kinetics lesson 7 catalysts & review
DESCRIPTION
Kinetics Lesson 7 Catalysts & Review. Commercial Catalysts H 2 SO 4 Many Organic Reactions PtSurface Catalyst for Diatomic Gases PdCatalytic Converter Automobile ConvertCO→CO 2 RhCatalytic Converter Automobile ConvertNO 2 →N 2 + O 2 Biological Catalysts - PowerPoint PPT PresentationTRANSCRIPT
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KineticsLesson 7 Catalysts
&Review
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Commercial Catalysts H2SO4 Many Organic Reactions
Pt Surface Catalyst for Diatomic Gases Pd Catalytic Converter Automobile
Convert CO → CO2
Rh Catalytic Converter Automobile
Convert NO2 → N2 + O2
Biological Catalysts Enzymes Biological Catalysts
Amylase
pepsin
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CuCO3(s) + 2HCl(aq) CuCl2(aq) + CO2(g) + H2O(l)
blueDescribe seven different ways to monitor the rate of the above reaction. State how each property would change as the reaction proceeds.
1. Mass of CaCO3(s) over time decreases
2. [HCl] over time decreases
3. [CuCl2] over time increases
4. Volume of CO2 over time increases
5. Mass of a open beaker over time decreases
6. Pressure of a closed beaker over time increases
7. Colour of the blue Cu2+ over time increases
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CuCO3(s) + 2HCl(aq) CuCl2(aq) + CO2(g) + H2O(l)
blue
Describe five different ways to increase the rate of the above reaction.
1. Increase the temperature
2. Increase [HCl]
3. Add a catalyst
4. Increase the surface area of CuCO3(s)
5. Agitate
We can't change the nature of the reactant because then we wouldn't have the same reaction. Replacing HCl with H2SO4 would be faster but a different reaction.
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl 1 min
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole 1 min 36.5 g
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole x 2 mole Al 1 min 36.5 g 6 mole HCl
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1 min 36.5 g 6 mole HCl 1 mole
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl 1 mole
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl 1 mole
Al remaining = 2.00 g - 1.48 g = 0.52 g
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2Al(s) + 6HCl(aq) → 3H2(g) + 2AlCl3(aq)
2.00 g of Al is placed in a beaker and allowed to react for 12.00 minuteswith 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170
12.00 min x 0.500 g HCl x 1mole x 2 mole Al x 27.0 g = 1.48 g Al 1 min 36.5 g 6 mole HCl 1 mole
Al remaining = 2.00 g - 1.48 g = 0.52 g
Beware subtraction- loss of 1 sig fig!
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
2.0 g NH3 min
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
2.0 g NH3 x 1 mole min 17.0 g
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
2.0 g NH3 x 1 mole x 3 mole H2 min 17.0 g 2 mole NH3
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
2.0 g NH3 x 1 mole x 3 mole H2 x 2.0 g min 17.0 g 2 mole NH3 1 mole
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N2 + 3H2 → 2NH3
The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.
2.0 g NH3 x 1 mole x 3 mole H2 x 2.0 g = 0.35 g min 17.0 g 2 mole NH3 1 mole min
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Label the Ea forward, Ea reverse, Activated complex, and ∆H.
Ea for
Energy of the Activated complex
Ea rev
∆H
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
Activated complex
reactants
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
Ea = PE activated complex - PE reactants
Activated complex
reactants
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
Ea =
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Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.
Ea = PE activated complex - PE reactants
Activated complex
reactants
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Why does this reaction have more than one step?
N2 + 3H2 → 2NH3
More than three reactant particles.
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A colllision that is not successful does not have:
Favourable geometry
Sufficient energy
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Define activated Complex
UnstableReaction IntermediateHigh PELow KEBonds forming and breaking
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Define Activation Energy
The minimum amount of energy required for a successful collision.
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Why does gasoline burn faster than wax?
Nature of the reactant.
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What happens to the KE, PE, and total energy as the activated complex forms?
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What happens to the KE, PE, and total energy as the activated complex forms?
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PE
What happens to the KE, PE, and total energy as the activated complex forms?
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What happens to the KE and PE as the activated complex turns into products?
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What happens to the KE, PE, and total energy as the activated complex turns into products?
products
Activated complex
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PE
products
Activated complex
What happens to the KE, PE, and total energy as the activated complex turns into products?
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What happens to the KE, PE, and total energy as the activated complex turns into products?
The PE decreases, the KE increases, and the total energy is constant.
PE
products
Activated complex
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If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ?
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If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ?
125 kJ
250 kJ
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If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ?
125 kJ
250 kJ
ΔH = -125 kJ
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What is the relationship between Ea and rate? Draw a graph.:
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What is the relationship between Ea and rate? Draw a graph.:
Inverse
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What is the relationship between Ea and rate? Draw a graph.:
Inverse
Ea
Rate
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According to the collision theory, how does increasing temperature increase the rate
More collisions and harder collisions
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According to the collision theory, how does increasing concentration increase the rate
More collisions
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According to the collision theory, how does a catalyst increase the rate
Lowers the Ea and allows low energy collisions to be successful.
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What is the formula of the activated complex in the reaction below:
CH3OH + HCl → CH3Cl + H2O
CH5OCl
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Why doesn't an unlit candle burn according to the collision theory?
The Ea is too high.
A match lights to candle
The match provides the Ea
The candle continues to burn.
Exothermic reaction