activation energy. review of exothermic reactants ep is higher than products ep. now, we must...
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Activation energy
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Review of Exothermic
• Reactants Ep is higher than Products Ep.
• Now, we must consider the activation energy (the energy needed so that the reactants bonds will break and reform to make product)
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Review of Endothermic
• Reactants Ep is lower than Products Ep.
• Need to add more energy to the system for the forward reaction to take place.
• Still need to consider activation energy
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Activated complex is an unstable chemical species containing partially formed bonds representing the maximum potential energy point in the change
-also known as the transition state
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Activated Complex
• Is the short-lived, unstable structure formed during a successful collision between reactant particles.
• Old bonds of the reactants are in the process of breaking, and new products are forming
• Ea is the minimum energy required for the activation complex to form and for a successful reaction to occur.
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Fast and slow reactions
• The smaller the activation energy, the faster the reaction will occur regardless if exothermic or endothermic.
• If there is a large activation energy needed, that means that more energy (and therefore, time) is being used up for the successful collisions to take place.
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Reaction Mechanism
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Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) 2NO2 (g)
N2O2 is detected during the reaction!
Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
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Elementary step: NO + NO N2O2
Elementary step: N2O2 + O2 2NO2
Overall reaction: 2NO + O2 2NO2
+
Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Reaction Intermediates
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Rate Laws and Rate Determining Steps
Writing plausible reaction mechanisms:
• The sum of the elementary steps must give the overall balanced equation for the reaction.
• The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
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Molecularity
• The number of molecules that participate as reactants in an elementary step
• Unimolecular: a single molecule is involved.– Ex: CH3NC (can be rearranged)
– Radioactive decay– Its rate law is 1st order with respect to that
reactant
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• Bimolecular: Involves the collision of two molecules (that form a transition state that can not be isolated)– Ex: NO + O3 NO2 + O2
– It’s rate law is 1st order with respect to each reactant and therefore is 2nd order overall.
– Rate =k[NO][O3]
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Termolecular: simultaneous collision of three molecules. Far less probable.
• Some possible mechanisms for the reaction; 2A + B C + D
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Unimolecular reaction A products rate = k [A]
Bimolecular reaction A + B products rate = k [A][B]
Bimolecular reaction A + A products rate = k [A]2
Rate Laws and Elementary Steps
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A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
Ea k
uncatalyzed catalyzed
ratecatalyzed > rateuncatalyzed
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Energy Diagrams
Exothermic Endothermic
(a) Activation energy (Ea) for the forward reaction
(b) Activation energy (Ea) for the reverse reaction
(c) Delta H
50 kJ/mol 300 kJ/mol
150 kJ/mol 100 kJ/mol
-100 kJ/mol +200 kJ/mol
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The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1: NO2 + NO2 NO + NO3
Step 2: NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate? Catalyst?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2
NO2
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Write the rate law for this reaction. Rate = k [HBr] [O2]
List all intermediates in this reaction.
List all catalysts in this reaction.
HOOBr, HOBr
None
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Reaction mechanisms are only educated guesses at the behaviour of molecules, but there are three rules that must be followed in proposing a mechanisms:
1. each step must be elementary
2. The slowest step must be consistent with the rate equation
3. The elementary steps must add up to the overall equation
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Temperature Dependence of the Rate Constant
k = A e -Ea/RT
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
ln k = --Ea
R1T
+ lnA
(Arrhenius equation)
Note: Changes to both the activation energy and temperature have exponential effects on the value k, hence rate of reaction
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• The Arrhenius equation show the effect of temperature on the rate constant, k
• It indicates that k depends exponentially on temperature
Arrhenius equation:
k = A e-Ea/RT
Ea – activation energyR – gas constant, 8.3145 J mol-1K-1
T - Kelvin temperatureA – Arrhenius constant (depends on collision rate and shape
of molecule)
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k = A e-Ea/RT
• As T increases, the negative exponent becomes smaller, so that value of k becomes larger, which means that the rate increases.
Higher T Larger k Increased rate
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ln k and 1/T is linear
• With R known, we can find Ea graphically from a series of k values at different temperatures.
• ln k2 = - Ea ( 1/T2 – 1/T1)
k1 R
• Ea = -R (ln k2 / k1) ( 1/T2 – 1/T1) -1
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Problem
• The decomposition of hydrogen iodide has rate constants of 9.51 x10-9 L/mol.s at 500.0 K and 1.10 x 10-5 L/mol.s at 600.0 K. Find Ea.
Ea = - (8.314)( ln 1.10 x10-5/ 9.51 x109)(1/600.00 – 1/500.0)
= 1.76 x 105 J/mol
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• If rearrange the equation and convert it to:lnk = - Ea . 1 + lnA
R T• A graph of ln k against 1/T will be linear with
a slope/gradient of –Ea/R and an intercept on the y-axis of lnA
lnk = - Ea . 1 + lnA R T y = m . x + b
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Plot lnk vs. 1/T = straight line
This is known as an Arrhenius plot