Kinetics Lesson 7 Catalysts & Review

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

CuCO3(s) + 2HCl(aq)  CuCl2(aq) + CO2(g) + H2O(l) blue Describe 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

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.

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 minutes with 2.00 M HCl. If the rate of consumption of HCl is 0.500 g/min, calculate the amount of Al remaining. 270.170

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 minutes with 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

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 minutes with 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

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 minutes with 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

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 minutes with 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

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 minutes with 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

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 minutes with 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

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 minutes with 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

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 minutes with 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!

N2 + 3H2 → 2NH3 The rate of formation of NH3 is 2.0 g/ min. Calculate the rate of consumption of H2 in g/min.

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

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

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

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

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

Label the Ea forward, Ea reverse, Activated complex, and ∆H. Energy of the Activated complex Ea for Ea rev ∆H

Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Activated complex reactants

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

Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.

Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants.

Describe the activation energy in terms of the PE of the activated complex and the PE of the reactants. Ea =

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

Why does this reaction have more than one step? N2 + 3H2 → 2NH3 More than three reactant particles.

A colllision that is not successful does not have: Favourable geometry Sufficient energy

Define activated Complex Unstable Reaction Intermediate High PE Low KE Bonds forming and breaking

Define Activation Energy The minimum amount of energy required for a successful collision.

Why does gasoline burn faster than wax? Nature of the reactant.

What happens to the KE, PE, and total energy as the activated complex forms?

What happens to the KE, PE, and total energy as the activated complex forms?

What happens to the KE, PE, and total energy as the activated complex forms?

What happens to the KE and PE as the activated complex turns into products?

What happens to the KE, PE, and total energy as the activated complex turns into products?

What happens to the KE, PE, and total energy as the activated complex turns into products?

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. Activated complex PE products

If the Ea (for) = 125 kJ and the Ea (rev) is 250 kJ, is the reaction exothermic or endothermic and what is ΔH ?

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

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

What is the relationship between Ea and rate? Draw a graph.:

What is the relationship between Ea and rate? Draw a graph.: Inverse

Rate Ea What is the relationship between Ea and rate? Draw a graph.: Inverse Ea Rate

According to the collision theory, how does increasing temperature increase the rate More collisions and harder collisions

According to the collision theory, how does increasing concentration increase the rate More collisions

According to the collision theory, how does a catalyst increase the rate Lowers the Ea and allows low energy collisions to be successful.

What is the formula of the activated complex in the reaction below: CH3OH + HCl → CH3Cl + H2O CH5OCl

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