Unit 9a: Kinetics and Energy Changes

Transcription

1 Unit 9a: Kinetics and Energy Changes Student Name: Key Class Period: Website upload 2015 Page 1 of 43 Unit 9a (Kinetics & Energy Changes) Key

2 Page intentionally blank Website upload 2015 Page 2 of 43 Unit 9a (Kinetics & Energy Changes) Key

3 Unit 9a Vocabulary: 1. Activated Complex: The species that are formed and decomposed during the mechanism; also called the intermediate. 2. Activation Energy: The energy that must be added to allow the reactants to start the reaction and form the activated complex. 3. Catalyst: A chemical that is added to a reaction to eliminate steps in the mechanism, increase the reaction rate, and decrease the activation energy without itself being consumed by the reaction. 4. Effective Collision: A collision between reactant particles that results in a chemical reaction taking place. 5. Enthalpy: The total amount of potential energy stored in a substance. 6. Endothermic: A reaction that absorbs and stores energy from the surrounding environment. 7. Entropy: A system s state of disorder. Entropy increases as temperature increases. Entropy increases as a substance goes from solid to liquid to gas. 8. Equilibrium: A system where the rate of forward change is equal to the rate of reverse change. At equilibrium there is no net change. 9. Exothermic: A reaction that releases stored energy into the surrounding environment. 10. Favored: A change in a thermodynamic property that contributes towards the reaction being spontaneous. 11. Free Energy: The total amount of energy available in a system to do work. Free Energy is a combination of both enthalpy and entropy. 12. Heat of Reaction: The net gain or loss of potential energy during a chemical reaction. 13. Inhibitor: A chemical that is added to a reaction to add steps to the mechanism, decrease the reaction rate, and increase the activation energy without itself being consumed by the reaction. 14. Kinetics: The study of reaction mechanisms and reaction rates. 15. Nonspontaneous: A reaction that requires a constant input of energy to occur, or the reaction will reverse or stop. 16. Reaction Rate: The amount of reactant consumed in a given unit of time. Website upload 2015 Page 3 of 43 Unit 9a (Kinetics & Energy Changes) Key

4 17. Spontaneous: A reaction that continues independently once started. 18. Thermodynamics: The study of heat flow during physical and chemical changes. 19. Unfavored: A change in a thermodynamic property that contributes towards the reaction being nonspontaneous. Website upload 2015 Page 4 of 43 Unit 9a (Kinetics & Energy Changes) Key

5 Notes page: Website upload 2015 Page 5 of 43 Unit 9a (Kinetics & Energy Changes) Key

6 Unit 9a Homework Assignments: Assignment: Date: Due: Website upload 2015 Page 6 of 43 Unit 9a (Kinetics & Energy Changes) Key

7 Topic: Kinetics Objective: How do reactions occur, and how fast do they occur? Chemical Kinetics: Chemical Kinetics are the study of: i. Reaction mechanisms (how reactions occur) and; ii. Reaction rates (how long the reaction takes to complete). Mechanism: A mechanism is the pathway the reaction takes. i. Each mechanism is a series of steps that leads from reactants to products; ii. Particles of reactant material must collide to react; iii. Collisions must occur with enough activation energy to react; iv. Molecules must be orientated (positioned) properly to react. Collisions that satisfy these requirements and lead to the initiation of the reaction are labeled effective collisions. The effective mechanism can occur in a series of steps, each involving electron shifts as old bonds are broken and new bonds are formed. How to speed up chemical reactions (and get a date) - Aaron Sams - YouTube - 4:56 Website upload 2015 Page 7 of 43 Unit 9a (Kinetics & Energy Changes) Key

8 Topic: Kinetic Mechanisms Objective: What types of steps may make a reaction mechanism? Reaction Example: CO (g) + NO 2(g) CO 2(g) + NO (g) This reaction does NOT go directly as stated above. It first starts with the decomposition of nitrogen dioxide. Step #1: Step #2: 2 NO 2(g) NO 3(g) + NO (g) CO (g) + NO 3(g) CO 2(g) + NO 2(g) Final step: CO (g) + NO 2(g) CO 2(g) + NO (g) Reaction Example (Reference Table I): H 2 + I 2 2 HI For the reaction above, a possible reaction might be: 1. Step #1: H 2(g) 2 H (g) (the covalent bond in the diatomic hydrogen molecule absorbs energy to break apart into hydrogen atoms; 2. Step #2: I 2(g) 2 I (g) (the covalent bonds in the diatomic iodine molecule absorb energy to break apart into iodine atoms; 3. Step #3: 2 I (g) + 2 H (g) 2 HI (g) (H would rather bond to I than H - electronegativity). New H-I bonds form, releasing energy; 4. Net Reaction: H 2(g) + I 2(g) 2 HI (g) (2 H and 2 I are intermediates (activated complex) formed in steps #1 & #2, which are then used in step #3. Website upload 2015 Page 8 of 43 Unit 9a (Kinetics & Energy Changes) Key

