Chemical Kinetics PDF

PHA 612: Pharmaceutical Inorganic Chemistry (with Qualitative Analysis) LECTURE: CHEMICAL KINETICS UNIVERSITY OF SANTO TOMAS - FACULTY OF PHARMACY (BS PHARMACY 2026 – 1BPH) [TRANS] CHEMICA...

PHA 612: Pharmaceutical Inorganic Chemistry (with Qualitative Analysis) LECTURE: CHEMICAL KINETICS UNIVERSITY OF SANTO TOMAS - FACULTY OF PHARMACY (BS PHARMACY 2026 – 1BPH) [TRANS] CHEMICAL KINETICS PART I CONTENT I Definition of Chemical Kinetics II Reaction Rate A. Reaction Rate and Stoichiometry B. Computations: Reaction Rate III Instantaneous Rate A. Computations: Instantaneous Rate IV Differential Law/Equations: Figure 1. Reaction rate of A A. Reaction and Reactant Order B. Reaction Order: Rate vs. Concentration Interpretation: Graphs A represents the reactant, B is the product, and t C. Computations: Order rate represents time. D. Reaction Order: Concentration vs. Time Delta (∆) means change. Graphs A is converted into B. We can express the rate for this equation through the formula given. CHEMICAL KINETICS Study of reaction rates and reaction mechanism Basically, the rate is just the same for reactants and o Answers how fast the reaction goes products. The reaction rate is equal. o Studies the reaction rates and reaction There is a negative sign before the reactant because reactant with respect to time decreases to mechanism by which a chemical process gets form the product. to its initial state to its final state ▪ Rate refers to the time for change to Reactant concentration decreases with time while occur product concentration increases with time. o In chemical kinetics, the primary concern is To look for the change in concentration, just simply the speed. subtract the final concentration to the initial concentration. 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 REACTION RATE Molarity = 𝐿𝑖𝑡𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 Unit: M/s or mol/L.s Change in reactant or product concentration with respect to time Rates at which reactants disappear or products appear o In a chemical reaction, reactants collide to form a product. Hence, the time at which the reactants disappear and products appear is called reaction rate. o The purpose of reaction rate is to be knowledgeable of the time as we want our reaction rates to be fast in order for it to be feasible for commercial use. o If the reaction rate is slow, it may not be possible for chemical production kasi magsslow down yung production. Figure 2. Representation of Reaction Time Interpretation: REACTION RATE AND STOICHIOMETRY Reactant concentration decreases in time and Stoichiometry is the quantitative relationship of product concentration increases in time. reactants and products in a chemical reaction (molar At the end of the chemical reaction, product concentration). concentration is already greater than the reactant The following are the possible balance chemical concentration. equation for a reactant-product formula: II. 2A → 1B I. A→B 1 ∆[𝐴] ∆[𝐵] ∆[𝐴] ∆[𝐵] 𝑅𝑎𝑡𝑒 = − 2 ∆𝑡 = ∆𝑡 𝑅𝑎𝑡𝑒 = − ∆𝑡 = ∆𝑡 Interpretation: 𝑚𝑜𝑙 ∆𝐴 = [𝐴]𝑓𝑖𝑛𝑎𝑙 − [𝐴]𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑢𝑛𝑖𝑡 𝑖𝑠 𝑀 𝑜𝑟 2 moles of A disappear for each mole of B formed. 𝐿 ∆𝑡 = [𝑡]𝑓𝑖𝑛𝑎𝑙 − [𝑡]𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑢𝑛𝑖𝑡 𝑖𝑠 𝑠𝑒𝑐 We need 2 moles of reactant in order to form 1 mole of the product. III. aA + bB → cC + dD 1 ∆[𝐴] 1 ∆[𝐵] 1 ∆[𝐶] 1 ∆[𝐷] 𝑅𝑎𝑡𝑒 = − 𝑎 ∆𝑡 = −𝑏 ∆𝑡 =𝑐 ∆𝑡 = 𝑑 ∆𝑡 Interpretation: BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS Small letters represent the coefficients that balance the equation. COMPUTATIONS: REACTION RATE EXAMPLE Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at a rate of 0.024 M/s. 