Kinetics Thermodynamics : spontaneity of reaction

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Presentation transcript:

Kinetics Thermodynamics : spontaneity of reaction G < 0 spontaneous 2H2(g) +O2(g)  2H2O(l) Go = -474 kJ Go = Ho - TSo a) low T b) high T Ho < 0 favorable So < 0 unfavorable Thermodynamically spontaneous very slow

Kinetics Kinetics: rates of reaction mechanism of reaction 2 H2(g) + O2(g) 2 H2(g) +  2 H2O(l) Greaction < 0 Greactants Gproducts thermodynamics kinetics

P.E.n-e Kinetics qn qe r 2H2(g) + O2(g) 2 H2O(l) Go = -474 kJ spontaneous + - + - P.E.n-e - + qn qe r

Low Temperature

High Temperature

High Temperature endothermic P.E. exothermic r (distance) + -

Kinetics rate of reaction [reactant] decrease increase [product] 1. Temperature (K.E.) 2. Concentration 3. Orientation

differential rate laws A + B  C a) + b) - - [A] = - [B] =  t [C]  t rate = +  t A + B  2 C - [A] =  t rate = - [B] =  t [C]  t 1 2  A + B  C +

[ ] 50 100 x t (min) 100 100 x 1 72 72 28 x 2 52 52 48 x x 3 37 37 63 x 4 27 27 73 time (min) 1 2 3 4 5 5 19 19 81 [ ] = t 81 - 0 = 16.2 min-1 ave. rate = 5 - 0 ave. rate = -[ ] = t - (19 ) - 100 = 16.2 min-1 5 - 0

x time (min) 1 2 3 4 5 [ ] 50 100 t (min) 100 1 72 28 2 52 52 48 3 37 37 63 4 27 27 73 5 19 19 81 ave. rate = - = 16.2 min-1 overall (19 - 100) 5 - 0 ave. rate = 1st minute - (72 ) - 100 = 28.0 min-1 1 - 0 ave. rate = - 27) = 8.0 min-1 5th minute - (19 5 - 4

slope of line tangent to curve = instantaneous rate x time (min) 1 2 3 4 5 [ ] 50 100 slope of line tangent to curve = instantaneous rate at t = 0 initial rate fastest rate

k = rate constant rate  [reactant] = k [reactant] t instantaneous (min) rate k = rate [reactant] 100 33.0 0.33 1 72 23.8 2 52 17.2 3 37 12.2 4 27 8.9 5 19 6.3 0.33 0.33 0.33 0.33 0.33 rate  [reactant] = k [reactant] k = rate constant

[A]a [B]b [C]c ... rate = (-D[A]) k = D t k = rate [reactant] k independent of [reactants] k dependent on Temperature [A]a [B]b [C]c ... rate = (-D[A]) k = D t k, a, b and c determined experimentally isolation method