Reaction Kinetics Honors Chemistry

What is Reaction Kinetics / Reaction kinetics is the study of rates of chemical processes. / We will look into: / how rxns happen / how to speed rxns up / Rxn mechanisms / Predicting the rate of a rxn. / Reaction kinetics is the study of rates of chemical processes. / We will look into: / how rxns happen / how to speed rxns up / Rxn mechanisms / Predicting the rate of a rxn.

Collision Theory / For a reaction to occur between two substances, the molecules of the 2 substances must collide at the right angles and with enough energy to break and/or remake bonds.

Consider this: / Which reaction should occur the fastest? Which is slowest? / CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2 H 2 O (g) / H 2 (g) + Cl 2 (g) --> 2HCl (g) / 3 Fe (s) + 2 I 2 (s) --> 2 FeI 3 (s) / Let’s look at them one by one. / Which reaction should occur the fastest? Which is slowest? / CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2 H 2 O (g) / H 2 (g) + Cl 2 (g) --> 2HCl (g) / 3 Fe (s) + 2 I 2 (s) --> 2 FeI 3 (s) / Let’s look at them one by one.

CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2 H 2 O (g) / For this rxn to occur quickly, one methane molecule must collide with two oxygen molecules simultaneously. The fact that the reactants are gases is good, because this means that they have more energy to break/make bonds.

H 2 (g) + Cl 2 (g) --> 2HCl (g) / For this reaction to occur quickly, one gaseous hydrogen molecule must collide one gaseous chlorine molecule. This seems pretty simple - and it is. This is the fastest of the 3 rxns.

2 Fe (s) + 3 I 2 (s) --> 2 FeI 3 (s) / For this reaction to occur quickly, 3 solid iron atoms must somehow come in contact with 2 solids iodine atoms. This is extremely unklikely, so this reaction is the slowest by far.

Reaction Mechanism / In the reaction: 2 NO 2 + F 2 --> 2NO 2 F 3 molecules must collide simultaneously. This is virtually impossible. What is more likely to happen is that two collide, make some product(s) which then collide with the third molecule. The possible steps for each collision is called a reaction mechanism. / In the reaction: 2 NO 2 + F 2 --> 2NO 2 F 3 molecules must collide simultaneously. This is virtually impossible. What is more likely to happen is that two collide, make some product(s) which then collide with the third molecule. The possible steps for each collision is called a reaction mechanism.

More on Reaction Mechanisms / 2 NO 2 + F 2 --> 2NO 2 F / Step 1: NO 2 + F 2 --> NO 2 F + F / Step 2: NO 2 + F --> NO 2 F / If we add these two steps we get the same rxn as above. The F atom seen in both of the two steps cancels out, so it is not part of the overall rxn; F would be called an intermediate. / 2 NO 2 + F 2 --> 2NO 2 F / Step 1: NO 2 + F 2 --> NO 2 F + F / Step 2: NO 2 + F --> NO 2 F / If we add these two steps we get the same rxn as above. The F atom seen in both of the two steps cancels out, so it is not part of the overall rxn; F would be called an intermediate.

Rate Determining Step / The Rate Determining Step is the slowest step in the reaction. / Why? Consider a funnel: the rate at which water moves through the funnel depends most on the stem and little on the rest, because it slows the water down the most. / The Rate Determining Step is the slowest step in the reaction. / Why? Consider a funnel: the rate at which water moves through the funnel depends most on the stem and little on the rest, because it slows the water down the most.

Which is the R.D.S? / So which step do you think is the RDS in: / 2 NO 2 + F 2 --> 2NO 2 F / Step 1: NO 2 + F 2 --> NO 2 F + F / Step 2: NO 2 + F --> NO 2 F / Step 1 requires a little more energy because the F 2 must be broken so it would be slower. / So which step do you think is the RDS in: / 2 NO 2 + F 2 --> 2NO 2 F / Step 1: NO 2 + F 2 --> NO 2 F + F / Step 2: NO 2 + F --> NO 2 F / Step 1 requires a little more energy because the F 2 must be broken so it would be slower.

So how can we speed up a rxn? / Let’s say that I have a ball of aluminum foil which I am trying to react with some hydrochloric acid. / The rxn is: / 2Al (s) + 6HCl (aq) --> 3H 2 (g) + 2AlCl 3 (s). / How can I make this reaction happen faster? / Let’s say that I have a ball of aluminum foil which I am trying to react with some hydrochloric acid. / The rxn is: / 2Al (s) + 6HCl (aq) --> 3H 2 (g) + 2AlCl 3 (s). / How can I make this reaction happen faster?

1. Use more of one reactant / If I add more HCl, then there is a much greater chance of HCl molecules colliding with Al atoms.