9 Topic: Kinetic Mechanisms Objective: What types of steps may make a reaction mechanism? Reaction mechanisms: A reaction mechanism is similar to different roads that lead to the same destination. Depending on which road taken, it requires a different amount of time and/or energy to reach your destination. The only way to learn exactly which mechanism is truly responsible for the reaction occurring is to complete an experiment. The overall rate of any mechanism is determined by the slowest step in the mechanism. o The slowest step is the rate-limiting step or rate-determining step. For any reaction, whichever step is the slowest is the part of the mechanism that controls the speed of the overall reaction. Watch Bozeman Chemistry The Rate-Limiting Step - YouTube - 6:46 Website upload 2015 Page 9 of 43 Unit 9a (Kinetics & Energy Changes) Key

10 Topic: Catalysts Objective: How might we increase the rate of a reaction? Catalysts: A catalyst is something that speeds up a reaction by modifying a step in the reaction mechanism with the result being a shortened mechanism. i. A catalyst generally lowers the activation energy required to initiate a mechanism, or a step in a mechanism, allowing the overall reaction to proceed at a faster rate. ii. Catalysts are not consumed by the reaction. A catalyst becomes a temporary part of the mechanism, then is released, and is able to work again within the mechanism. iii. When you dissolve an ionic solid (crystal) in water to form an aqueous ionic solution before undergoing a double replacement reaction, the water acts as a catalyst as the moving water molecules increase the rate that the dissolved ions will collide. Faster collisions = faster reactions! Website upload 2015 Page 10 of 43 Unit 9a (Kinetics & Energy Changes) Key

11 Topic: Inhibitors Objective: How might we decrease the rate of a reaction? Inhibitor: An inhibitor is something that slows down a reaction by adding additional steps to the mechanism with the result generally lengthening the amount of time needed for the reaction to complete. i. Inhibitors generally increase the activation energy; if you require more energy to initiate a reaction, it will take longer to add that extra energy. ii. Inhibitors are not consumed by the reaction. iii. Certain non-ferrous metals form a patina (a combination of oxides, carbonates, or sulfides) that will inhibit additional oxygen from gaining access to the metal and causing further oxidation. The Statue of Liberty is made of copper sheets; the green color we see is the result of oxygen reacting with the copper and forming the greenish patina. Website upload 2015 Page 11 of 43 Unit 9a (Kinetics & Energy Changes) Key

12 Topic: Reaction Rates Objective: How do we measure the amount of a reaction over time? Reaction Rate: The speed at which a reactant is chemically changed into a product is known as the Reaction Rate. i. The rate of reaction is generally measured in terms of amount of reactants consumed over a period of time. ii. Reaction rate is measured by the number of effective molecular collisions that occur per unit of time. The greater the number of effective collisions in a period, the faster the reaction rate. iii. Any action that speeds up the rate of effective collisions will increase the rate of the reaction. Website upload 2015 Page 12 of 43 Unit 9a (Kinetics & Energy Changes) Key

13 Topic: Rate and Reactant Nature Objective: How may the type of reactants change reaction rate? Ionic Reactions: 1. Ionic reactions occur quickly; 2. Ionic reactions occur in liquid water, and the nature of liquid water allows a large number of collisions in aqueous solution; 3. Aqueous solutions of ions have the ionic bonds already broken, and they can readily make new ionic bonds; 4. Precipitate reactions occur almost immediately upon mixing solutions. Covalent reactions: 1. Covalent reactions occur slowly; 2. Covalently bonded molecules have many different types of bonding, and that can make some covalent reactions very slow; 3. Covalently bonded molecules may have single, double, or even triple bonds that require large amounts of energy to break; 4. Covalent molecules used in cellular processes use biological catalysts, known as enzymes, to speed up the process of breaking down these complex bonds more efficiently and more quickly. Website upload 2015 Page 13 of 43 Unit 9a (Kinetics & Energy Changes) Key

