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DETERMINING THE RATE
EXPONENT
DETERMINING THE RATE EXPONENT
How would you set up
an experiment to
determine the rate
exponent for the
following reaction?
C6H12O6(s) + H2SO4(aq)
DETERMINING THE RATE EXPONENT
Exponents for a rate law expression
must be determined experimentally
(empirically).
Concentrations of each reactant must be
changed one at a time to see how
much the overall reaction rate does or
doesn’t change.
Related to how the reaction takes place
at the atomic level, which we cannot
see.
DETERMINING THE RATE EXPONENT
HANDOUT - Table 1
Concentration
Factor Change
Reaction Rate
Factor Change
Rate Law Exponent
on the
Concentration
2x no change 0
3x no change 0
4x no change 0
2x 21
= 2 1
3x 31
= 3 1
4x 41
= 4 1
DETERMINING THE RATE EXPONENT
HANDOUT - Table 1
Concentration
Factor Change
Reaction Rate
Factor Change
Rate Law Exponent
on the
Concentration
2x 22
= 4 2
3x 32
= 9 2
4x 42
= 16 2
2x 23
= 8 3
3x 33
= 27 3
4x 43
= 64 3
DETERMINING THE RATE EXPONENT
HANDOUT - Example 1
Initial Concentration
(mol / L)
Initial Rate of
Formation of
Products (mol/L•s)
[A] [B]
0.10 0.10 0.20
0.20 0.10 0.40
0.30 0.10 0.60
0.30 0.20 2.40
0.30 0.30 5.40Order for [A]:
Order for [B]:
Order of overall rxn:
k =
1
2
3
2.0 x 102
L2
mol-2
s-1
r=k[A][B]2
DETERMINING THE RATE EXPONENT
HANDOUT - Example 2
1
1
2.2 x 10-5
s-1
Initial Concentration
(mol/L)
Initial Rate of Formation of
Products
(mol/L•s)
[SO2Cl2]
0.100 2.2 x 10-6
0.200 4.4 x 10-6
0.300 6.6 x 10-6
Order for [SO2Cl2]:
Order of overall rxn:
k =
r=k[SO2Cl2]
DETERMINING THE RATE EXPONENT
HANDOUT - Example 3
Initial Concentration
(mol/L) Initial Rate of
Formation of
Products
(mol/L•s)
[A] [B]
0.40 0.30 1.0 x 10-4
0.80 0.30 4.0 x 10-4
0.80 0.60 1.6 x 10-3
Order for [A]:
Order for [B]:
Order of overall rxn:
k =
2
2
4
6.9 x 10-3
L3
mol-3
s-1
r=k[A]2
[B]2
DETERMINING THE RATE EXPONENT
HANDOUT - Example 4
Order for [NO]:
Order for [H2]:
Order of overall rxn:
k =
2
1
3
1.2 L2
mol-2
s-1
r=k[NO]2
[H2]
Initial Concentration (mol/L)
Initial Rate of
Formation of
Products (mol/L•s)
[NO] [H2]
0.10 0.10 1.23 x 10-3
0.10 0.20 2.46 x 10-3
0.20 0.10 4.92 x 10-3
DETERMINING THE RATE EXPONENT
Ex. 1 The decomposition of N2O5 is:
2 N2O5  4 NO2 + O2
This is a first order rxn with respect to N2O5. Given
initial rate of 2.1 x 10-4
mol/L•s and initial [N2O5] =
0.40 mol / L, predict the rate when [N2O5] = 0.80 M.
2.1 x 10-4
mol L-1
s-1
= k [0.40mol/L]
2.1 x 10-4
mol L-1
s-1
= k
0.40mol/L
0.000525 s-1
= k
r = k [N2O5]
DETERMINING THE RATE EXPONENT
Ex. 1 The decomposition of N2O5 is:
2 N2O5  4 NO2 + O2
This is a first order rxn with respect to N2O5. Given
initial rate of 2.1 x 10-4
mol/L•s and initial [N2O5] =
0.40 mol / L, predict the rate when [N2O5] = 0.80 M.
r = k [N2O5]
= 0.000525 s-1
[0.80mol/L]
= 4.2x10-4
mol L-1
s-1
Notice how this is a first order reaction, so doubling
the concentration DOUBLES the rate of reaction
Therefore the rate is 4.2x10-4
mol L-1
s-1
DETERMINING THE RATE EXPONENT
Since this is a second order reaction, doubling the concentration
quadruples the rate.
2 x [ ] = r x 4
Therefore the rate of reaction is 1.3x102
mmol/L•min
Ex. 2 Given 2C4H6  C8H12 where r = k[C4H6]2
.
If the initial rate was 32 mmol C4H6/L•min at a certain
[C4H6], what would be the initial rate if the
concentration were doubled?
DETERMINING THE RATE EXPONENT
half-life - the time required for half the
concentration of a reactant to be used up in a rxn
Ex. 3 If the mass of an antibiotic is 2.464 g, what
mass will remain after 6.0 h, if the half-life is 2.0 h,
and no more antibiotic is added.
mfinal = minitial (0.5)n n=# of half lives
mfinal = 2.464g (0.5)6.0/2.0
= 2.464g (0.5)3
= 0.308g
= 0.31g
Therefore 0.31g of antibiotic remains
DETERMINING THE RATE EXPONENT
Equation relating half-life and the rate constant:
kt =0.6931
2
The radioisotope lead-212 has half-life of 10.6 h.
