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O.A # 46

O.A # 46. Reaction rate. O.A # 47. Activation Energy. O.A # 48. Catalyst. O.A # 49. Collision Theory. O.A # 50. Inhibitor. Page 27-. C h 17: Rates of Reaction. Rate : measure of changes in a chemical rxn over time Can be measured by: Time to complete rxn

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O.A # 46

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  1. O.A # 46 Reaction rate

  2. O.A # 47 Activation Energy

  3. O.A # 48 Catalyst

  4. O.A # 49 Collision Theory

  5. O.A # 50 Inhibitor

  6. Page 27- Ch 17: Rates of Reaction Rate: measure of changes in a chemical rxn over time Can be measured by: • Time to complete rxn • Rate of disappearance of reactants • Rate of appearance of products

  7. Rate Law Rate = K [R]x K = rate constant. ↑ temp ↑ K [R] = concentration of reactants x = coefficients in balanced equation The [R] is raised to power of the appropriate coefficients in balanced equation

  8. Don’t include SOLIDS (s) and LIQUIDS (L) • GASES (g) and AQUEOUS SOLUTIONS (aq) only! • Left side of reaction • Rate = K [R]x Write the rate expression for the following: 1. __Na(s) + ___Cl2(g) → ___NaCl(s) 2 2 Rate = K [Cl2] 2. ___NO(g) + ___Cl2(g) → ___NOCl(g) 2 2 Rate = K [NO]2 [Cl2]

  9. a) Write rate expression Rate = K [HC2H3O2 ] b) Calculate rate if K = 3.5 x 10-5 /min and [HC2H3O2 ]=4.0M 3. HC2H3O2(aq) + H2O(l) → H3O+(aq) + C2H3O2(aq) Rate = 3.5 x 10-5/min x 4.0M = 0.00014M/min c) If you start with 2.0 M, how long will it take 0.30M of HC2H3O2 to react? Rate = 3.5 x 10-5/min x 2.0M= 0.00007M/min 1 min_____ 0.00007M X = (0.30 M x 1 min) 0.00007M = X____ 0.30 M X = 4300 min

  10. Page 28- Factors Affecting Rate • Nature of Reactants: shape, surface area, size, electronegativity • Concentration of Reactants (M): - particles must meet to collide - ↑ [ ] ↑ collisions

  11. Temperature: - must have enough KE to react - usually ↑ temp. ↑ rate • Catalyst: substance that ↑ rate of reaction without being consumed by the rxn

  12. Activation Energy (Ea): Minimum amount of energy colliding molecules must have in order to react

  13. Activated Complex H2 + I2→ 2 HI Activated complex:- middle step - Short-lived molecule -High PE, but unstable Original molecules Forms two new molecules of HI

  14. O.A # 51 Reversible Reaction

  15. O.A # 52 Chemical Equilibrium

  16. O.A # 53 Law of chemical equilibrium

  17. O.A # 54 Le Chatelier’s Principle

  18. O.A # 55 Equilibrium Constant

  19. Conditions of EQ Page 29- Ch 18: Equilibrium (EQ) • Closed system (forward and reverse rxn ↔ ) • Both rxns at same time and rate • Temp, color, pressure and [ ] are constant

  20. KEQ: Equilibrium Constant • Constant value for rxns at same temp. KEQ = [P]x [R]y KEQ > 1: Products are favored KEQ < 1: Reactants are favored

  21. N2(g) + H2(g) ↔ NH3(g) 3 2 1. Calculate KEQ: KEQ = [P]x = [R]y [N2] = 2.12 M, [H2]= 1.75 M, [NH3]= 84.3 M (84.3)2 [NH3]2__ = [N2] [H2]3 (2.12) (1.75)3 KEQ = 625 2. Products or reactants are favored? KEQ > 1: Products are favored

  22. 3. Find [NH3] if [N2] = 5.31M, [H2]= 2.51M KEQ = 625 KEQ = [NH3]2__ [N2] [H2]3 625 = X2 (5.31) (2.51)3 X2 = (625)(5.31) (2.51)3 =52480.22676 √X2 = √52480.22676 X = [NH3]= 229 M

  23. Page 30- Le Chatelier’s Principle If a stress is added to a system at EQ, the system counteracts that change and reaches a new state of EQ. Shifts right • Products are favored • Forward rxn speeds up Shifts left • Reactants are favored • Reverse rxn speeds up Favored = greater [ ] concentration

  24. Pressure:only gases are affected H2 (g) + I2(S) 2 HI (g) ↑ pressure: shifts toward the lesser amount of moles (COEFFICIENTS) to relieve pressure ↓ pressure: is the opposite 1:2 Catalyst: Increases rate of both rxns (↔), favors neither rxn, doesn’t create a stress

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