14 Topic: Rate and Temperature Determining Objective: How Molecular may the Polarity: temperature change reaction rate? Temperature: You should recall that temperature is the measure of the average kinetic energy of particles in a system. Kinetic energy is moving energy; if you increase the temperature, you increase the kinetic (moving) energy of the system. i. If you increase the kinetic (moving) motion, you will increase the number of possible effective collisions. ii. We add heat energy (Bunsen burner; hotplate) to increase the temperature (kinetic energy) so we can complete a reaction in a reasonable amount of time. Lower Temperature Higher Temperature Website upload 2015 Page 14 of 43 Unit 9a (Kinetics & Energy Changes) Key

15 Topic: Rate and Concentration Objective: How may concentration affect the rate of reaction? Concentration: Concentration of reactants affects the orientation of colliding particles. The more particles there are in a given volume, the greater the chance they will collide. i. As particles will only react if they collide in an orientation that allows them to bond, having more particles increases the random odds of the orientation being correct. ii. If the greater concentration allows for more random orientations, a greater number of possible effective collisions may result. iii. Think of cars in a parking lot; if more cars are moving around, the greater the chance for a collision than if the parking lot is empty. Lower Concentration Higher Concentration Website upload 2015 Page 15 of 43 Unit 9a (Kinetics & Energy Changes) Key

16 Topic: Rate and Surface Area Objective: How may surface area affect the rate of reaction? Surface Area: Surface area of the reactants affects the rate of reaction much like concentration. If you increase the surface area, you increase the number of properly orientated collision sites. i. This property only works with solids, as liquids and gases have maximum surface area already. ii. Crushing a solid will increase the surface area while the volume stays the same, effectively increasing the collision locations. iii. This property you already learned in Earth Science in regards to weathering; larger rocks weather slower than smaller rocks, as the surface area to volume ratio for larger rocks is lower. Larger Particles (less area) Smaller particles (more area) Larger particles have a lower surface area : volume ratio Smaller particles have a higher surface area : volume ratio Website upload 2015 Page 16 of 43 Unit 9a (Kinetics & Energy Changes) Key

17 Notes page: Website upload 2015 Page 17 of 43 Unit 9a (Kinetics & Energy Changes) Key

18 Kinetics Practice Regents Questions: (ungraded) 1. In most aqueous reactions as temperature of the system increases, the effectiveness of particle collisions a) Increases b) Decreases c) Remains the same 2. Given the reaction of Mg (s) + 2 H 2 O (l) Mg(OH) 2(aq) + H 2(g), at which temperature will the reaction occur at the greatest rate? a) 25 C b) 50 C c) 75 C d) 100 C 3. Given the reaction of A (s) + B (aq) C (aq) + D (s), which change below would increase the rate of this reaction? a) A decrease in pressure c) A decrease in temperature b) An increase in pressure d) An increase in temperature 4. As the concentration of reacting particles increases, the rate of reaction generally a) Decreases b) Increases c) Remains the same 5. Given the reaction of A 2(g) + B 2(g) 2 AB (g) + heat, an increase in the concentration of A 2(g) will a) Increase the production of B 2(g) b) Decrease the production of AB (g) c) Decrease the frequency of collisions between A 2(g) and B 2(g) d) Increase the frequency of collisions between A 2(g) and B 2(g) 6. At STP, which 4.0 gram sample of zinc metal will react the fastest with dilute hydrochloric acid? a) Bar c) Powdered b) Flat sheet d) Irregular lump Website upload 2015 Page 18 of 43 Unit 9a (Kinetics & Energy Changes) Key

19 Student name: Key Class Period: _3, 5, & 10_ Kinetics homework Please carefully remove this page from your packet to hand in. 1. Which statement below explains why the speed of a chemical reaction is increased when the surface area of a reactant is increased? a) This change increases the concentration of the reactant. b) This change increases the density of the reactant particles. c) This change exposes more reactant particles to a possible collision. d) This change alters the electrical conductivity of the reactant particles. 2. A catalyst works by a) Increasing the energy released during a reaction. b) Increasing the potential energy of the reactants. c) Decreasing the potential energy of the reactants. d) Decreasing the activation energy of the reaction. 3. An increase in the surface area of reactants in a reaction will result in a) A decrease in the heat of reaction. b) An increase in the heat of reaction. c) A decrease in the rate of the reaction. d) An increase in the rate of the reaction. 4. For a given reaction mechanism (A to E) and the table of observed durations, which step is the rate-determining step? Step Duration Rate Determining step? A secs B secs C secs D secs E secs Website upload 2015 Page 19 of 43 Unit 9a (Kinetics & Energy Changes) Key