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔) → 2𝑁2𝑂5 (𝑔) A. Write the rate expression for the reaction 1 ∆[𝑁𝑂 ] ∆[𝑂 ] 1 [𝑁2 𝑂5 ] Figure 4. 𝑅𝑎𝑡𝑒 = − 4 ∆𝑡 2 = − ∆𝑡2 = 2 ∆𝑡 ∆[𝐵𝑟 ] ∆[𝐵𝑟]2𝑓𝑖𝑛𝑎𝑙 − ∆[𝐵𝑟]2𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑅𝑎𝑡𝑒 = − ∆𝑡 2 = 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 B. At what rate is 𝑁2 𝑂5 being formed? 𝑅𝑎𝑡𝑒 = 0.0101 𝑀 − 0.0120 𝑀 = 3.80 𝑥 10 𝑀/𝑠 −5 ∆[𝑁2 𝑂5 ] −0.024 𝑂2 2𝑚𝑜𝑙 𝑁2 𝑂5 50 sec − 0 𝑠𝑒𝑐 ∆𝑡 = 𝑠 𝑥 −1𝑚𝑜𝑙 𝑂2 = 0.048 𝑀/𝑠 o Positive yung final answer kasi dapat Positive value because it’s in the product side. discrete yung value ng rate. C. At what rate is 𝑁𝑂2 reacting? ∆[𝑁𝑂2 ] −0.024 𝑂2 −4𝑚𝑜𝑙 𝑁𝑂2 EXAMPLE #2 ∆𝑡 = 𝑠 𝑥 −1𝑚𝑜𝑙 𝑂 = −0.096 𝑀/𝑠 2 At some time, we observe that the reaction: Negative value because it’s in the reactant side. 2𝑁2 𝑂5 (𝑔) → 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔) is forming 𝑁𝑂2 at the rate if 0.072 𝑚𝑜𝑙/𝐿. 𝑠 INSTANTANEOUS RATE A. Write the rate expression for the reaction 1 ∆𝑁 𝑂 1 ∆𝑁𝑂 ∆𝑂 𝑅𝑎𝑡𝑒 = − 2 ∆𝑡2 5 = 4 ∆𝑡 2 = ∆𝑡2 B. Provided that the rate of 𝑁𝑂2 is 0.072 𝑚𝑜𝑙/𝐿. 𝑠, what is the rate of change of 𝑂2 at this time? ∆[𝑂2 ] 0.072 𝑚𝑜𝑙 𝑁𝑂2 1 𝑚𝑜𝑙 𝑂2 ∆𝑡 = 𝐿.𝑠 𝑥 = 0.018 𝑚𝑜𝑙/𝐿. 𝑠 𝑂2 4 𝑚𝑜𝑙 𝑁𝑂2 C. What is the rate of change of 𝑁2 𝑂5 when the rate of 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔) is produced at 0.072 𝑚𝑜𝑙/𝐿. 𝑠? ∆[𝑁2 𝑂5 ] 0.072 𝑚𝑜𝑙 𝑁𝑂2 −2 𝑚𝑜𝑙 𝑁2 𝑂5 − ∆𝑡 = 𝐿.𝑠 𝑥 4 𝑚𝑜𝑙 𝑁𝑂2 = −0.036 𝑚𝑜𝑙/ Figure 3. 𝐿. 𝑠 𝑁2𝑂5 Concentration vs. Time Graph Rate for specific instance Summary: o Instantaneous rate depends kung asaang 2𝑁2 𝑂5 has a rate of −0.036 𝑚𝑜𝑙/𝐿. 𝑠 time na yung chemical reaction 𝑂2 has a rate of 0. 018 𝑚𝑜𝑙/𝐿. 𝑠 Determined from concentration vs. time graph by 𝑁𝑂2 has a rate of 0.072 𝑚𝑜𝑙/𝐿. 𝑠 drawing a line tangent to the curve at the point that corresponds to a particular time. Question: Negative slope of line tangent to the curve at time (t) Bakit magkakaiba yung rate nila kung binanggit kanina na o Negative slope because it follows a linear equal silang lahat? equation Take note na yung “rate of change” is different from o 𝑦 = 𝑚𝑥 + 𝑏 (m is the slope) the “rate of reaction.” Same value for same concentration of reactants at Once it is already the "rate of reaction” being asked, constant Temperature. that’s the time lang na magiging equal yung mga o Provided na constant yung temperature, yung rates. value ng concentration of reactants will remain the same. D. What is the rate of reaction at this time? 1 ∆𝑁2 𝑂5 1 0.036 𝑚𝑜𝑙 𝑁2 𝑂5 𝑅𝑎𝑡𝑒 = − =− = 0.018 𝑚𝑜𝑙/𝑙. 𝑠 2 ∆𝑡 2 𝐿. 𝑠 COMPUTATIONS: INSTANTANEOUS RATE 1 ∆𝑁𝑂2 1 0.072 𝑚𝑜𝑙 𝑁𝑂2 EXAMPLE #1 𝑅𝑎𝑡𝑒 = = = 0.018 𝑚𝑜𝑙/𝑙. 𝑠 4 ∆𝑡 4 𝐿. 𝑠 Using the data provided in the table, calculate the average rate over the 1st 50 seconds time interval. ∆𝑂2 𝐵𝑟2 (𝑎𝑞) + 𝐻𝐶𝑂𝑂𝐻 (𝑎𝑞) → 2𝐵𝑟 − (𝑎𝑞) + 2𝐻+ (𝑎𝑞) + 𝐶𝑂2 (𝑔) 𝑅𝑎𝑡𝑒 = ∆𝑡 = 0.018 𝑚𝑜𝑙/𝑙. 𝑠 RATE LAW/ EQUATIONS Expresses the relationship of reaction rate to the rate constant (k) & reactant concentrations raised to some powers Always determined experimentally BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS REACTION AND REACTANT ORDER REACTION ORDER: RATE VS. CONCENTRATION dependent GRAPHS Reaction rate Rate constant (k) & Reactant concentration Reaction order (1st, 2nd, Zeroth) – is always defined/dependent in terms of reactant concentrations o Product concentration do not appear in the rate law because the reaction rate is being studied under conditions where reverse reaction does not contribute to the overall rate Reactants Products ; not reversible Reactant order is not related to the stoichiometric coefficient in the balanced chemical equation; not connected or related to each other Figure 6. Rate vs. Concentration Graphs o The value of the exponent n (reactant order) Interpretation: – [x & y = 0, 1, 2] must