Chapter 12 Review - Kinetics:  the study of the speed of a reaction

1. Section 12.1:  Rates of Reaction
1. Kinetics is the study of the speed or rate of a reaction.  We care because the speed of the reaction is involved in bombs, explosions, hazards, food spoiling, drug response and industry. 
2. Average rate (average between two times) = DM / Dt = (M_f - M_i) / (t_f - t_i)
3. A rate's units are always M/t such as M/s, M/min, M/yr, etc.
4. As time passes, the [reactant] decreases so the rate is a negative number with respect to reactants (See figure 12.1)
5. As time passes, the [product] increases so the rate is a positive number with respect to products
6. As a rxn proceeds, the rate slows down so the rate is fastest at the beginning of a reaction.
7. The larger the [reactant], the faster the reaction. Rates depend on [reactant] NOT [product] 
8. For this simple generic pretend reaction ( 1X g 1Y)
1. Every time one X reacts, one Y is formed. So the rate of X disappearance is equal to the rate of Y appearance.   So -D[X] / Dt =  D[Y] / D t
2. The rate constant k = rate / [X] in this example.  k is a constant so as the rate increases so must [X].  Rearranging we get rate = k[X] which takes the form of y = mx + b (straight line equation where y is what is plotted on the Y axis, x is the x axis, m is the slope and b is the y intercept which is zero in this example.) 
3. When we plot rate versus [X] the slope is the rate constant k. 
4. Finding the units for k is important.  In this example rate = k[X].  The units for a rate are M/s.  The units for [X] is M.  So the units for k in this example must be 1/s or s^-1.
9. Note that there is a difference between average rate and instantaneous rate, which is called "the rate."  The average rate is the average between two different times.  The instantaneous rate is the rate at that one moment in time.  Unless the word average is used, we are talking about an instantaneous rate. (See figure 12.2)
10. Rate expressions / rate equalities 
1. For this pretend reaction: 1 C  g 2D    Every time one C disappears, two D's appear.  So rate = -D[C] / Dt =  D[D] / 2D t.  Note the two in the denominator. 
2. For this pretend reaction:    5F + 2G  g 3H + 4K   write the rate expressions:   rate = -D[F] / 5Dt =  - D[G] / 2D t  = D[H] / 3Dt =  D[K] / 4D t
3. What are the rate expressions for this reaction?  SiO₂ + 3 C  g  SiC + 2 CO    Answer:  rate =  -D[SiO₂ ] / Dt =  - D[C] / 3D t  = D[SiC] / Dt = D[CO] / 2Dt
4. For the reaction above, calculate the rate with respect to carbon if the rate with respect to carbon monoxide is 0.78 M/s.  Answer:   - D[C] / 3D t  = D[CO] / 2Dt.  Now plug in and rearrange to get:  D[C] / Dt = -3/2 (0.78M/s) = -1.17 M/s
5. Example 12.1, problems 12.1-2
2. Section 12.2:  the Rate Law
1. The rate law is a math expression that relates the rate in M/s to the reactant concentrations and it looks like this:  rate = k[react]^x[react]^y[react]^z
2. Remember that k is the rate constant.  The exponents xyz are called the "order" of the reaction.  The overall order is the sum of the exponents. The exponents can only be found from an experiment and are usually 0, 1 or 2.  The order is NOT the coefficients from the balanced reaction!!!  We can actually have fractions and negative numbers for orders (1/2, -2, 3/4...) but in this class the orders will be 0. 1, or 2.
3. What is the rate law and overall order for this reaction:  2 SO₂ + O₂ g  2 SO₃   Answer:  Without any more information we can only say that the rate law is:  rate = k[SO₂]^x[O₂]^y and the overall order is x + y. Example 12.2 and problem 12.3
3. Section 12.3 Rate Law Calculations
1. So how do we find the order?  We use experimental data and compare trials where only one concentration changes.  We need only one variable to be changing and then we can find the exponents x and y. In general, if [ ] doubles and the rate doubles it is first order.  If [ ] doubles and the rate quadruples, it is second order.  If [ ] doubles but the rate stays the same, it is zero order. 
