6.5 Rate Law.

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Presentation transcript:

6.5 Rate Law

Video CrashCourse on Kinetics https://www.youtube.com/watch?v=7qOFtL3VEBc

Key Words Rate Law Equation Rate Constant Order of Reaction Total Order of Reaction

Rate Equations and Order of Reactions

rate law or rate equation For the reaction aA + bB  cC + dD Rate  k[A]x[B]y rate law or rate equation

Rate  k[A]x[B]y For the reaction aA + bB  cC + dD where x and y are the orders of reaction with respect to A and B x  y is the overall order of reaction.

Rate  k[A]x[B]y For the reaction aA + bB  cC + dD x = 0  zero order for A x = 1  first order for A x = 2  second order for A y = 0  zero order for B y = 1  first order for B y = 2  second order for B

The reaction is zero order for A and second order for B. For the reaction aA + bB  cC + dD Describe the reaction with the following rate law. Rate  k[B]2 The reaction is zero order for A and second order for B.

Rate  k[A]x[B]y k is the rate constant For the reaction aA + bB  cC + dD Rate  k[A]x[B]y k is the rate constant Temperature-dependent Can only be determined from experiments

mol dm3 s1/(mol dm3)x+y or, mol dm3 min1 /(mol dm3)x+y For the reaction aA + bB  cC + dD Rate  k[A]x[B]y units of k : - mol dm3 s1/(mol dm3)x+y or, mol dm3 min1 /(mol dm3)x+y

Rate  k[A]0[B]0 units of k = mol dm3 s1/(mol dm3)0+0 For the reaction aA + bB  cC + dD Rate  k[A]0[B]0 units of k = mol dm3 s1/(mol dm3)0+0 = mol dm3 s1 = units of rate

Rate  k[A][B]0 units of k = mol dm3 s1/(mol dm3)1+0 = s1 For the reaction aA + bB  cC + dD Rate  k[A][B]0 units of k = mol dm3 s1/(mol dm3)1+0 = s1

Rate  k[A][B] units of k = mol dm3 s1/(mol dm3)1+1 = mol1 dm3 s1 For the reaction aA + bB  cC + dD Rate  k[A][B] units of k = mol dm3 s1/(mol dm3)1+1 = mol1 dm3 s1 The overall order of reaction can be deduced from the units of k

mol dm3 s1/(mol dm3)x+y+z+… For the reaction aA + bB + cC + …  products Rate  k[A]x[B]y[C]z… units of k : - mol dm3 s1/(mol dm3)x+y+z+…

Determination of rate equations To determine a rate equation is to find k, x, y, z,… Rate  k[A]x[B]y[C]z… Two approaches : - Initial rate method Graphical method

Determination of Rate Equations by Initial Rate Methods

rate  k[Cl(aq)]x[ClO3(aq)]y[H+(aq)]z 5Cl(aq) + ClO3(aq) + 6H+(aq)  3Cl2(aq) + 3H2O(l) Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 rate  k[Cl(aq)]x[ClO3(aq)]y[H+(aq)]z

From experiments 1 and 2, = 2z 4 = 2z  z = 2 Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 From experiments 1 and 2, = 2z 4 = 2z  z = 2

From experiments 2 and 3, = 2y 2 = 2y  y = 1 Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 From experiments 2 and 3, = 2y 2 = 2y  y = 1

From experiments 1 and 4, = 2x 2 = 2x  x = 1 Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 From experiments 1 and 4, = 2x 2 = 2x  x = 1

rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2 Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2 From experiment 1, 1.0105  k(0.15)(0.08)(0.20)2 k = 0.02 mol3 dm9 s1

rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2 Expt [Cl(aq)] / mol dm3 [ClO3(aq)] [H+(aq)] Initial rate / mol dm3 s1 1 0.15 0.08 0.20 1.0105 2 0.40 4.0105 3 0.16 8.0105 4 0.30 2.0105 rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2 From experiment 2, 4.0105  k(0.15)(0.08)(0.40)2 k = 0.02 mol3 dm9 s1

(a) rate  k[C]x[D]y[E]z Q.15 2C + 3D + E  P + 2Q Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (a) rate  k[C]x[D]y[E]z

(a) rate  k[C]x[D]y[E]z Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (a) rate  k[C]x[D]y[E]z From experiments 1 and 2, = 2x 8 = 2x  x = 3

(a) rate  k[C]x[D]y[E]z Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (a) rate  k[C]x[D]y[E]z From experiments 1 and 3, = 2y 1 = 2y  y = 0

(a) rate  k[C]x[D]y[E]z Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (a) rate  k[C]x[D]y[E]z From experiments 1 and 4, = 3z 9 = 3z  z = 2

(a) rate  k[C]3[D]0[E]2 = k[C]3[E]2 Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (a) rate  k[C]3[D]0[E]2 = k[C]3[E]2

(b) rate  k[C]3[E]2 From experiment 1, 3.0103  k(0.10)3(0.10)2 Expt [C] / mol dm3 [D] [E] Initial rate / mol dm3 s1 1 0.10 3.0103 2 0.20 2.4102 3 4 0.30 2.7102 (b) rate  k[C]3[E]2 From experiment 1, 3.0103  k(0.10)3(0.10)2 k = 300 mol4 dm12 s1

Initial concentration Q.16 H+ CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I(aq) Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 (a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z

Initial concentration Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 (a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z From experiments 1 and 2, = 1.67x 1 = 1.67x  x = 0

Initial concentration Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 (a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z From experiments 1 and 4, = 2y 2 = 2y  y = 1

Initial concentration Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 (a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z From experiments 3 and 4, = 2z 2 = 2z  z = 1

Initial concentration Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)] = k[CH3COCH3(aq)][H+(aq)]

Initial concentration Initial rate / mol dm3 s1 Initial concentration / mol dm3 [I2(aq)] [CH3COCH3(aq)] [H+(aq)] 3.5 105 2.5104 2.0101 5.0103 1.5104 1.4 104 4.0101 1.0102 7.0 105 (b) Rate = k[CH3COCH3(aq)][H+(aq)] From experiment 1, 3.5105  k(2.0101)(5.0103) k = 0.035 mol1 dm3 s1

Task Review Questions 1, 3 and 4 on page 382.