20 5. Explain how the following would affect the reaction rate for the below examples. (Increase, Decrease, Remains the same) a) Adding additional N 2(g) to the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) : I b) Removing H 2(g) from the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) : D c) Increasing the pressure on the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) : I d) Use powdered NaCl (s) and not large crystals of NaCl (s) in the reaction: I e) Adding water as a catalyst to a double replacement reaction: I f) Adding CuSO 4(aq) inhibitor to a reaction of acid and metal: D g) Increasing pressure on Na (s) + ZnSO 4(aq) Na 2 SO 4(aq) + Zn (s) : R 6. A student places three separate samples of sugar in three different insulated cups each containing 50.0 ml of distilled water at 50.0 C. Sample A is a single cube of sugar, Sample B is granulated sugar, and Sample C is powered sugar. Which sample will dissolve the slowest? Explain your answer in terms of effective collisions. Sample A is a single cube; it has the lowest surface area: volume ratio. More molecules of the sugar are hidden from effective collisions inside 7. A student places three identical 5.0 g gram samples of sugar into three different insulated cups each containing 50.0 ml of distilled water. Cup A contains water at 10 C, Cup B contains water at 20 C, and Cup C contains water at 30 C. In which cup will the sugar dissolve the fastest? Explain your answer in terms of effective collisions. Cup C has the highest average kinetic energy; therefore, higher kinetic energy would have a higher rate of effective collisions Website upload 2015 Page 20 of 43 Unit 9a (Kinetics & Energy Changes) Key

21 Topic: Energy Changes & Enthalpy Objective: How does energy change during chemical reactions? Energy Changes in Reactions: In chemical reactions, reactants form products with either an associated release or an associated absorption of energy. Enthalpy is the total amount of Potential Energy (PE) stored in a substance. i. By changing the amount of PE involved in a chemical reaction, we can change the enthalpy of the system. a) Exothermic reactions release energy, so the enthalpy of the system lowers. b) Endothermic reactions store energy, so the enthalpy of the system rises. Enthalpy decreases Enthalpy increases A + B C + D Reactants Energy Energy A + B C + D Products Website upload 2015 Page 21 of 43 Unit 9a (Kinetics & Energy Changes) Key

22 Topic: Endothermic Reactions Objective: How is energy absorbed during a reaction? Endothermic Reactions: During an endothermic reaction, energy is absorbed as a reactant. A + B + Energy C + D i. Energy can be considered as an additional reactant; ii. Heat energy is absorbed by the reactants; iii. This heat energy is absorbed from the environment around the reaction, and the heat in the surroundings decreases; iv. The products of an endothermic reaction have more energy than did the starting reactants; v. Kinetic energy is stored in the created bonds of the products; vi. The excess stored energy makes the products of endothermic reactions chemically unstable and highly reactive; vii. Examples of endothermic reactions include nitroglycerine, trinitrotoluene (TNT), and other explosives that have high amounts of energy stored within their chemical bonds and are easily reactive. Website upload 2015 Page 22 of 43 Unit 9a (Kinetics & Energy Changes) Key

23 Endothermic Reaction example: For the General Reaction of A + B C: i. If H A (Heat Energy of Reactant A ) is equal to 40 kj, and H B = 20 kj, then the reactants together have a total heat energy of 60 kj. If the given H C = 110 kj, then (110 kj 60 kj) = 50 kj of heat energy was absorbed by the reactants from the surroundings to form the product. ii. This reaction can therefore be rewritten to show the energy change as: A (40kJ) + B (20 kj) + Absorbed Energy (50 kj) C (110 kj) (40 kj + 20 kj + 50 kj = 110 kj) 110 kj iii. Note that there is a total of 110 kj of heat energy on both sides of the equation. Equal energy amounts on both sides of the equation satisfies the Law of Conservation of Energy. 110 kj 110 kj Website upload 2015 Page 23 of 43 Unit 9a (Kinetics & Energy Changes) Key