be determined by Under zero order: experiment o Rate is constant regardless the increase/decrease in concentration. EXAMPLE #1 o Does not depend on the amount of the 𝐹2 (𝑔) + 2𝐶𝑙𝑂2 → 2𝐹𝐶𝑙𝑂2 (𝑔 remaining concentration. Rate Law Equation: 𝑘 [𝐹2]1 [𝐶𝑙𝑂2 ]1 or 𝑘 [𝐹2][𝐶𝑙𝑂2 ] o Unaffected by the change in concentration. Interpretation: Under 1st order: aA + bB → cC + dD o They are directly proportional to each Rate = 𝑘 [𝐴] 𝑥 [B] 𝑦 (add x & y for overall order other. reaction) Under 2nd order: o Reaction is 𝑥 𝑡ℎ order in A o Same as the 1st order and vice versa but is o Reaction is 𝑦 𝑡ℎ order in B more drastic in the increase of rate. o Reaction is (𝑥 + 𝑦)𝑡ℎ order overall (so to get the reaction order overall, just add x & y) SUMMARY OF REACTION ORDER CONCENTRATION RATE 0th order (0) Increase/Decrease Constant EXAMPLE #2 1st order (1) Double (ADD) Doubles 𝐹2 (𝑔) + 2𝐶𝑙𝑂2 → 2𝐹𝐶𝑙𝑂2 (𝑔) concentration Rate law equation: 𝑘 [𝐹2] 𝑥 [𝐶𝑙𝑂2 ] 𝑦 2nd order (2) Double concentration Quadruples COMPUTATIONS: ORDER RATE EXAMPLE #1 For a certain reaction where x = 1 & y = 2, the rate law for the reaction is: Rate = 𝒌 = [𝑨][𝑩]𝟐 Figure 5. Rate Data for the Reaction between F2 and ClO2 Assuming initially: [A] = 1.0 M Interpretation: [B] = 1.0 M Double [𝐹2] with [𝐶𝑙𝑂2 ] constant, o Rate doubles A. If we double the concentration of A at constant B, o x = 1 (first order rate) determine by what factor the rate (𝑟𝑎𝑡𝑒2 ) will increase. Quadruple [𝐶𝑙𝑂2 ] with [𝐹2] constant, 𝑅𝑎𝑡𝑒2 = k [1 + 1] 2 = k = 2 (rate doubles) o Rate quadruples o y = 1 (first order rate) B. If the concentration of B is doubled at constant A, determine by what factor the rate (𝑟𝑎𝑡𝑒3 ). Rate law = 𝑘 [𝐹2]1 [𝐶𝑙𝑂2 ]1 or 𝑘 [𝐹2][𝐶𝑙𝑂2 ] 𝑅𝑎𝑡𝑒3 = k [1+1]2 = k 2 = 4 (rate quadruples) ∴ If you double the concentration and if the rate doubles, it is in the first order rate EXAMPLE #2 For a hypothetical reaction: A + B + C → Products, the rate law is determined to be Rate = 𝒌 = [𝑨][𝑩]𝟐 [𝑪]𝟎 – cancel out What happens to the reaction rate when we make each of the following concentration changes? A. We double the concentration of A without changing the concentration of B or C. (DOUBLES) BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS B. We double the concentration of B without changing B. Rate Constant the concentration of A or C. (QUADRUPLES) You can use any data from the experiment. (1, 2 or C. We double the concentration of C without changing 3) the concentration of A or B. (NO EFFECT – because 𝑟𝑎𝑡𝑒 5.0 𝑥 10−5 𝑀 𝑘 = [𝑁𝑂]2 [𝐻 ] = (10.0 𝑥 10−3 𝑀)2 (2.0𝑠 𝑥 10−3 𝑀) the rate is constant) 2 D. We double all three concentrations simultaneously. 102 Final rate constant: 2.5 𝑥 𝑀 2.𝑠 Answer: 8x (8 time the original value) C. Rate of the Reaction when: Rate = [1+1] [1+1]2 0 = 20 = 8x [NO] = 12.0 x 10-3 M [H2] = 6.0 x 10-3 M EXAMPLE #3 Multiply the rate constant to the given concentrations Determine the rate law and calculate the rate constant for 𝑅𝑎𝑡𝑒: (2.5 𝑥 102𝑀 2. 𝑠)(12.0 𝑥 10−3𝑀)2(6.0 𝑥 10−3𝑀) the following reaction from the following data. Final rate of reaction: 2.2 𝑥 10−4 𝑀/𝑠 REACTION ORDER: CONCENTRATION VS. TIME GRAPHS TABLE 1 Figure 7. Experiments of S2O82 AND I- Solution for rate law: The Reactants will be the one that will be used. Rate law expression: k[S2O82-]x[I-]Y o Double [S2O82-] at constant [I-], rate doubles (Expt 2 and 3): x = 1 o Double [I-] at constant [S2O82-], rate doubles (Expt 1 and 2): y = 1 Final rate law: k[S2O82-][I-] Solution for constant rate (k): 10−4 𝑀 2.2 𝑥 Interpretation: 𝑘 = [𝑆 𝑅𝑎𝑡𝑒 = 𝑠 = 0.08 𝑀 2− 2 𝑂8 ] [0.08 𝑀][0.034 𝑀) 𝑠 Zero Order – As the time increases, the o You can use any data from the experiment. concentration decreases. It decreases at a constant (1, 2 or 3) rate. Half-life decreases with time (dependent on the concentration). 