What is the rate constant for this isotope?
r = k[Pb-212]1
kt = 0.6931
2
and
k = 0.693
t1
2
k = 0.693
10.6h
k = 0.0654h-1
.: the rate
constant is
0.0654h-1
Can ONLY be
applied to a
FIRST ORDER
reaction

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Tang 07 determining the rate exponent

  • 2. DETERMINING THE RATE EXPONENT How would you set up an experiment to determine the rate exponent for the following reaction? C6H12O6(s) + H2SO4(aq)
  • 3. DETERMINING THE RATE EXPONENT Exponents for a rate law expression must be determined experimentally (empirically). Concentrations of each reactant must be changed one at a time to see how much the overall reaction rate does or doesn’t change. Related to how the reaction takes place at the atomic level, which we cannot see.
  • 4. DETERMINING THE RATE EXPONENT HANDOUT - Table 1 Concentration Factor Change Reaction Rate Factor Change Rate Law Exponent on the Concentration 2x no change 0 3x no change 0 4x no change 0 2x 21 = 2 1 3x 31 = 3 1 4x 41 = 4 1
  • 5. DETERMINING THE RATE EXPONENT HANDOUT - Table 1 Concentration Factor Change Reaction Rate Factor Change Rate Law Exponent on the Concentration 2x 22 = 4 2 3x 32 = 9 2 4x 42 = 16 2 2x 23 = 8 3 3x 33 = 27 3 4x 43 = 64 3
  • 6. DETERMINING THE RATE EXPONENT HANDOUT - Example 1 Initial Concentration (mol / L) Initial Rate of Formation of Products (mol/L•s) [A] [B] 0.10 0.10 0.20 0.20 0.10 0.40 0.30 0.10 0.60 0.30 0.20 2.40 0.30 0.30 5.40Order for [A]: Order for [B]: Order of overall rxn: k = 1 2 3 2.0 x 102 L2 mol-2 s-1 r=k[A][B]2
  • 7. DETERMINING THE RATE EXPONENT HANDOUT - Example 2 1 1 2.2 x 10-5 s-1 Initial Concentration (mol/L) Initial Rate of Formation of Products (mol/L•s) [SO2Cl2] 0.100 2.2 x 10-6 0.200 4.4 x 10-6 0.300 6.6 x 10-6 Order for [SO2Cl2]: Order of overall rxn: k = r=k[SO2Cl2]
  • 8. DETERMINING THE RATE EXPONENT HANDOUT - Example 3 Initial Concentration (mol/L) Initial Rate of Formation of Products (mol/L•s) [A] [B] 0.40 0.30 1.0 x 10-4 0.80 0.30 4.0 x 10-4 0.80 0.60 1.6 x 10-3 Order for [A]: Order for [B]: Order of overall rxn: k = 2 2 4 6.9 x 10-3 L3 mol-3 s-1 r=k[A]2 [B]2
  • 9. DETERMINING THE RATE EXPONENT HANDOUT - Example 4 Order for [NO]: Order for [H2]: Order of overall rxn: k = 2 1 3 1.2 L2 mol-2 s-1 r=k[NO]2 [H2] Initial Concentration (mol/L) Initial Rate of Formation of Products (mol/L•s) [NO] [H2] 0.10 0.10 1.23 x 10-3 0.10 0.20 2.46 x 10-3 0.20 0.10 4.92 x 10-3
  • 10. DETERMINING THE RATE EXPONENT Ex. 1 The decomposition of N2O5 is: 2 N2O5  4 NO2 + O2 This is a first order rxn with respect to N2O5. Given initial rate of 2.1 x 10-4 mol/L•s and initial [N2O5] = 0.40 mol / L, predict the rate when [N2O5] = 0.80 M. 2.1 x 10-4 mol L-1 s-1 = k [0.40mol/L] 2.1 x 10-4 mol L-1 s-1 = k 0.40mol/L 0.000525 s-1 = k r = k [N2O5]
  • 11. DETERMINING THE RATE EXPONENT Ex. 1 The decomposition of N2O5 is: 2 N2O5  4 NO2 + O2 This is a first order rxn with respect to N2O5. Given initial rate of 2.1 x 10-4 mol/L•s and initial [N2O5] = 0.40 mol / L, predict the rate when [N2O5] = 0.80 M. r = k [N2O5] = 0.000525 s-1 [0.80mol/L] = 4.2x10-4 mol L-1 s-1 Notice how this is a first order reaction, so doubling the concentration DOUBLES the rate of reaction Therefore the rate is 4.2x10-4 mol L-1 s-1
  • 12. DETERMINING THE RATE EXPONENT Since this is a second order reaction, doubling the concentration quadruples the rate. 2 x [ ] = r x 4 Therefore the rate of reaction is 1.3x102 mmol/L•min Ex. 2 Given 2C4H6  C8H12 where r = k[C4H6]2 . If the initial rate was 32 mmol C4H6/L•min at a certain [C4H6], what would be the initial rate if the concentration were doubled?
  • 13. DETERMINING THE RATE EXPONENT half-life - the time required for half the concentration of a reactant to be used up in a rxn Ex. 3 If the mass of an antibiotic is 2.464 g, what mass will remain after 6.0 h, if the half-life is 2.0 h, and no more antibiotic is added. mfinal = minitial (0.5)n n=# of half lives mfinal = 2.464g (0.5)6.0/2.0 = 2.464g (0.5)3 = 0.308g = 0.31g Therefore 0.31g of antibiotic remains
  • 14. DETERMINING THE RATE EXPONENT Equation relating half-life and the rate constant: kt =0.6931 2 The radioisotope lead-212 has half-life of 10.6 h. What is the rate constant for this isotope? r = k[Pb-212]1 kt = 0.6931 2 and k = 0.693 t1 2 k = 0.693 10.6h k = 0.0654h-1 .: the rate constant is 0.0654h-1 Can ONLY be applied to a FIRST ORDER reaction