24 Endothermic Reaction example: For a reaction from the NYS Chemistry Reference Table I: N 2(g) + O 2(g) 2 NO (g) H = kj i. The plus (+) sign in front of the kj indicates that this reaction is endothermic, and that energy was absorbed. You can add the energy for the reaction listed in the Reference Table I to the reactants side, as it becomes a part of the products. N 2(g) + O 2(g) kj 2 NO (g) ii. This new reaction equation shows that 1 mole of N 2(g) and 1 mole of O 2(g) absorb kj of energy during the formation of 2 moles of NO (g). iii. 2 moles of NO (g) therefore have kj more energy stored in bonds than 1 mole of N 2(g) and 1 mole of O 2(g) have in their combined bonds. iv. NO (g) is more unstable than N 2(g) and O 2(g). Website upload 2015 Page 24 of 43 Unit 9a (Kinetics & Energy Changes) Key

25 Topic: Exothermic Reactions Objective: How is energy released during a reaction? Exothermic Reactions: During an endothermic reaction, energy is released as a product. A + B C + D + Energy i. Energy can be considered as an additional product; ii. Heat energy is released from the reactants; iii. This heat energy is released into the environment around the reaction, and the heat in the surroundings increases; iv. The products of an exothermic reaction have less energy than did the starting reactants; v. Potential energy in the bonds of the reactants is lost from the products; vi. The lost energy makes the products of exothermic reactions chemically more stable and less reactive than their reactants; vii. An example of an exothermic reaction is the burning of paper or wood. After they have burned, the ash that remains has less stored energy and is more stable (inflammable). Website upload 2015 Page 25 of 43 Unit 9a (Kinetics & Energy Changes) Key

26 Exothermic Reaction example: For the General Reaction of A + B C: i. If H A = 60 kj and H B = 40 kj, then the reactants together have a total heat energy of 100 kj. If H C = 30 kj, then (100 kj - 30 kj) = 70 kj of heat energy was released from the reactants to the surroundings while forming the products. ii. This reaction can therefore be rewritten to show the energy change as: A (60kJ) + B (40 kj) C (30 kj) + Released Energy (70 kj) (60 kj + 40 kj = 100 kj) (30 kj + 70 kj = 100 kj) iii. Note that there is a total of 100 kj of heat energy on both sides of the equation. Equal energy amounts on both sides of the equation satisfies the Law of Conservation of Energy. 100 kj 100 kj Website upload 2015 Page 26 of 43 Unit 9a (Kinetics & Energy Changes) Key

27 Exothermic Reaction example: For a reaction from the NYS Chemistry Reference Table I: C (s) + O 2(g) CO 2(g) H = kj i. The minus (-) sign in front of the kj indicates that this reaction is exothermic, and that energy was released. You can add the energy for the reaction listed in the Reference Table I to the products side, as it was lost from the reactants. C (s) + O 2(g) CO 2(g) kj ii. This new reaction equation shows that C (s) + O 2(g) together release kj of energy during the mechanism that forms CO 2(g). iii. 1 mole of CO 2(g) therefore has kj less energy stored in its bonds than 1 mole of C (s) and 1 mole of O 2(g) have in their combined bonds. iv. CO 2(g) is more stable than N 2(g) and O 2(g) are alone. Watch Energy & Chemistry: Crash Course Chemistry #17 - YouTube - 9:25 Website upload 2015 Page 27 of 43 Unit 9a (Kinetics & Energy Changes) Key

28 Topic: Molar Heat Energy Objective: How could you calculate molar heat energy? The reaction on the previous page shows the energy released during the SYNTHESIS of one mole of CO 2(g). The mechanism for this synthesis reaction releases 393.5kJ of energy into the environment. We can use this quantitative amount of energy for the synthesis of one mole of CO 2(g) to calculate the energy released by any molar ratio. Example: i. How much energy is released during the mechanism to synthesize 2.3 moles of CO 2(g)? 2.3 moles of CO 2(g) x = kj kj ii. The above equation states that 910 kj of energy are released when 2.3 moles of CO 2(g) are synthesized. iii. What of the decomposition of CO 2(g)? Decomposition is the reverse process of synthesis, so the same given energy would be used, but the sign would be the OPPOSITE. Example: kj mole of CO 2(g) kj mole of CO 2(g) 2.3 moles of CO 2(g) x = kj kj i. The above equation states that 910 kj of energy are absorbed when 2.3 moles of CO 2(g) are decomposed. H of Decomposition = (opposite sign) H of Synthesis Website upload 2015 Page 28 of 43 Unit 9a (Kinetics & Energy Changes) Key