1st Order – With respect to time, the concentration EXAMPLE #4 decreases, although the concentration halves in equal time intervals. Half-Life is constant (independent of the initial concentration). 2nd Order – Concentration decreases rapidly, but the rate of decreases slows down. Half-Life increases with time. Additional information: Half-life is the time required for the reactant concentration to decrease by ½. Through this, you will be able to determine the expiration of medicines. Figure 8. Experiments of NO and H2 A. Rate Law TABLE 2 Rate law expression: 𝑘[𝑁𝑂] 𝑥 [𝐻2] 𝑦 o Double [NO] with constant [H2]: Rate Quadruples approximately (Expt 1 and 2): x=2 o Double [H2] with constant [NO]: Rate Doubles (Expt 2 and 3): y = 1 Final rate law: k[NO]2[H2]1 Reaction Order Overall: o Add xth and yth order o (x + y)th order o (2+1) = 3rd order reaction overall Interpretation: The slope is -k because it follows the linear equation (y=mx+b). BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS The plot will be concentration vs time. that are formed, there should be an effective collision The 1st order - ln(concentration) between the atoms or molecules. The 2nd order – Slope = k; has a reciprocal equation Two requirements for an effective collision: showing why the trend is upwards and that 1. The orientation of colliding molecules must be concentration is increasing with time. favorable for making and breaking bonds. 2. Activation energy (Ea) is the minimum amount of energy required to initiate a TABLE 3 chemical reaction.  Ea =  Collision of molecules =  Reaction rate =  Reaction Time 1 1 Rate (𝑎) Ea 𝑎 collision of molecules 𝑎 time o Reaction rate is inversely related with activation energy but directly related to the collision of molecules, and inversely related with time. However, not all collision of molecules will result to a chemical change. Refer to example below. Table of Formulas and Units per order. TABLE 4 Figure 1. Molecular Collision Theory Interpretation: a.) In the first illustration, there is an effective orientation of collision between the reactants. Thus, a reaction is taking place and the products 𝑁𝑂2 and 𝑁2 are produced. b.) In the second illustration, there is an ineffective Interpretation: orientation of collision between the reactants. Thus, Zero order is dependent on the concentration; the no reaction is taking place and the products were higher the concentration gets, the half-life increases. not produced. Directly proportional. c.) In the third illustration, though there is a correct 1st order – not affected; half-life is constant orientation of collision between the reactants, still 2nd order – the higher the concentration, the lower there is no reaction taking place and the products the half-life. Inversely proprtional. will not be produced. PART II CONTENT I Collision Theory of Reaction Rate CONCLUSION ACCORDING TO THE COLLISION II Factors affecting Reaction Rates THEORY III Temperature in terms of the Arrhenius Equation For a chemical reaction to occur: IV Reaction Mechanisms 1. Molecules must collide with sufficient energy, V Catalysis known as the activation energy, so that chemical bonds can break 2. Molecules must collide with the proper A LOOK AHEAD orientation. Rate of a reaction usually increases with temperature. Meeting these 2 criteria will result in a chemical Activation energy, which is the minimum amount of reaction known as a successful collision or an effective energy required to initiate a chemical reaction, also collision. influences rate. The collision theory provides an explanation regarding You’ll be able to examine mechanism of a reaction in the interaction of molecules to cause a reaction and be terms of the elementary steps and see that we can able to form new products. determine the rate law from the slowest or rate- determining step. This is important in verifying FACTORS AFFECTING REACTION RATE mechanisms in experiments. You’ll be able to know the effect of catalyst on the rate of a reaction. 