Mark Important
Points in Margin Rate Laws
Rate Laws
Unit: Kinetics
Knowledge/Understanding:
· how the rate law and rate constant relate to experimental data
· what the rate constant means
Skills:
· calculate rate law & rate constant from initial concentration experiment data
· calculate activation energy from the Arrhenius equation
12.1
Reaction rates: Show how the concentration changes over time, usually the change in concentration (mol/L) per second: mol/L·s. This shows how the concentration changes over time.
For most reactions, the rate decreases as the reaction occurs.
For this reaction 2NO(g) + O2(g)à 2NO2(g)
The rate that NO is consumed is twice as fast as the rate oxygen is consumed, because the mole ratio of NO to O2 is 2:1.
initial rate: the rate at which reactants are converted to products at the instant when the reaction begins.
Instantaneous rate: The rate at a specific time during a reaction, so the initial rate is usually the fastest.
2NO2 à 2NO +O2
Time (s) / [NO2]0 / 1.0 x 10-1
100 / 5.0 x 10-2
200 / 2.5 x 10-2
Average rate: Calculate with the following equation, but use the greatest time range given.
Average rate = = (1.0 x 10-1 mol/L– 2.5 x 10-2 mol/L)/200 sec =
12.2
rate law: an equation that relates the rate of a chemical reaction to the concentrations of (zero or more of) the reactants. It is in this form:
Rate = k[A]n
k is a constant – the units depend on the order of reaction.
A is a reactant
n is the order—a whole number or a zero.
Example: if n = 1, the reaction is “1st order.” If n = 2, “second order”
12.3
Method of Initial Rate experiments can be used to determine which reactants affect the rate of a reaction.
How do you design an experiment? Increase the concentration of one reactant, while keeping the other reactant concentrations the same. Then observe the rate. Does it increase? Repeat for all reactants.
Consider the reaction:
2NO(g) + O2(g)↓ 2NO2(g)
The following initial rate data were collected:
[NO]/ [O2]
/ rate
0.010 / 0.010 / 2.5×10−3
0.010 / 0.020 / 5.0×10−3
0.030 / 0.020 / 4.5×10−2
Comparing the first vs. second experiments, we see that [NO] stays the same, but [O2] doubles. When [O2] doubles, the rate also doubles. Therefore, the rate is proportional to [O2]1 or just [O2].
Comparing the second vs. third experiments, we see that [O2] stays the same, but [NO] goes up by a factor of 3. When [NO] goes up by a factor of 3, the rate goes up by a factor of 9, which equals 32. Therefore, the rate is proportional to [NO]2.
The rate law is therefore:
rate = k[NO]2[O2]
To determine the value of k, plug in the data from any of the three experiments. Arbitrarily selecting the second one:
5.0×10−3=k [0.010 mol/L]2[0.020 mol/L]
5.0×10−3=k (2×10−6 )
Note that the units for k are whatever is necessary to make the rate come out in . In this case, k is multiplied by concentration to the 3rd power, or by . In order to end up with the rate in , we need to cancel two of the moles from the numerator (hence the mol2 in the denominator), cancel two of the liters from the denominator (hence the ℓ2 in the numerator), and also get seconds into the denominator.
x =
As a second example, consider the following reaction and initial rate data:
(CH3)3CBr(aq)+OH−(aq)à(CH3)3COH(aq)+Br−(aq)
[(CH3)3CBr]/ [OH−]
/ rate
1.) / 0.50 / 0.050 / 0.005
2.) / 1.0 / 0.050 / 0.010
3.) / 1.5 / 0.050 / 0.015
4.) / 1.0 / 0.10 / 0.010
5.) / 1.0 / 0.20 / 0.010
Looking at experiments #1 vs. #2, and #3, [OH−] is constant, [(CH3)3CBr] is increasing by a factor of 2, and the rate is also increasing by a factor of 2. Looking at #1 vs. #3, [OH−] is constant, [(CH3)3CBr] is increasing by a factor of 3, and the rate is also increasing by a factor of 3. These show that the rate is proportional to [(CH3)3CBr].
Looking at experiments #2, #4, and #5, [(CH3)3CBr] is constant, [OH−] is increasing, and the rate is constant. This means the rate is not related to [OH−], which means [OH−] will not appear in the rate law.
The rate law is therefore:
rate = k[(CH3)3CBr]
Now, plugging in to find the value of k, let’s choose experiment #2.
0.010=k(1.0)
This time, because k is multiplied by only one concentration, we already have the units of . This means k only needs to put seconds in the denominator to get the desired rate units of .
Reaction Mechanism
Goal: Evaluate a mechanism based on rate law to determine if it is acceptable or not.
In an elementary reaction, the reaction happens in one step. For these reactions, multiply the concentration of each reactant raised to the power of its coefficient. Example:
A + B à C
Rate law = [A][B]
2A + B à C
Rate law = [A]2[B]
Most reactions do not happen in one step.
Mechanism: A reaction mechanism describes the steps in a chemical reaction.