29 Kinetics Practice Regents Questions: (ungraded) 1. Given the reaction of A + B C + D + heat, which statement best describes this reaction? a) The forward reaction is exothermic, and the reverse reaction is always exothermic. b) The forward reaction is exothermic, and the reverse reaction is always endothermic. c) The forward reaction is exothermic, and the reverse reaction can be either exothermic or endothermic. d) The forward reaction is endothermic, and the reverse reaction can be either endothermic or exothermic. Salt A and Salt B were dissolved in separate 100-mL samples of water. The water temperatures were measured and recorded as shown in the table below. Salt A Salt B Initial water temperature: 25.1 C 25.1 C Final water temperature: 30.2 C 20.0 C 2. Which statement is a correct interpretation of these results? a) The dissolving of only salt B was exothermic. b) The dissolving of only salt A was endothermic. c) The dissolving of both salt A and salt B was endothermic. d) The dissolving of salt A was exothermic and the dissolving of salt B was endothermic. 3. Given the reaction of Fe (s) + S (s) FeS (s) + energy, which statement about this reaction is true? a) It is exothermic. b) It is endothermic. c) The potential energy of the reactants is the same as the potential energy of the product. d) The potential energy of the reactants is lower than the potential energy of the product. Website upload 2015 Page 29 of 43 Unit 9a (Kinetics & Energy Changes) Key

30 Topic: Potential Energy Diagrams Objective: How do we depict the flow of energy within a reaction? Potential Energy Diagrams: A Potential Energy (PE) Diagram is a graphical depiction of the flow of potential energy as a reaction goes from start to finish. A basic Potential Energy flow process consists of these steps: i. All reactions start with reactants, therefore the PE begins at the Heat of Reactants (H R ) level; ii. Activation Energy (E A ) must be added to the reactants to initiate the reaction mechanism: a) E A is the Energy needed to have effective collisions between reactant particles; b) E A raises the level of PE in the reaction to the Heat of Activated Complex (H AP - intermediate step) level; iii. The Activated Complex is VERY short-lived and breaks down quickly to form the products of the reaction. This final step lowers the level of PE to the Heat of Products (H P ) level. Watch Bozeman Chemistry Activation Energy - YouTube - 4:51 Website upload 2015 Page 30 of 43 Unit 9a (Kinetics & Energy Changes) Key

31 Topic: Creating PE Diagrams Objective: How do we make a Potential Energy Diagram? Creating a Potential Energy Diagram: Seven items must be included and labeled in a PE Diagram: A. The axes: 1. y-axis is PE in kj; 2. x-axis is the Reaction Coordinate (time from start to finish; this time is unmeasured, therefore it has NO UNITS! B. The Potential Energy Levels: 3. Heat of Reactants; 4. Heat of Products; 5. Heat of Activated Complex C. The Energy Changes during the reaction: 6. Activation Energy (E A ) as an arrow from H R to H AC ; 7. H as an arrow from H R to H P ( H shows NET energy change) Website upload 2015 Page 31 of 43 Unit 9a (Kinetics & Energy Changes) Key

32 Topic: Interpreting PE Diagrams Objective: How do we read a Potential Energy Diagram? Interpreting Endothermic Potential Energy Diagrams: Endothermic Reactions: Energy is absorbed by the reactants as they form products, so the NET amount of PE will increase. For the PE Diagram for reaction A + B + 50 kj C as shown below: Experiments have determined that the combined heats of reactants A and B is 60 kj. This is the combined Heat of Reactants (H R ). Experiments have also determined that the heat of product C is 110 kj. This is the Heat of Products (H P ). Additional experiments show that A and B require 70 kj of energy to be input before A and B will react. This 70 kj is the Activation Energy (E A ). For this Website upload 2015 Page 32 of 43 Unit 9a (Kinetics & Energy Changes) Key

33 reaction, adding 70 kj to reactants A and B yields the Activated Complex, which has its own Heat of Activated Complex (H AC ), which is ALWAYS the highest energy level in ANY reaction. This reaction started with reactants totaling 60 kj and a product having 110 kj, which means we have a NET increase of 50 kj from the reactants. The + 50 kj is called the Heat of Reaction, and shown as H. This is taking energy from the surroundings and storing WITHIN the bonds of product C, meaning it is an endothermic reaction. i. The bonds in C have more energy and are less stable than bonds in A and B ; ii. The temperature of the surroundings has DECREASED as surrounding kinetic energy was stored within the bonds of C. If we added a catalyst to this reaction, the activation energy (E A ) would decrease by removing step(s) from the mechanism. As a result, the H AC would also decrease. The opposite would happen should an inhibitor be added to the reaction. Website upload 2015 Page 33 of 43 Unit 9a (Kinetics & Energy Changes) Key

