Rate Law 5-2

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Rate Law5-2

• an expression which relates the rate to the
  concentrations and a specific rate constant

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• For a general reaction between reactant A and B
  at a constant temperature the reaction can be
  represented by
• aA bB -------gt products
• Rate law is expressed as
• Rate kAm Bn

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• The rate law equation expresses the relationship
  between the concentration of the reactants and
  the rate of the equation.

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Rate constant (k) Exponents m and n
1 Varies with temperature Constant with temperature
2 Constant under constant conditions Can only be determined experimentally
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Exponents and Orders of reactions.

• The values of exponents (m or n) can only be
  determined through experimentation and may or may
  not be the values of the coefficients in the
  balanced chemical equation.

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• The value of the exponents determines the order
  of the reaction. If a reaction has a single
  reactant and the value of the exponent is one
  then it is first order.
• If the exponent were two then it is second order.
  
• If more than one reactant is present in a
  reaction, the sum of the exponents (m n) is
  called the overall reaction order.

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• 2 N2O5 (g) ? 4 NO2 (g) O2 (g)
• Experimentally the rate was found to be first
  order for N2O5 (g).
• Rate k N2O5 (g) 1

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• The value of exponent is not the same as the
  balance for N2O5 in the equation.
• The overall reaction order would be first order.

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• NO2 (g) CO (g) ? NO (g) CO2 (g)
• Experimentally the rate was found to be second
  order for NO2 (g) and zero order for CO (g). The
  rate law can be written
• Rate k NO2 2 CO 0
• or Rate k NO2 2

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• The value of the exponents is not the same as the
  coefficients.
• The overall reaction order is second order ( 2
  0 ).

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• 2 HI (g) ? H2 (g) I2 (g)
• Experimentally the rate was found to be second
  order for HI (g).
• Rate k HI 2

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Assignment 1 Rate Law Equations
Reaction Rate dependent on Rate law Reaction order
1 A B -----gt products A 2 B1
2 2NO(g) O2 (g) ----gt NO2 (g) NO1 O21
kA 2 B1
3
kNO1O21
2
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3 2NO(g) 2H2 (g) ----gt N2(g) 2 H2O(g) NO2 H21
4 2NO2 (g) Cl2 (g) ------gt 2NO2Cl(g) NO21 Cl21
5 2ClO2 (aq) 2OH - ------gt ClO3- (aq) ClO2- (aq) H2O (l) ClO20 OH1
6 C D ------gt products C2 D0
kNO2 H21
3
kNO1 Cl21
2
kClO0 OH1
1
kC 2 D0
2
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• Determination of Rate Exponents using Initial
  rates of reactions.
• Chemists can examine the change in the initial
  rates of the reaction when concentrations of
  reactants are changed to determine the order of
  reactants (for a specific equation).

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Order of Reaction
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Order of reactants Change in intial rate when concentration is doubled



Doubles the rate of reaction
First order
Quadruples the rate of the reaction
second order
No change in the rate of the reaction
zero order
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• In the following questions determine
• the order of each reactant
• and the overall rate
• the value of the rate constant
• All the experiments were conducted under
  conditions of constant temperature.

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a.. Single reactant A -----gt B C
the following data was collected
Exp Initial A (mol/L) Initial rate (mol/Ls)
1 0.01 4.8 x 10-6
2 0.02 9.6 x 10-6
3 0.03 1.4 x 10-5
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Finding the rate law and the constant

• Choose the data from the table to solve for the
  exponent and the constant.
• Rate law k A m
• Choose two experimental data and then divide one
  by the other to find the exponent
• Calulate K by putting the experimental data into
  the rate law including the value of the expnent
  and solve for K.

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• Rate law k A m
• Exp1 4.8 x 10-6 k 0.01m
• Exp 2 9.6 x 10-6 k 0.02m
• Exp1 1 1m
• Exp 2 2 2m
• m 1

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Finding k

• Chose one set of data and solve
• Exp 1
• Exp1 4.8 x 10-6 k 0.011
• 4.8 x 10-6 k 0.011
• 0.01 0.01
• 4.8 x 10-4 k

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• Rate law 4.8 x 10-4 A1

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b. Two reactants
exp Initial A Initial B Initial rate
1 0.01 0.03 2.4 x 10-4
2 0.03 0.03 7.2 x 10-4
3 0.01 0.06 2.4 x 10-4
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• Choose the data such that the concentration of
  one of the reactants, A, is the same in both the
  experimental data.
• In this case exp 1 and exp 2

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Rate law k AmBn

• Exp1 2.4 x 10-4 k 0.01m 0.03n
• Exp 3 2.4 x 10-4 k 0.01m0.06n
• 1 ( 1/2 ) n
• n 0

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Solve for m now

• Exp1 2.4 x 10-4 k 0.01m 0.03n
• Exp 2 7.2 x 10-4 k 0.03m0.03n
• 1/3 ( 1/3 ) m
• m 1

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Rate law k A1B1

• Solve for k
• Exp1 2.4 x 10-4 k 0.011 0.030
• 0.01 0.01
• 2.4 x 10-2 k
• Rate law 2.4 x 10-2 A1B0

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2 ICl H2 ? I2 2 HCl
exp ICl H2 Initial rate
1 0.10 0.01 0.002
2 0.20 0.01 0.004
3 0.10 0.04 0.008
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Rate law k IClmH2n

• Exp1 0.002 k 0.10m 0.01n
• Exp 2 0.004 k 0.20m0.01n
• 1/2 (1/2)m
• m 1

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• Exp 1 0.002 k 0.10m 0.01n
• Exp 3 0.008 k 0.10m0.04n
• 1/4 (1/4)n
• n 1

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Solve for k

• Exp1 0.002 k 0.10 1 0.01 1
• Exp1 0.002 k 0.10 1 0.01 1
• 0.001 0.001
• 2 k

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• Rate law 2 ICl 1 H2 1

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• 1. For the general reaction
• A B -----gt C
• the following data was collected

Exp A (mol/L) B (mol/L) Initial rate (mol/Ls)
1 0.002 0.05 2 x 10-5
2 0.004 0.05 8 x 10-5
3 0.002 0.10 2 x 10-5
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Answer

• Rate law 5 A 2 B 0

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• 2. For the reaction
• H2O2 (aq)2HI (aq)? 2H2O (l)I2 (aq)the following
  data was collected

exp H2O2 (mol/L HI (mol/L) Initial rate (mol/Ls)
1 0.05 0.05 0.002
2 0.05 0.10 0.004
3 0.10 0.05 0.004
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answer

• Rate law 0.8 H2O2 1 HI1

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• 3. For the reaction
• (CH3)3Br OH- ?(CH3)3COH Br-
• the following data was collected

exp (CH3)3CBr (mol/L) OH- (mol/L) Initial rate (mol/Ls)
1 0.03 0.04 1.2 x 10-3
2 0.03 0.08 1.2 x 10-3
3 0.06 0.04 2.4 x 10-3
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answer

• Rate law
• 0.04 (CH3)3CBr 1 OH 0