Notes from 1-30-2001 (Tuesday)

Review

Rate Law – Experimentally determined equation that expresses how the rate of reaction changes with reactants concentration.

A + 2B +...® products

Rate = - D [A] = - D [B] = k [A]^m [B]ⁿ...

Dt 2 Dt

m & n are experimentally measured and are usually 0, 1, 2, ... (usually integers, 3 and higher is rare, can be negative in the case of a poison, can be a fraction such as 1/2).

k is the rate constant. k will increase with temperature and the rate law could change with temperature.

For the following reaction data, taken at 540 K

a. determine the rate law with respect to each reactant and

b. determine the rate constant with proper units.

CO + NO₂ ® CO₂ + NO

Exp #	[CO] M	NO2	Initial Rate M/hr
1	5.1 X 10^-4	.35 X 10^-4	3.4 X 10^-8
2	“	.18 X 10^-4	1.7 X 10^-8
3	1.0 X 10^-3	.35 X 10^-4	6.8 X 10^-3

This would have a rate law of

Rate = k [CO] [NO₂]

This is said to be first order in carbon monoxide, first order in nitrogen dioxide and second order overall. k = 1.9 / (M hr) (note how the units of k cancel one of the M from [CO][NO₂] so the rate has the correct units.

If the rate law was found to be

Rate = k [CO]²

The reaction would be 2^nd order in carbon monoxide, zero order in nitrogen dioxide and second order overall. k would have the same units.

Time Dependence of Rate for 1^st and 2^nd Order Reaction

1^st Order Reaction

rate = -d[A]/dt = k [A]

With a little bit of calculus it is easily shown that

- LN [A]_t/[A]₀ = kt

Which is equivalent to LN [A]_t = -kt + LN [A]_o

Remembering the equation of a line is y = mx + b, if we plot time on the x-axis and LN [A] on the y-axis, we should get a line with a slope of -k if the reaction is first order.

The above graph shows how [N₂O₅] decomposition appears to follow first order kinetics.

That is Rate = .0289/min [N₂O₅]

Now we can calculate a concentration after some amount of time, or how long it will take to reach a certain concentration.

How much N₂O₅ will be left at 40.0 minutes? (.315 of what you started with).

How long till half of the initial N₂O₅ is gone? At this time A_t = A₀/2 and

For a 2^nd Order Reaction

rate = -d[A]/dt = k [A]² Here doubling the [A] will make the reaction go four times as fast and tripling the [A] will make it go 9 times as fast (because 2² = 4 and 3²=9). A little bit of calculus shows that

1/[A]_t = k t + 1/[A]₀

If the data is second order, a plot of time on the x-axis and 1/[A_t] on the y-axis should give a straight line with a slope of k.

For a Zero Order Reaction

rate = -DA/Dt = k

Rearrangement of variables shows that

-DA = k Dt

[A]_t = -kt + [A]₀

Consider the sample exercise from Brown, Lemay, and Bursten for the gas-phase decomposition of nitrogen dioxide at 300.ºC

NO₂ → NO + O₂

Sample Exercise 14.7
Time / s	[NO2]	LN [NO2]	1/[NO2]
0	0.01	-4.605170186	100
50	0.00787	-4.844697217	127.0648
100	0.00649	-5.037492748	154.0832
200	0.00481	-5.337058195	207.9002
300	0.0038	-5.572754212	263.1579

The above data shows how a 2^nd order data looks if plotted the three different ways we discussed above. The “Zero order and “1^st order” plots does not look linear, while the second order plot does, so apparently this reaction is not following zero or 1^st order kinetics.

Half- Life

For a first order reaction

- LN [A]/[A_o] = kt

The time it takes for half of a reactant is called the half life. At this time [A]_t = .5 [A]₀, so

LN .5[A]₀ /[A_o] = - LN .5 = LN 2 = kt_1/2

and t_1/2 = LN 2 / k (note t_1/2 is independent of initial concentration). This is used a lot to describe radioactive decay of different isotopes. A short half life is a substance that decays (or reacts) quickly and will quickly disappear.

The above reaction shows how the cancer chemotherapy drug cisplatin reacts in water. This reaction follows pseudo first order kinetics with a rate constant for cisplatin at 37ºC to be 1.87 X10^-3 /minute. What is the half-life of cisplatin? (371 minutes or 6.1 hours).

How long till the concentration is the body reaches 1/100 of an initial amount? (41 hours) Pseudo means “false” and this reaction has a rate law of

Rate = k [cisplatin] [water]

If the water is a solvent where its concentration is much larger than cisplatin and thus it has a very small change during the reation, we can consider it a constant and include it in the rate constant.

Rate = k’[cisplatin]

Temperature and Rate

Collision Theory – For reactants to react they must meet two requirements.