2. Increase surface area / If I spread the aluminum out, or shred it, the HCl molecules will collide easier with Al atoms.

3. Increase the concentrations / If I use a stronger concentration of acid (for example 10.0 M HCl instead of.100 M HCl) there are more HCl molecules, so collisions are more likely to happen. (For a gas this might mean increasing pressure.)

4. Add energy / When energy is added, the reactants have more energy, and thus collide more often and stronger. Of course heating the Al or the HCl would be dangerous, but effective.

5. Add a catalyst / A catalyst is a chemical that speeds up a rxn without being consumed in the rxn. This will be explained more later on. / In the case of our rxn water is a catalyst.

Enthalpy Diagram / An enthalpy diagram is an energy graph. / This enthalpy diagram is of an exothermic reaction. How do we know this?

Activated Complex / A complex is an unstable molecule formed as an intermediate. / The activated complex is the complex requiring the most energy to form. / A complex is an unstable molecule formed as an intermediate. / The activated complex is the complex requiring the most energy to form.

Activation Energy / One of the stipulations of the collision theory is that the collision must have enough energy. This energy, called activation energy, is the energy needed to convert the reactants into the activated complex. / What is the activation energy in this rxn?

Some questions to think about / What is the activation energy?  What is  H? / Is the reaction endothermic or exothermic? / If the reaction were reversed, how would these answers change?

Effect of a catalyst / A catalyst speeds up a rxn by lowering the activation energy. This often occurs by offering a different mechanism.

Rate Law Expression / We have said that the rate of a reaction depends on a number of factors, but the most important is the the amount of reactants. / In a Rate Law Expression, we show that the rate of a reaction depends on the concentrations of reactants. Specifically, the rate law is a constant times the concentrations of the reactants. / Since solids and liquids are not compressible, we cannot change their concentrations or densities like we can with gases or solutions. So, solids and liquids are not used in Rate Law Expressions. / We have said that the rate of a reaction depends on a number of factors, but the most important is the the amount of reactants. / In a Rate Law Expression, we show that the rate of a reaction depends on the concentrations of reactants. Specifically, the rate law is a constant times the concentrations of the reactants. / Since solids and liquids are not compressible, we cannot change their concentrations or densities like we can with gases or solutions. So, solids and liquids are not used in Rate Law Expressions.

Theoretical Rate Law / Let’s start simple with a simple rxn. / H 2 (g) + Cl 2 (g) --> 2 HCl / The rate of this reaction depends largely on the amount of hydrogen and the amount of chlorine gases present. / The Theoretical rate law expression would be: rate = k*[H 2 ]*[Cl 2 ] / K is a constant based on the temperature and the nature of the chemicals. It is a different constant for any rxn. Let’s not worry about that now. / Let’s start simple with a simple rxn. / H 2 (g) + Cl 2 (g) --> 2 HCl / The rate of this reaction depends largely on the amount of hydrogen and the amount of chlorine gases present. / The Theoretical rate law expression would be: rate = k*[H 2 ]*[Cl 2 ] / K is a constant based on the temperature and the nature of the chemicals. It is a different constant for any rxn. Let’s not worry about that now.

Why is it “Theoretical” / We already said that a reaction mechanism depends most on the Rate Determining Step. Without seeing the mechanism or experimental data, we can’t know for sure to what extent each reactant actually contributes. / We’ll move on to actual rate laws later. / We already said that a reaction mechanism depends most on the Rate Determining Step. Without seeing the mechanism or experimental data, we can’t know for sure to what extent each reactant actually contributes. / We’ll move on to actual rate laws later.

Try these: / Write the Theoretical Rate Law Expressions for: / H 2 (g) + I 2 (s) --> 2 HI (s) / 2 H 2 (g) + O 2 (g) --> 2 H 2 O / 2C 2 H 6 (g) + 7O 2 (g) --> 4CO 2 (g) + 6H 2 O (g) / Write the Theoretical Rate Law Expressions for: / H 2 (g) + I 2 (s) --> 2 HI (s) / 2 H 2 (g) + O 2 (g) --> 2 H 2 O / 2C 2 H 6 (g) + 7O 2 (g) --> 4CO 2 (g) + 6H 2 O (g)

The answers are: / H 2 (g) + I 2 (s) --> 2 HI (s) / Rate = k*[H 2 ] / 2 H 2 (g) + O 2 (g) --> 2 H 2 O / Rate = k*[H 2 ] 2 *[O 2 ] / 2C 2 H 6 (g) + 7O 2 (g) --> 4CO 2 (g) + 6H 2 O (g) / Rate = k*[C 2 H 6 ] 2 *[O 2 ] 7 / H 2 (g) + I 2 (s) --> 2 HI (s) / Rate = k*[H 2 ] / 2 H 2 (g) + O 2 (g) --> 2 H 2 O / Rate = k*[H 2 ] 2 *[O 2 ] / 2C 2 H 6 (g) + 7O 2 (g) --> 4CO 2 (g) + 6H 2 O (g) / Rate = k*[C 2 H 6 ] 2 *[O 2 ] 7

Actual Rate Law / In order to determine the actual (nontheoretical rate law) we need experimental data of the concentrations of reactants and the rate of the reaction.