34 Topic: Interpreting PE Diagrams Objective: How do we read a Potential Energy Diagram? Interpreting Exothermic Potential Energy Diagrams: Exothermic Reactions: Energy is released from the reactants as they form products, so the NET amount of PE will decrease. For the PE Diagram for reaction X + Y Z + 40 kj as shown below: Experiments have determined that the combined heats of reactants (H R ) for X and Y is 70 kj. Experiments have also determined that the heat of product (H P ) of Z is 30 kj. Additional experiments show that X and Y require 80 kj of energy to be input before X and Y will react. This 80 kj is the Activation Energy (E A ). For this reaction, adding 80 kj to reactants X and Y yields the Activated Website upload 2015 Page 34 of 43 Unit 9a (Kinetics & Energy Changes) Key

35 Complex, which has its own Heat of Activated Complex (H AC ), which is ALWAYS the highest energy level in ANY reaction. This reaction started with reactants totaling 70 kj and a product having 30 kj, which means we have a NET decrease of 40 kj from the reactants. The - 40 kj is called the Heat of Reaction, and shown as H. This is taking energy from the bonds in X and Y and releasing that energy into the surroundings as kinetic energy, meaning it is an exothermic reaction. i. The bonds in Z have less energy and are more stable than bonds in X and Y ; ii. The temperature of the surroundings has INCREASED as kinetic energy was released from the bonds of X and Y. If we added a catalyst to this reaction, the activation energy (E A ) would decrease by removing step(s) from the mechanism. As a result, the H AC would also decrease. The opposite would happen should an inhibitor be added to the reaction. Website upload 2015 Page 35 of 43 Unit 9a (Kinetics & Energy Changes) Key

36 Catalysts and Inhibitors: For the PE diagram to the LEFT, Curve 1 would be the control (normal) PE curve of Activation Energy. Curve 2 would be the catalyzed (lower) PE curve of Activation Energy. For the PE Diagram to the RIGHT, Curve 2 would be the control (normal) PE curve of Activation Energy. Curve 1 would be the inhibited (higher) PE curve of Activation Energy. Website upload 2015 Page 36 of 43 Unit 9a (Kinetics & Energy Changes) Key

37 Topic: Reversing Chemical Reactions: Chemical reactions are almost all reversible. Synthesis reactions (putting together) may be reversed as decomposition (breaking down) reactions. With enough kinetic energy, almost any chemical reaction may be reversed. Reversing Reactions Objective: What can be done to reverse a chemical reaction? When a reaction is reversed, the products become the reactants. The reaction coordinate goes from right to left instead of left to right. Note that the forward (exothermic) reaction has a very small E A, but the reverse (endothermic) reaction has a much greater E A. On the diagram below the forward (exothermic) reaction should have a downward H arrow, as the forward reaction goes from higher PE (Point H) to lower PE (Point E). On this diagram, the H arrow for the reverse (endothermic) reaction goes upwards instead. Forward (exothermic) H ^ E ^ Reverse Website upload 2015 Page 37 of 43 (endothermic) Unit 9a (Kinetics & Energy Changes) Key

38 Endothermic Reaction: Exothermic Reaction: Website upload 2015 Page 38 of 43 Unit 9a (Kinetics & Energy Changes) Key

39 Potential Energy Practice Regents Problems: (ungraded) 1. Which information about a chemical reaction is provided by a potential energy diagram? a) The change in solubility of the reacting substances. b) The oxidation states of the reactants and the products. c) The average kinetic energy of the reactants and the products. d) The energy released or absorbed during a chemical reaction. The potential energy diagram below represents a chemical reaction. 2. Which arrow represents the activation energy of the forward reaction above? a) A b) B c) C d) D 3. Given the reaction of S (s) + O 2(g) SO 2(g) + energy, which diagram shown below best represents the potential energy changes for this reaction? 4. The activation energy required for a chemical reaction may be decreased by a) Adding more reactant to the mechanism. b) Increasing the surface area of the reactant. c) Increasing the temperature of the reactant. d) Adding a catalyst to the reaction mechanism. Cont d next page Website upload 2015 Page 39 of 43 Unit 9a (Kinetics & Energy Changes) Key