1. The colliding species must be oriented correctly.

2. The particle or particles colliding must have enough kinetic energy. The higher the temperature, the more particles have this minimum energy. This energy is called the activation energy, E_a. E_a can vary from 0 to several hundred kJ/mol.

Consider the gas-phase reaction of nitrogen dioxide with carbon monoxide to produce nitrogen monoxide and carbon dioxide. An oxygen atom is transferred from the NO₂ to the CO. The first pair of molecules shown crashing together could react by breaking the NO bond and forming a bond between the C and the O. The second collision shown is less likely to result in a reaction because the two oxygen atoms are pointed away from the carbon. The first collision is more likely than the second to produce the products shown.

For the first collision shown above to produce products, the collision must have enough energy to break the nitrogen oxygen bond. Otherwise the two molecules would just bounce and travel on their separate ways. This minimum energy is called the activation energy, E_a. The energy profile below shows a reaction where the reactants must have at least 40 kJ/mol in order to react. The molecules that do react produce products that have less potential energy so this reaction is an exothermic reaction, where the net change of energy going from reactants to products is –50 kJ/mol. Also note that if this reaction went backward from products to reactants, the backward reaction would have an activation energy of 50 + 40 = 90 kJ/mol. So an exothermic reaction will have a lower activation energy than the backward reaction. An endothermic forward reaction, will have a higher forward activation energy than the backward reaction.

The Arrhenius Equation is an empirical equation that can be thought to include these two factors.

k = A exp [-E_a/RT]

A is the frequency factor and it can be thought of as the number of effective collisions per unit of time. The units will be the same as k and for a first order reaction A is typically a large number such as 10¹⁰ to maybe 10¹⁶ /s.

The exponential term (which for very, very, large T or very, very, small E_a is at maximum 1) can be thought of as the fraction of molecules that have enough energy to react. T needs to be in Kelvin and the gas constant needs to be in energy units such as

8.314 X 10^-3kJ/K mol.

Of the two factors, the one that is more important to know is the activation energy. We usually assume neither the Ea or A change with temperature, (because this makes something complicated, much, much more complicated.)

Comparison of the exponential term e^-E/RT, for different E_a and Ts.

	300 K	310 K
100 kJ/mol	3.9 X 10^-18	14. X 10^-18
110 kJ/mol	7 X 10^-20	29 X 10^-20

Note two things, ok three things.

1. An increase in T increases the fraction of molecules that have enough energy to react, (over a factor of 3 for both E_a).

2. Note how an increase in E_a greatly reduces the fraction of molecules that have enough energy to react.

3. Note how small the numbers are.

If you want to compare reaction rates at different temperatures you could do something like this where I divide one equation by a second equation

k₁/k₂ = A exp [-E_a/RT₁]

A exp [-E_a/RT₂]

This will simplify to

k₁/k₂ = exp [-E_a/R (1/T₁ – 1/T₂) ]

If you take the LN of both sides

LN (k₁/k₂) = E_a/R (1/T₂-1/T₁)

Since k is proportional to the rate, if k doubles, the rate doubles.

Two examples. A reaction has an activation energy of 70 kJ/mol. How much faster will the reaction go at 50°C (323 K) compared to 25 °C (298 K). The answer will be the ratio of k₁/k₂ where k1 corresponds to the rate constant at 50°C and k2 at 25°C.

LN (k₁/k₂) = E_a/R (1/T₂-1/T₁) = 2.186 Take e^x of both sides and you get

e^(LN (k₁/k₂)) = e^2.186

k₁/k₂ = 8.9

So the reaction goes 8.9 times faster at 50°C.

A second example. A rule of thumb is that many reactions that go at a reasonable speed at room T go twice as fast with a 10°C rise in T. Estimate the E_a of this type of reaction.

Ok, so k₁/k₂ = 2 and T₂ = 298 K and T₁ = 308 K. Solve for E_a = 53 kJ/mol.

Often times, the Arrhenius equation is used as follows to determine the activation energy.

k = A exp [-E_a/RT]

Taking the LN of both sides

LN k = LN exp [-E_a/RT] + LN A

LN k = -E_a/RT + LN A Which can be thought of as looking like our equation of a line

y = mx + b Where m is the slope = -E_a/R, x is 1/T, and b is the y intercept = LN A.

For example

Using the equation of the line, -E_a/R = -30317 K or E_a = 252 kJ/mol and LN A = 29.346 or A = e^(29.346) = 5.6 X 10¹² s^-1.

Often it is easier to measure something that is proportional to the rate constant, such as a color change or pressure and make an Arrhenius plot of LN (color change/time) versus 1/T(K). Our slope will give us a good E_a, but the y intercept is meaningless if we do not know how the color change relates to concentrations.