1. THE NATURE OF REACTANTS AND STATE OF You’ll also be able to know the characteristics of SUBDIVISION heterogeneous catalysis, homogeneous catalysis, and Octet rule states that elements having an atomic enzyme catalysis. number smaller than 20 with less than 8 electrons in the outermost shell, tend to react—either to lose or gain electrons—for them to be able to attain stability, which MOLECULAR COLLISION THEORY is having 8 electrons in the outermost shell. Molecules and atoms must collide in order for them to chemically react and to be able to produce the product BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS o An exception to the octet rule is Helium. o Molecules will be moving faster so there will Helium having 2 valence electrons is be greater collision. This is true for the considered to be stable already. gaseous reactions. Metals that will have a faster reaction rate: Group IA or o Activation energy needed will be lesser. Alkali Group (K, Na, Li) & IIA or Alkali Earth Group (Ba, To react: KE ≥ Ea Sr, Ca, Mg) o In order for the reaction to take place, kinetic o Because of the elements’ reactivity, they will energy should be greater or equal to the readily lose electrons and acquire positive amount of the activation energy, charges (cations). In cases when the activation energy is not enough, o Since they are highly reactive, they will only molecules will remain intact and no change will result need a minimum Ea. Thus, they will have from the collision. Thus, no product is formed. faster reaction rate and the collision when With the collision of molecules, there is a formation of they enter the chemical reaction, will also be temporary species called Activated Complex. faster. Non-metals that will have a faster reaction rate: Group ACTIVATED COMPLEX VIIA or Halogens (F, Cl, Br, I) These are the temporary species formed by the o Similar with the previously mentioned metal reactant molecules as a result of the collision before groups, they will have faster reaction rate they form the products. because of their reactivity. They are considered to be highly unstable species o They will readily gain electrons and acquire because of the very high potential energy negative charges for them to be able to attain stability (anions). o They also need minimum Ea. Thus, reaction THERMOCHEMICAL REACTIONS rate will be faster and collision will be greater. Compounds that will have a faster reaction rate: Ionic Compounds a.k.a. Electrovalent Compounds o Best examples are salts. Particle size that will have a faster reaction rate: Finer or smaller particle size o The finer the particle size, the greater the surface area that will be exposed in the reaction. Thus, the reaction rate will be faster. TWO TYPES OF COMPOUNDS Figure 3. A+B → C+D These are the chemical reactions that would include the amount of heat that may either be released or absorbed in the system. Two types: Exothermic and Endothermic reaction Exothermic Reaction: Reaction where heat is released. Upon reacting, there is heating. o Graph: Started with an unstable reactant and the amount of activation energy is minimal with the formation of activated Figure 2. Comparison of the Two Types of Compounds complex, you can measure the amount of Notes: activation energy that is needed in the Ammonia or ammonium hydroxide are made up of exothermic reaction. Also, the formation of non-metals. This