2. Example:  Determine the rate law, the rate constant, and the overall order for this reaction:  2 NO + Cl₂ g 2 NOCl.  Given - the following data

experiment	[NO]	[Cl2]	rate (M/s)
1	0.10M	0.10M	0.117
2	0.10M	0.20M	0.468
3	0.10M	0.30M	1.054
4	0.20M	0.30M	2.107
5	0.30M	0.30M	3.161

1. First what do we know about the rate law?  That it is rate = k[NO]^x[Cl₂]^y so we need to find x and y 
2. Pick two experiments where only [Cl₂] changes such as experiments 1 and 2.
3. Divide the rate law using experiment 1 numbers by experiment 2 numbers:  [0.117 / 0.468] = [(k)(0.10)^x(0.10)^y ] /  [(k)(0.10)^x(0.20)^y ]
4. After canceling out we are left with:   [0.117 / 0.468] = [(0.10)^y ] /  [(0.20)^y ]
5. Simplify to:   0.25 = (0.10 / 0.20)^y
6. This simplifies to 0.25 = 0.50^y
7. We can see that y must be 2 because 0.50 squared is 0.25.  But to solve mathematically we take the log of both sides and get ylog0.5 = log0.25 thus y = log0.25 / log0.5 = 2.
8. So we found y and must now find x
9. Pick two experiments where only [NO] changes such as experiments 4 and 5.
10. Dividing rate law 4 by rate law 5 we get  [2.107 / 3.161] = [(k)(0.20)^x(0.30)^y ] /  [(k)(0.30)^x(0.30)^y ]
11. Simplify to:   0.666 = (0.20 / 0.30)^x
12. Which is 0.666 = 0.666^x
13. We see that x must be 1.
14. The rate law is rate = k[NO][Cl₂]²
15. The overall order is just the sum of the exponents:  1 + 2 = 3.  Third order overall. 
16. Now to find the value of the rate constant k.  We can pick any experiment and plug the numbers into the rate law since we know it now.
17. 0.117 M/s = k(0.10M)(0.10M)²
18. Solving for k = (0.117M/s) / 0.0010M³
19. So k = 117 M^-2s^-1
20. Here is one more question:  What is the rate if [NO] = 0.15M and [Cl₂] = 0.40M?
21. We answer this by plugging into the rate law now that we know k.
22. rate = (117 M^-2s^-1)(0.15M)(0.40M)²
23. rate = 2.81 M/s
2. Example 12.3-4 and problem 12.4-5.  
3. The units for a rate are always M/time but the units for k vary with the order. Being able to get the correct units for k is important.
3. Section 12.4  Reactant concentration and time for First order overall reactions. 
1. If the overall order is one, the rate law is:  rate = k[A] where A is the reactant in a reaction such as A g B. 
1. We also know from section 12.1 that rate =  -D[A] / Dt
2. Because the rate is the same we can say  k[A]  =  -D[A] / Dt
3. Through the power of calculus  , we can integrate k[A]  =  -D[A] / Dt  with the result of:   ln ( [A]_t / [A]_o) = -kt
4. We call this the first order integrated rate law and it is very important.  [A]_t is the concentration of A at time t.  [A]_o is the original concentration of A at time zero.
5. Math note:  ln 5 / ln 7 IS NOT = ln (5/7).   ln (5/7) = ln 5 - ln 7
6. We can rearrange the first order integrated rate law so that it fits the equation for a straight line (y = mx + b)
7.  ln [A]_t  = -kt + ln [A]_o) where y = ln[A]_t, x = t, slope m = -k, and b = ln[A]_o
8. So remember that a graph of ln[A]_t versus time produces a straight line IF the reaction is first order.  Remember that the slope of the line is the negative rate constant.  So the rate constant can be found graphically from the slope of the line. Figure 12.6
2. Example:  If a first order reaction has a starting reactant concentration of 2.4M, what is the rate constant if after 7.5 minutes the reactant concentration is 1.8M?
1. ln (1.8 / 2.4) = -k(7.5 min)
2. k = (ln 0.75) / 7.5 min = 0.038 min^-1
3. Example:  If a first order reaction has a rate constant of 4.73 x 10² s^-1, how long would it take for only 75% to react?
1. First realize that the starting amount is 100% and if 75% reacted into product, 25% remains as reactant
2. ln (25 / 100) = -4.73 x 10² s^-1 (t)
3. t = - (ln 0.25) / 4.73 x 10² s^-1 = 2.93 x 10^-3 s
4. Note that these equations also work for pressures in a gas phase reaction.  Instead of using molarities, use pressures.
5. Example 12.5-6 and problem 12.7-8
4. Section 12.5  The half-life for first order reactions 
1. The half life (t_1/2) is defined as the time it takes for 1/2 of the reactant to react.  In other words, the time it takes for 50% to turn into product which means that 50% remains as reactant. .
2. How much of a sample would remain after 3 half lives?  Answer:  Starting with 100%, after one half life there would be 50%, after two half lives there would be 25%, after three half lives there would be 12.5% reactant remaining. Divide by two again to get 4 half lives, etc...