Determining the Rate Law / 2 H 2 + O 2 --> 2H 2 O / The data on the right depicts how the rate of water production varies with differing concentration of H 2 and O 2. / 2 H 2 + O 2 --> 2H 2 O / The data on the right depicts how the rate of water production varies with differing concentration of H 2 and O 2. [H 2 ][O 2 ]rate.25 M 1 M/s M2 M/s.50 M 4 M/s

Isolate your variables / Let’s look at the 1st two rows. / The [H 2 ] changes, but the [O 2 ] doesn’t. That means we can compare the [H 2 ] directly to the rate. / Let’s look at the 1st two rows. / The [H 2 ] changes, but the [O 2 ] doesn’t. That means we can compare the [H 2 ] directly to the rate. [H 2 ][O 2 ]rate.25 M 1 M/s M2 M/s.50 M 4 M/s

Isolate your variables / / / Compare the H 2 and the rate. /.50/.25 = 2 / 2/1 = 2 / As the hydrogen doubled, the rate doubled. This means that the hydrogen is 1st order. / Compare the H 2 and the rate. /.50/.25 = 2 / 2/1 = 2 / As the hydrogen doubled, the rate doubled. This means that the hydrogen is 1st order. [H 2 ][O 2 ]rate.25 M 1 M/s M2 M/s.50 M 4 M/s

Now isolate the other reactant / In the last two rows, the [H 2 ] doesn’t change, but the [O 2 ] does. / The [O 2 ] doubles (.50/.25=2) and the rate doubles (4/2=2). / This means that [O 2 ] is 1st order. / In the last two rows, the [H 2 ] doesn’t change, but the [O 2 ] does. / The [O 2 ] doubles (.50/.25=2) and the rate doubles (4/2=2). / This means that [O 2 ] is 1st order. [H 2 ][O 2 ]rate.25 M 1 M/s.50 M.25 M2 M/s.50 M 4 M/s

So…. / Since the [H 2 ] is first order and the [O 2 ] is first order, the rate law is now: / Rate = k*[H 2 ]*[O 2 ] / In the theoretical rate law, the hydrogen was second order, yet we see in the actual rate law that it is really first order. / Since each of the 2 reactants are first order, the reaction is second order. / Since the [H 2 ] is first order and the [O 2 ] is first order, the rate law is now: / Rate = k*[H 2 ]*[O 2 ] / In the theoretical rate law, the hydrogen was second order, yet we see in the actual rate law that it is really first order. / Since each of the 2 reactants are first order, the reaction is second order.

Try this: N 2 + 3H 2 --> 2 NH 3 [N 2 ][H 2 ]Rate.125 atm.0333 atm1.75 atm/hr.125 atm.1332 atm28.0 atm/hr.375 atm.0333 atm1.75 atm/hr

Rate = k*[H 2 ] 2 / In the 1st two rows, the [N 2 ] is constant. / The [H 2 ] went up by 4 times, the rate went up by 16 times. 4 x =16, so x=2. / Hydrogen is second order. / In the 1st and 3rd rows, the [H 2 ] is constant. / The [N 2 ] went up by 2 times, the rate didn’t change. 2 x =1, so x=0. / nitrogen is zero order, meaning it does not affect the rate. / In the 1st two rows, the [N 2 ] is constant. / The [H 2 ] went up by 4 times, the rate went up by 16 times. 4 x =16, so x=2. / Hydrogen is second order. / In the 1st and 3rd rows, the [H 2 ] is constant. / The [N 2 ] went up by 2 times, the rate didn’t change. 2 x =1, so x=0. / nitrogen is zero order, meaning it does not affect the rate.

Determining the value of K / In the last example, we found a rate law of rate=k*[H 2 ] 2 for the reaction N 2 + 3H 2 --> 2NH 3. / So let’s substitute in any row of data: / In the last example, we found a rate law of rate=k*[H 2 ] 2 for the reaction N 2 + 3H 2 --> 2NH 3. / So let’s substitute in any row of data: [N 2 ][H 2 ]Rate.125 atm.0333 atm 1.75 atm/hr.125 atm.1332 atm 28.0 atm/hr.375 atm.0333 atm 1.75 atm/hr

First Row / Rate=k*[H 2 ] 2 / 1.75 atm/hr = k*[.0333 atm] 2 / 1.75 atm/hr = k* atm 2 / /atm*hr = k / Since the data for H 2 and the rate are the same in the first and 3rd row, k must be the same for both rows. / Rate=k*[H 2 ] 2 / 1.75 atm/hr = k*[.0333 atm] 2 / 1.75 atm/hr = k* atm 2 / /atm*hr = k / Since the data for H 2 and the rate are the same in the first and 3rd row, k must be the same for both rows.