kind of base is considered as a the product is stable. covalent compound. Endothermic Reaction: Reaction where heat is In general, organic compounds, acids, and gases absorbed. Upon reacting, there is cooling. are considered covalent compounds because they o Graph: Started with stable reactant and the are non-metals. amount of activation energy needed is a lot because of the cooling effect in endothermic reaction. Also, the formation 2. CONCENTRATION of products is unstable. Concentration refers to the amount of reactants. Conclusion: Exothermic reaction is the ideal type of The effect of concentration will depend on the reaction reaction because we must always start with an order which is also known as the differential rate law: unstable reaction and should end with stable o Zero order:  Reactant concentration = no products. The activation energy needed should also effect on reaction rate be minimal. o First and Second order:  Reactant For every 10th rise in temperature, regardless of concentration =  Reaction rate =  Ea =  whether the reaction is exothermic or endothermic, Collision of molecules reaction rate is approximately doubled or tripled. 3. TEMPERATURE  Temperature (Hot) =  Kinetic energy (KE) =  Vibrational Energy (VE) =  Reaction rate BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS TEMPERATURE DEPENDENCE OF THE RATE However, there is a great amount of activation CONSTANT energy necessary for the reaction to occur. Arrhenius equation: In a catalyzed reaction, there is a catalyst present k = Ae (-Ea/Rt) in the reaction. We started with an unstable reactant and ended up with stable products. But compared to Where: the uncatalyzed reaction, there is lesser amount of K = Reactant constant activation energy necessary for the reaction to A = Collision frequency factor occur. Therefore, there is faster reaction rate in a Ea = Activation energy (J/mol) catalyzed reaction. R = Gas constant (8.314 J/mol.K) T = Absolute temperature (Kelvin) REACTION MECHANISM This is the overall progress of a chemical reaction. Figure 4. It is the sequence of elementary steps that will lead to Correlation: the product formation. The rate constant is directly proportional to the Information gathered in the reaction mechanism will be collision frequency factor but inversely related to discussed below. activation energy. Therefore, the rate constant will depend on the temperature of the reaction because 1. INTERMEDIATES the rate constant is directly related to the These are the species that normally appear in the temperature. sequence of elementary steps or reaction mechanism but not in the overall balanced equation. o They are completely removed or deleted and 4. PRESSURE should not appear visible or present in the overall reaction.  Pressure =  Reaction rate =  Collision of molecules What we usually see is the overall reaction and we do =  Ea =  Reaction Time not see the sequence of elementary steps. o As the pressure is increased, especially for the reaction vessel, the reaction rate will also They are formed in the early elementary steps but in be increased. The collision of molecules will the latter steps, they are consumed and they do not be greater, the needed activation energy is appear anymore in the overall reaction. minimal and time required for the reaction will take place will be shorter. EXAMPLE This is only true for the gases. What is the intermediate product of the overall reaction? 