Consider this reaction: NO2(g) + CO(g) à NO(g) + CO2(g)
Experiments show the rate law is: Rate = k[NO2]2
Chemists think the reaction happens in two steps:
(1) NO2(g) + NO2(g) à NO3(g) + NO(g)
(2) NO3(g) +CO(g) à NO2(g) + CO2(g)
NO3 is called an intermediate. It is in the mechanism, but it is not written in the overall equation (NO2(g) + CO(g) à NO(g) + CO2(g)).
Why do chemists think this is the mechanism? To determine if a mechanism is acceptable, you must check two things. First, does it add up to the overall equation?
(1) NO2(g) + NO2(g) à NO3(g) + NO(g)
(2) NO3(g) +CO(g) à NO2(g) + CO2(g)
NO2(g) + NO2(g) + NO3(g) +CO(g) à NO3(g) + NO(g) + NO2(g) + CO2(g)
NO2(g) + CO(g) à NO(g) + CO2(g)
Note that when you add them and cancel the molecules that appear on both sides, you get the overall reaction
When there is more than one step in a reaction, the slowest step determines the rate, so only the reactants in the slowest step are in the rate law.
Because the rate law is Rate = k[NO2]2 we can conclude that 2 NO2 molecules are in the slow step, so this mechanism agrees with the rate law:
(1) NO2(g) + NO2(g) à NO3(g) + NO(g) SLOW
(2) NO3(g) +CO(g) à NO2(g) + CO2(g) FAST
The ‘slow’ step is called the rate determining step.
Collision Model
Learning Goals:
· Explain how each of the following influence reaction rate:
o Activation energy
o Temperature
o Frequency of collisions
o Orientation of collisions
· Draw a PE vs. reaction progress graph for endo and exo reactions. Label: ∆E and Ea
· Will the forward or reverse reaction would be faster if it is exothermic? Endothermic?
Vocabulary check:
· Particles = molecules or atoms.
· Collision (noun) = crash, hit
The collision model explains reaction rates. It states that chemicals must collide to react.
Remember that kinetic molecular theory of gases explains that…
· At higher temperatures, molecules move faster and thus collide more frequently.
This is why at higher temperatures, reaction rates are faster. Also, I sweat faster. Eeew.
Not all collisions cause a reaction. WHY? WwwwwhhhHHhhhyYyyy?
In order to react, colliding particles must have enough energy. The minimum energy needed to react is called the activation energy.
Remember the lightstick video? Draw the Potential Energy graph of the reactants and products in the lightstick reaction. Label the reactants, products, ∆E and Ea.
reaction progress ®
When the P.E. drops, energy is given off in the form of ______.
Between the reactants and products there is an energy barrier (like climbing a hill). The energy needed to go from the reactants to the top of the barrier is called the: ______.
Not all collisions are strong enough to react (less than activation energy). Why?
· Some particles have high KE, some have low KE.
· Only particles with enough KE can collide with energy greater than or equal to the activation energy.
Sketch the KE distribution of particles:
Number of particlesKinetic Energy ®
· Draw the curve.
· Mark the “threshold energy.”
· Color in the particles that will react.
· Using two different colors, add a curve to #4 showing the same chemicals at a higher temperature and a curve at a lower temperature.
HOWEVER, even particles with enough KE do not always react. WHY?
· Molecules must have the correct orientation.
Review:
o particles must collide to react
o collisions must be hard enough to break bonds
o Only particles with enough KE will collide hard enough
o some particles have high KE, some have low
o temperature is a measure of the average kinetic energy of the particles
o Increasing temperature increases frequency of collisions with enough KE to react.
o Particles must have the correct orientation to react
Activation Energy
Recall that the rate of a chemical reaction increases as the activation energy decreases. In 1889, the Swedish chemist Svante Arrhenius determined that the rate constant k obeys the equation:
where A is a constant, Ea is the activation energy, R is the gas constant, and T is the temperature (in Kelvin). e is the number from the natural log 2.71…
Taking the logarithm of both sides of the equation gives the Arrhenius equation in logarithmic form:
Because A is a constant, so is ln A. Therefore, the Arrhenius equation is a straight line plot in y=mx+b form:
This means that if we graph lnk vs. , the slope will be .
Multiplying the slope by the gas constant R gives the activation energy Ea for the reaction.
If you have two values for k at two different temperatures, you can calculate Ea as follows:
The equation is usually given as:
Notice that we reversed the order of & in order to get rid of the negative sign in front of . Note also that if you use , then Ea comes out in .
Catalysts
A catalyst is a chemical that speeds up a reaction but is not “used up.”
A catalyst lowers activation energy, which increases reaction rate.
A homogeneous catalyst is in the same phase as the reacting molecules. Example: a gas catalyst with gas reactants.
A heterogeneous catalyst is present in a different phase, usually a solid. Example: Solid platinum speeds up the reaction of gaseous ethylene and hydrogen.
Draw an energy distribution curve of a. uncatalyzed reaction b. catalyzed reaction. Label Ea. Shade the area of particles with enough energy to react.
Number of particlesKinetic Energy ®
Use this space for additional notes.
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