3. Plug into the integrated rate law:   ln (50/100) = -kt_1/2
4. Solving for t_1/2 = 0.693 / k
5. Note that as the half life gets longer, the rate constant gets smaller.  So a teeny tiny rate constant would have a long half life and vice versa. (See figure 12.7)
6. Example:  If the half life for a first order reaction is 144 seconds, how long would it take for a starting 3.50M reactant to reach a concentration of 2.75M?  
1. First we use the half life to find k.  144 s = 0.693 / k
2. k = 4.8125 x 10^-3 s^-1
3. Now we use the first order integrated rate law.  ln [2.75M / 3.50M] = -(4.8125 x 10^-3 s^-1)t 
4. Solve for t = -0.24116 = -(4.8125 x 10^-3 s^-1)t 
5. t = 50.1 seconds
7. Example 12.7 and problems 12.9-10
5. Section 12.6  Second order overall reactions
1. If the order is second overall, the reaction could be A g B with a rate law of rate = k[A]²  OR the reaction could be A + B g C with a rate law of rate = k[A][B].  Both add up to an overall order of 2.  We will only work with the first option as the second option is too complex for this class.
2. What are the units of k?  Well rate M/s = k M²    so    k = M / sM² = s^-1M^-1
3. Using this reaction A g B with a rate law of rate = k[A]²  we also know that the rate = -D[A] / Dt
4. So this must be true:  k[A]² = -D[A] / Dt
5. Integrating this (calculus magic  ) gets us the second order integrated rate law:  1 / [A]_t = kt + 1 / [A]_o
6. We can also get the half life equation for second order:  t_1/2 = 1 / k[A]_o
7. Note that is we graph 1 / [A]_t on the y axis and time t on the x axis we get a slope of k and an intercept of 1 / [A]_o.  Thus a second order reaction gives a straight line for this graph. 
8. Example:  A second order reaction has a half life of 699 seconds.  If the initial reactant concentration is 0.0335M, what will the reactant concentration be after 855 seconds? 
1. First find k:  699 s = 1 / k(0.0335M)
2. k = 0.0427 M^-1s^-1
3. Now plug into the second order integrated rate law:  1 / [A]_t = (0.0427M^-1s^-1)(855 s) + 1 / (0.0335M)
4. 1 / [A]_t = 36.5127 M^-1 + 29.851 M^-1
5. [A]_t = 0.0151 M
9. Example 12.8 and prob 12.11
6. Section 12.7  Zero Order
1. See Table 12.4
2. Note that zero order reactions do exist, but are rare.   rate = k[A]^o = k.  
3. When the reactant concentration changes the rate remains the same and equals the rate constant
7. Section 12.8  Reaction Mechanisms
1. Many reactions actually occur in several steps called elementary steps.
2. All these steps together are called the reaction mechanism and sum up to the overall balanced reaction.
3. Example:  reaction:  A + 2B  g  2C
1. step one A + B  g D
2. step two D + B g  2C
3. Notice the D cancels out - it is an intermediate and does not appear in the overall reaction.
4. Usually one of the steps is slower than the others - it is the slow step and the reaction rate depends on it.  It is like a bottle-neck step.  The rate law will be for that step.  For step one above the rate law is rate = k[A][B].  For step two above the rate law is rate = k[D][B].  Rate laws for elementary steps always have their exponents (order) as 1.  If the real rate law is rate = k[A][B] then step one is the slow step.  Step two is faster.
5. Molecularity = # reactants in an elementary step  Table 12.5
1. A g B has one reactant, it is unimolecular, the rate law is rate = k[A] which is first order
2. A + B g C has two reactants, it is bimolecular, the rate law is rate = k[A][B] which is second order
3. A+ B + C g D has three reactants, it is termolecular, the rate law is rate = k[A][B][C] which is third order
4. Example 12.9 problem 12.12
8. Section 12.9  Determining Mechanisms
1. How do you determine which elementary step in a mechanism is the slow step (the rate determining step)? 
1. Look at all the elementary steps in the mechanism and write down their rate law
2. Look at experimental data and see which rate law matches the real data.  That step is the rate determining step
3. Example:  reaction is CO₂ + NO  g  CO + NO₂  and the real rate law is rate = k[CO₂]²
4. the mechanism is given as:  step one CO₂ + CO₂ g  CO₃ + CO
5. and step two:                                  CO₃ + NO  g  CO₂ + NO₂
6. the rate law for step one is rate₁ = k[CO₂][CO₂] = k[CO₂]²
7. the rate law for step two is rate₂  = k [CO₃][NO]
8. Since rate₁ matches the real rate law the slow step must be step one.
9. If rate₁ didn't match the real rate law by default step two must be the slow step.  Don't worry about more than this - there are some complicated math steps for when step two is slow.  Just remember if step one is not the slow step it must be step two.
