Saturday Study Session 2 Theme of the day: Molecules Session 4 – Kinetics: How to turn your hoopty reaction into a Ferrari reaction

Initial Rate of Formation of Z (mol L-1 sec-1) Experiment [X]0 [Y]0 Initial Rate of Formation of Z (mol L-1 sec-1) 1 0.40 0.10 R 2 0.20 ? 1. The table above shows the results from a rate study of the reaction X + Y→Z. Starting with known concentrations of X and Y in experiment 1, the rate of formation of Z was measured. If the reactions was first order with respect to X and second order with respect to Y, the initial rate of formation of Z in experiment 2 would be A) 4R B) C) R D) 2R

Clue: [X] cut in half but [Y] doubled. Question 1 Answer D Clue: [X] cut in half but [Y] doubled. If x is cut in half and Y is doubled x is 1st order so has a halving affect, the Y is 2nd order so has a 4x effect so ½ times 4 = 2

Rate = [X]1[Y]2 Since concentration of X is cut in half we’ll put in ½ for X and the concentration of Y is doubled we’ll put in 2 for Y Rate = [½ ]1[2]2 And ½ x 4 = 2 so the overall rate should double.

A) 15.5 sec B) 124 sec C) 41.3 sec D) 62.0 sec 2. Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressure of cyclobutene in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature? A) 15.5 sec B) 124 sec C) 41.3 sec D) 62.0 sec

Clue: How many half lives? Question 2 Answer C Clue: How many half lives? First order means the half life is constant. 1/8 remaining mean’s 3 half lives occurred. 124 seconds in 3 half lives means 41.3 seconds per half life.

1st order so half life is constant In the stem it specifies that 1/8 of the original reactant is left which means the amount of reactant gets cut in half 3 times ½ x ½ x ½ = 1/8 If 3 half lives occurs in 124 seconds then 1 half life occurs in 41.3 seconds.

3. Factors of reactant molecule collisions that affect the rate of a chemical reaction include which of the following? A) Kinetic energy of collisions B) Frequency of collisions AND Kinetic energy of collisions C) Frequency of collisions AND Orientation of particles during collisions. D) Kinetic energy of collisions AND Orientation of particles during collisions AND Frequency of collisions

Clue: basic definitions Question 3 Answer D Clue: basic definitions

For a reaction to occur molecules must collide, collide with enough energy (kinetic) and collide in the right orientation.

4. X + 2Y® Z + 3Q For the reaction represented above, the initial rate of decrease in [X] was 2.8 x 10-3 mol L-1 s-1. What was the initial rate of decrease in [Y]? A) 5.6 x 10-3 mol L-1 s-1 B) 1.4 x 10-3 mol L-1 s-1 C) 2.8 x 10-3 mol L-1 s-1 D) 7.0 x 10-4 mol L-1 s-1

Clue: Not really kinetics, more stoichiometry. Question 4 Answer A Clue: Not really kinetics, more stoichiometry. Sometimes problems are really simple. Remind them how AP writes units in the mol L-1 format

According to the equation 1X + 2Y® Z + 3Q however much X disappears 2x much Y should disappear.

H2 gas and N2 gas were placed in a rigid vessel and allowed to reach equilibrium in the presence of a catalyst according to the following equation.

5. Which of the following was true for the system between time t1 and time t2 ? The concentration of N2 decreased. B) The rate of formation of NH3 molecules was equal to the rate of disappearance of H2 molecules. C) The number of effective collisions between H2 and N2 was zero. D) The rates of the forward and reverse reactions were equal.

Clue: The graph is the key. Question 5 Answer D Clue: The graph is the key.

The graph shows between t1 and t2 the concentrations are constant The graph shows between t1 and t2 the concentrations are constant. However there are still collisions going on. It is just that the reaction is happening at the same rate forward and backward. B looks promising but the balanced equation has different coefficients for N2 and NH3. That is AP trying to see if you’ll bite at an answer that looks good but isn’t really true. Remember to read all the choices and to think critically.

6. Which of the following statements best explains why an increase in temperature of 5-10 Celsius degrees can substantially increase the rate of a chemical reaction? A) The activation energy for the reaction is lowered. B) The number of effective collisions between reactant particles is increased. C) The rate of the reverse reaction is increased. D) DH for the reaction is lowered.

Clue: What does temperature describe about molecules? Question 6 Answer B Clue: What does temperature describe about molecules? It might be helpful to

Increasing the temperature means the molecules have more kinetic energy or are moving faster. When they move faster they will collide more meaning a better chance of hitting hard enough in the right orientation. Activation energy and DH would remain unchanged.

7. A reaction mechanism for the destruction of ozone, O3(g), is represented above. In the overall reaction, NO(g) is best described as ? A) an inhibitor B) a catalyst C) a reactant D) an intermediate E) a product

Clue: Look at NO at the beginning and the end. Question 7 Answer A Clue: Look at NO at the beginning and the end.

Something that occurs at the beginning, changes in the middle and yet at the end is returned to normal is called a catalyst. Catalysts do not take change in the chemical reaction and are left behind to catalyze more reactions. .

8. The rate law for the reaction of nitrogen dioxide and chlorine is found to be rate = k [NO2]2[Cl2]. By what factor does the rate of the reaction change when the concentrations of both NO2 and Cl2 are doubled? A) 6 B) 8 C) 4 D) 2

Clue: Like the Glade commerical says, plug it in, plug it in. Question 8 Answer C Clue: Like the Glade commerical says, plug it in, plug it in.

rate = k [NO2]2[Cl2] And both reactant concentrations are doubled so rate = k [2]2[2] So the rate will go up by a factor of 4x2 or 8.

Free Response 1 (8 pts possible) This was question 5 from 2013. Mean of 3.70 and SD of 2.04 parts a and b are more to do with KMT and nature of molecules but a good time to bring it up with students one more time. Skip a and b if you want.

Part b is another example of question that doesn’t really have to do with kinetics so skip it if you want.

Please assure the students that most students did not get the two step curve. They drew the traditional one hill graph. The last point is a gimme everyone should know.

Short Free Response 1 (3-5 points possible) a. b. This is a part of question #3 on the 2012 exam. On the entire question there were 9 points possible with an average of 3.20. Feel free to give one part a, b, or c or any combination of those. Most of the short questions come in at 3 points. If you only give them b then be sure to give them the value for k. c.

Short Free Response 2 (4 points possible) a. This is a part of question #6 on the 2011 exam. On the entire question there were 8 points possible with an average of 3.98. Be sure to have them reference the equation sheet and rearrange the integrated rate laws to put them in slope intercept formula so they can recognize a graph and how to match it up with order.