40 In the diagram below, which letter represents the activation energy for the reverse reaction? a) A b) B c) C d) D 5. In a potential energy diagram, the difference between the potential energy of the products and the potential energy of the reactants is equal to the a) Heat of reaction b) Entropy of reaction c) Activation energy of the reverse reaction d) Activation energy of the forward reaction The potential energy diagram below shows the reaction for X + Y Z: 6. When a catalyst is added to the reaction, it will change the value of a) 1 and 2 c) 1 and 3 b) 2 and 3 d) 3 and 4 Website upload 2015 Page 40 of 43 Unit 9a (Kinetics & Energy Changes) Key

41 Student name: Key Class Period: _3, 5, & 10_ Please carefully remove this page from your packet to hand in. Potential Energy Diagrams Homework: Below are three partially completed reactions. Find the appropriate Heat of Reaction in Reference Table I and write the Heat of Reaction in the proper space, leaving the other space BLANK. State if the reaction is EXOTHERMIC or ENDOTHERMIC based on Table I. Identify the products as being STABLE or UNSTABLE when compared to the reactants for that reaction. 9 pts. Given Reaction: EXOthermic or ENDOthermic? Products STAble or UNSTAble? 1. N 2(g) + 3 H 2(g) + 2 NH 3(g) kj Exo Stable 2. N 2(g) + 2 O 2(g) kj 2 NO 2(g) + Endo Unstable 3. H 2(g) + I 2(g) kj 2 HI (g) Endo Unstable Answer questions #4 through #13 using Reference Table I for the formation of NO (g) from its elements. 1 pt. ea. except where noted. 4. How many moles of NO (g) are formed in the reaction on Table I? 2 moles 5. What is the Heat of Reaction for the formation of NO (g) per mole? _+91.3_ kj/mole 6. Show your calculations for the above answer: kj/2 moles NO (g) = 91.3 kj/1 mole NO (g) 7. How much energy is absorbed as 4.5 moles of NO (g) are formed? +410 kj 8. Show your calculations for the above answer: kj/mole NO (g) x 4.5 moles NO (g) = = +410 kj 9. How much energy is released as 2.7 moles of NO (g) are decomposed? -250 kj 10. Show your calculations for the above answer: kj/mole NO (g) x 2.7 moles NO (g) = = -250 kj Samples of N 2(g) and O 2(g) are reacted to form moles of NO (g). 11. How much energy will be absorbed by the reaction given above? kj 12. Show your calculations for the above answer: kj/mole NO (g) x moles NO (g) = = 0.91 kj Cont d next page Website upload 2015 Page 41 of 43 Unit 9a (Kinetics & Energy Changes) Key

42 13. If the reaction in question #11 was carried out in a calorimeter in 50.0 g of H 2 O at an initial temperature of 20.0 C, what would the final water temperature be? Show ALL work. 3 pts. q=mc T q/mc = T -910 J / [(50.0 g) x (4.18J/g C)] = -4.4 C T = 20.0 C = 15.6 C final water temperature (water LOST -910 J to the reaction) kj x 1000 J/1 kj = -910 J Using the Potential Energy Diagram below, answer the following questions. 14. What is the heat of the reactants? a) 30 kj b) 110 kj c) 140 kj d) 160 kj 15. What is the heat of the products? a) 30 kj b) 110 kj c) 140 kj d) 160 kj 16. What is the heat of the activated complex for the uncatalyzed reaction? a) 30 kj b) 110 kj c) 140 kj d) 160 kj 17. What is the activation energy for the catalyzed reaction? a) 30 kj b) 110 kj c) 140 kj d) 160 kj 18. What is the H of this reaction? a) +80 kj b) -80 kj c) +130 kj d) -130 kj 19. What type of reaction is shown on this diagram? a) Exothermic b) Endothermic c) Both types d) Neither type 20. Using the dashed catalyzed reaction line as a guide, draw a dotted ( ) curved line on the above diagram to indicate what the PE curve would look like if a catalyst were added. Website upload 2015 Page 42 of 43 Unit 9a (Kinetics & Energy Changes) Key

43 Notes page: Website upload 2015 Page 43 of 43 Unit 9a (Kinetics & Energy Changes) Key