealed container at 125 degrees celsius and reach equilibrium concentrations of
H2 = 0.10M, O2 = 0.10M and H2O = 5.0M. Calculate Keq.
2H2 (g) + O2 (g) <—> 2H2O (g) + heat
work/answer:
Keq = (H2O)2/(H2)2(O2) = (5.0)2/(0.10)2(0.10) = 25/0.001 = 2.5x104 >>> 1
^very exothermic reaction
6. ex: if the equilibrium concentration of H2 is 0.010M and O2 is 0.0040M, what is [H2O]eq in M? Keq =
2.5x104 at 125 degrees celsius.
2H2 (g) + O2 (g) <—> 2H2O (g) + heat
work/answer:
Keq = (H2O)2/(H2)2(O2)
2.5x104 = (H2O)2/(0.010)2(0.0040) = 0.1M
Le Chatelier’s Principle
• *Removing H+ will shift the reaction right
• A<—> B
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Document Summary
Rate of reaction - (change in concentration)/(change in time, reaction mechanism, 1. Reactant particles must collide with a specific orientation: ex: ab + cd > ac + bd, a c, b d, collide in this orientation, 3. C6h12o6 (s) + 6o2 (g) > 6co2 (g) + 6h2o (g) Sublimation of dry ice: co2 (s) > co2 (g) N2 (g) + 3h2 (g) > 2nh3 (g) G = h - t( s) direct relationship between g and h. The effect of h and s on g. Sign of h / sign of s / sign of g. (exothermic) / - (increase in entropy) / - (exergonic) = spontaneous. + (endothermic) / - (decrease in entropy) / + (endergonic) = nonspontaneous. + (endothermic) / + (increase in entropy) / more info needed = more info needed. (exothermic) / - (decrease in entropy) / more info needed = more info needed.