2. Example 12.10, problem 12.13-14
9. Section 12.10  How Activation Energy and Temperature effect Reaction Rates
1. Collision Theory
1. Reactants must collide in order to react
2. Increase the number of collisions and the reaction rate will increase
3. Most collisions do NOT result in a reaction  
4. Collision success is higher when reactants collide at an 180 degree angle (head on)
5. Collision energy must be at least = to activation energy (E_a), the minimum energy necessary for reaction to occur (break reactant bonds and form new product bonds)  Remember a chemical reaction is all about the electrons being shared and transferred to make stable bonds.
6. Reactants must be oriented correctly for any chance of reaction to occur
2. So if we increase the concentration, there are more reactants to collide thus increasing number of collisions so the reaction rate will increase
3. So if we raise the temperature, the reactants are moving faster with more energy, so the chances of meeting the E_a is greater thus increasing the reaction rate.  Skip all the math stuff.
4. 
5. The above is an E_a diagram for an exothermic reaction.  E_a is the activation energy and is the difference between the top of the reaction path and the reactant energy.  DE is the change in energy for the reaction, it is the difference between the reactant and product energy, it is a negative number for exothermic reactions because the product energy is lower than the reactant energy.  TS is the transition state which is where bonds from the reactants are partially broken while bonds for the products are partially formed.  The TS is the highest point in the reaction path thus having the highest energy.  
6. 
7. The above is an E_a diagram for an endothermic reaction.  E_a is the activation energy and is the difference between the top of the reaction path and the reactant energy.  DE is the change in energy for the reaction, it is the difference between the reactant and product energy, it is a positive number for endothermic reactions because the product energy is higher than the reactant energy.  TS is the transition state which is where bonds from the reactants are partially broken while bonds for the products are partially formed.  The TS is the highest point in the reaction path.  
8. As the activation energy lowers, the reaction rate increases because the energy barrier is smaller and easier to get over.  Problem 12.15
10. Section 12.11  The Arrhenius Equation
1. k = Ae^-Ea/RT
2. A = collision frequency factor (constant, don't worry about it), Ea = activation energy is J/mol, R = 8.314 J/molK, and T is temperature in Kelvin
3. As Ea increases, k and the rate decrease
4. As T increases, k and the rate increase
5. This can be arranged to fit a straight line graph:  lnk = -(Ea/R)(1/T) + lnA
6. The rate constant k changes with temperature.  So for any reaction we have different k values at different temperatures.  So by running a reaction at different temperatures, you can solve for the different rate constants.  Then take 1/T for all temperatures in Kelvin and plot on the x axis.  Take ln of all rate constants and plot on the y axis.  (Plot lnk versus 1/T)  The slope of the resulting line is -Ea / R.  Since we know R you can solve for Ea, the activation energy.  Example 12.11
7. If you know two temperatures and 2 k values you can use this equation to find Ea:  ln (k₁ / k₂) = Ea / R (1/T₂ - 1/T₁)
8. Example:  The rate constant for a reaction at 20.0^oC is 0.0785 s^-1.  At 80.0^oC it is 1.03 s^-1.  Find the activation energy.

1. change Celsius to Kelvin then plug in all data
2. ln (0.0785 / 1.03) = Ea / 8.314 J/molK (1/353K - 1/293K)
3. Solve carefully - it is easy to make math errors here
4. Ea = 3.69 x 10⁴ J/mol
8. Problem 12.16
11. Section 12.12  Catalysis
1. A catalyst is a substance that increases the rate of reaction by lowering the reaction pathway between reactants and products without itself being consumed.  So it is reusable!
2. 
3. In the diagram above we see the normal reaction path with a higher Ea and the new path when a catalyst is used. Because the new path has a lower activation energy, it is the preferred path and the rxn rate increases. 
12. Section 12.13
1. Heterogeneous catalyst = catalyst in a different state from the reactants.  Usually a solid catalyst with liquid, gas, or aqueous (dissolved in water) reactants.  It is easy to remove the catalyst after the reaction by filtration so it can be used again.
2. Homogeneous catalyst = catalyst in the same state as the reactants.  Usually an aqueous catalyst with liquid or aqueous reactants.  It is harder to remove the catalyst after the reaction.
3. Enzymes are our bodies catalysts and they are huge proteins that work for a specific reaction.

Good Job - This is a long chapter.  Study hard!