Second Row / Rate=k*[H 2 ] 2 / 28.0 atm/hr = k*[.1332 atm] 2 / 28.0 atm/hr = k* atm 2 / /atm*hr = k / Rate=k*[H 2 ] 2 / 28.0 atm/hr = k*[.1332 atm] 2 / 28.0 atm/hr = k* atm 2 / /atm*hr = k

Rate Stoichiometry / Rates can be compared stoichiometrically, because they are based on concentrations, not on masses. / Given a rxn: xA + yB  zC / Rate rxn = Rate A =Rate B = Rate C / x b c / Rates can be compared stoichiometrically, because they are based on concentrations, not on masses. / Given a rxn: xA + yB  zC / Rate rxn = Rate A =Rate B = Rate C / x b c

An example / Given the reaction: / H O 2 + N 2 --> 2 HNO 3 / If the rate of RXN is 1.5 atm/hr, what is the rate of consumption of Oxygen? / 1.5 atm/hr * -3 = -4.5 atm/hr / What is the rate of production of ammonia? / 1.5 atm/hr * +2 = 3.0 atm/hr / Given the reaction: / H O 2 + N 2 --> 2 HNO 3 / If the rate of RXN is 1.5 atm/hr, what is the rate of consumption of Oxygen? / 1.5 atm/hr * -3 = -4.5 atm/hr / What is the rate of production of ammonia? / 1.5 atm/hr * +2 = 3.0 atm/hr

I.C.E, I.C.E. Baby / When we are comparing starting values to final values, we have a three step process, called I.C.E. / Let’s do an example to show this concept: / When we are comparing starting values to final values, we have a three step process, called I.C.E. / Let’s do an example to show this concept:

C 3 H O 2 --> 3 CO 2 + 4H 2 O / If we begin the reaction with 10 atm of propane and 10 atm of oxygen, we find that carbon dioxide is made at a rate of.04 atm/ min. If the rate stays reasonably constant for one hour, what is the partial pressure of each gas?

I stands for initial / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / We know we started with 10 atm of each reactant, and we can safely infer that we do not start with products. / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / We know we started with 10 atm of each reactant, and we can safely infer that we do not start with products.

C. Stands for Change / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / C:-.8 atm -4 atm +2.4 atm +3.2atm / We know that carbon dioxide is made at.04 atm/min, so in 60 min 2.4 atm will be made. Now we do stoichiometry to compare CO 2 to the other chemicals. / 2.4 atm CO 2 * (-1 C 3 H 8 /3 CO 2 ) = -.8 atm C 3 H 8 / 2.4 atm CO 2 * (-5 O 2 /3 CO 2 ) = -4 atm O 2 / 2.4 atm CO 2 * (+4 H 2 O /3 CO 2 ) = +3.2 atm H 2 O / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / C:-.8 atm -4 atm +2.4 atm +3.2atm / We know that carbon dioxide is made at.04 atm/min, so in 60 min 2.4 atm will be made. Now we do stoichiometry to compare CO 2 to the other chemicals. / 2.4 atm CO 2 * (-1 C 3 H 8 /3 CO 2 ) = -.8 atm C 3 H 8 / 2.4 atm CO 2 * (-5 O 2 /3 CO 2 ) = -4 atm O 2 / 2.4 atm CO 2 * (+4 H 2 O /3 CO 2 ) = +3.2 atm H 2 O

E. Stands for End / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / C:-.8 atm -4 atm +2.4 atm +3.2atm / E:9.2 atm 6 atm 2.4 atm 3.2 atm / To determine the end results, we just add up our initial with our change. So the answers are: / [C 3 H 8 ] = 9.2 atm[O 2 ] = 6.0 atm / [CO 2 ] = 2.4 atm[H 2 O] = 3.2 atm / C 3 H O 2 --> 3 CO 2 + 4H 2 O / I:10 atm 10 atm 0 atm 0 atm / C:-.8 atm -4 atm +2.4 atm +3.2atm / E:9.2 atm 6 atm 2.4 atm 3.2 atm / To determine the end results, we just add up our initial with our change. So the answers are: / [C 3 H 8 ] = 9.2 atm[O 2 ] = 6.0 atm / [CO 2 ] = 2.4 atm[H 2 O] = 3.2 atm