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This topic cuts across the molecular reaction taking place in the biochemistry of living systems and the rates and conditions in which they occur
Typology: Thesis
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Learning goals and key skills:
Understand the factors that affect the rate of chemical reactions Determine the rate of reaction given time and concentration Relate the rate of formation of products and the rate of disappearance of reactants given the balanced chemical equation for the reaction. Understand the form and meaning of a rate law including the ideas of reaction order and rate constant. Determine the rate law and rate constant for a reaction from a series of experiments given the measured rates for various concentrations of reactants. Use the integrated form of a rate law to determine the concentration of a reactant at a given time. Explain how the activation energy affects a rate and be able to use the Arrhenius Equation. Predict a rate law for a reaction having multistep mechanism given the individual steps in the mechanism. Explain how a catalyst works.
C (^) (diamond) → C (^) (graphite) ∆G°rxn = -2.84 kJ spontaneous!
C (^) (graphite) + O2 (g) → CO2 (g) ∆G°rxn = -394.4 kJ spontaneous!
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.
Measured from the change in concentration of reactants or products per unit time.
Reaction rates are affected by
C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq )
C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq )
rate = change in conc. / change in time
aA + bB → cC + dD
The rate can be related to the concentration of the reactants or products. Reactants will have a negative sign. Don’t forget to use the stoichiometric coefficients.
Ammonia reacts with oxygen to form nitrogen monoxide and steam. Relate the reactant and product reaction rates.
Rate laws show the relationship between the reaction rate and the concentration of reactants (and sometimes products).
aA + bB cC + dD
rate equation:
Rate = R = k [A]x^ [B]y
x and y are experimentally determined. They are not the stoichiometric coefficients!
Rate = R = k [A]x^ [B]y
k is the rate constant for this reaction.
x = order of the reaction with respect to the concentration of reactant A
y = order of the reaction with respect to the concentration of reactant B
x+y = overall reaction order or (total) order of the reaction.
t 0
st (^) order
The half-life, t1/2 is the time needed for the concentration of a reactant to decrease to half of its initial value.
[A]t1/2 = ½[A] 0
t
A A k
A A k k k k
t
1 2
0
(^12) 0 0
1 2 2 0 693 /
ln [ ] [ ] ln [ ] [ ] ln^ ln. = − = − = − = =
Example
The 1st^ order rate constant for the decomposition of N 2 O 5 to NO 2 and O 2 at 70 °C is 6.82 × 10 -3^ s-1. What is the half-life of this reaction at 70 °C?
Radioactive decay follows first-order kinetics. Rate of decay of radioactive isotopes are given in terms of a half-life.
(^238) U → 234 Th + α 4.5 x 10 (^9) years (^14) C → 14 N + β 5730 years (^131) I → 131 Xe + β 8.05 days
Element 106 - seaborgium (^263) Sg 0.9 s
isotope half-life uses C-11 20.33 min PET scans Na-24 14.951 h cardiovascular system tracer K-40 1.248 × 109 yr dating of rocks Fe-59 44.495 d red blood cell tracer Co-60 5.2712 yr radiation therapy Tc-99 6.006 h biomedical imaging Ra-226 1600 yr radiation therapy Am-241 432.2 yr smoke detectors
for a reaction to proceed,
Svante Arrhenius developed a mathematical relationship between k and Ea :
where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.
Taking the natural logarithm of both sides, the equation becomes
ln k = - Ea/R ( 1 ) + ln A T
Example
For the reaction CO + NO 2 → CO 2 + NO k =0.220 M-1^ s-1^ at 650. K and k =1.30 M- s-1^ at 700. K. Show that the activation energy for this reaction is 134 kJ/mol.
Most reactions have more than one ELEMENTARY step. The NET reaction is the sum of all the elementary steps.
Most rxns. involve a sequence of elementary steps. 2 I-^ + H 2 O 2 + 2 H+^ → I 2 + 2 H 2 O Rate = k [I-] [H 2 O 2 ] NOTE
Step 1: NO 2 + NO 2 → NO 3 + NO (slow) Step 2: NO 3 + CO → NO 2 + CO 2 (fast)
determining step , it does not appear in the rate law.
Step 2: NOBr 2 + NO → 2 NOBr (slow)
Step 1 includes the forward and reverse reactions.
Step 1: NO + Br 2 NOBr 2 (fast) k 1 k- k 2
k 2 k 1 Rate = (^) k − [NO] [Br 2 ] [NO]
Rate = k [NO]^2 [Br 2 ]
Rate of the reaction is controlled by the slow step — the rate-determining step (rds) or rate-limiting step. Elementary Step 1 is bimolecular and involves I-^ and H 2 O 2. Therefore, this predicts the rate law should be Rate ∝ [I-] [H 2 O 2 ] — as observed The species HOI and OH-^ are reaction intermediates.
2 I-^ + H 2 O 2 + 2 H+^ → I 2 + 2 H 2 O Rate = k [I-] [H 2 O 2 ]
Step 1 — slow H 2 O 2 + I-^ → HOI + OH- Step 2 — fast HOI + I-^ → I 2 + OH- Step 3 — fast 2 OH-^ + 2 H+^ → 2 H 2 O