2𝑁𝑂(𝑔) + 𝑂2 (𝑔)→ 2𝑁𝑂2 (𝑔) 5. CATALYST Two elementary steps: First: 𝑁𝑂 + 𝑁𝑂→ 𝑁2 + 𝑂2 It speeds up the rate of the chemical reaction by Second: 𝑁2 𝑂2 + 𝑂2 → 2𝑁𝑂2 lowering the EA. It is not consumed in the reaction. It does not Explanation: participate in the chemical reaction. The elementary steps are related because the May either be speed up or slow down without being product formed in the first step will become the reactant in the second step. consumed in the reaction Two types: It’s not always 2 elementary steps in reaction 1. Accelerators (+catalyst): speeds up the mechanism, there could be three and four. chemical reaction because of the lowering of If the species is on the same side like the 𝑁𝑂 + 𝑁𝑂, Ea they must be added so there will be two moles of the 2. Inhibitors (-catalyst): slows down the reaction nitrogen monoxide in the reaction mechanism and or they will retard the chemical reaction we cannot cancel. What we have to cancel are those species that will POTENTIAL ENERGY AND REACTION PROGRESS appear on opposite sides like the 𝑁2 𝑂2 on the product side of the first elementary step and the other 𝑁2 𝑂2 on the reactant side of the second elementary step. Therefore, the intermediate product in the equation given is 𝑁2 𝑂2. It is the one cancelled and should not appear in the overall reaction. In cases when there are two moles of 𝑁2 𝑂2 on the product and one mole of 𝑁2 𝑂2 on the reactant side, you have to get their difference so there will be 1 mole of 𝑁2 𝑂2 that will remain on the product side. They will be present on the overall reaction. Thus, it is not considered an intermediate. Check if the equation is balanced because if the overall reaction is not balanced, then it is incorrect. Figure 5. Interpretation: 2. REACTION MOLECULARITY This is based on the number of reacting molecules in In an uncatalyzed reaction, there is no catalyst the elementary step: present in the reaction. We started with an unstable reactant and ended up with stable products. It can be further classified into the following: BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS Step 1: 𝑟𝑎𝑡𝑒 = 𝑘[𝑁𝑂2 ]2 a.) UNIMOLECULAR REACTION There is only 1 molecule present in the elementary 2. Any catalyst? step. In step 1, the catalyst must be a reactant while in the A → Products | Rate = k[A] final step, it has to be a product. b.) BIMOLECULAR REACTION In the overall reaction, it should not appear There are 2 molecules present in the elementary anymore. step. Considering the two conditions above, we can A + B → Products | Rate = k[A][B] therefore say that there is no catalyst used here. A + A → Products | Rate = k [A]2  We cannot consider the catalyst as an intermediate c.) TERMOLECULAR REACTION because a catalyst is not consumed in the reaction. There are 3 molecules present in the elementary It does not participate the reaction as it only hastens step. or retards the chemical reaction. 2A + B → Products 3. What is the molecularity of steps 1 and 2? Steps 1 and 2 are both bimolecular reactions. WRITING PLAUSIBLE REACTION MECHANISMS Plausible means acceptable or valid. INFORMATION GATHERED FROM REACTION The sum of elementary steps must give the overall balanced equation for the reaction. MECHANISM The rate-determining step should predict the same Overall reaction for the reaction mechanism rate law that is determined experimentally. Intermediates are the species present in the o The rate determining step, which is the elementary steps but not in the final or overall reaction. slowest step in the sequence of elementary Molecularity refers to the number of molecules present steps before the formation of the product, in each elementary step (unimolecular, bimolecular, should be the same with the rate law which termolecular, polymolecular) is determined experimentally. Rate-determining step is the slowest step which can be derived from the rate law. 3. RATE-DETERMINING OR LIMITING STEP Rate law: if in any case that the rate law is not given in It is the slowest step in the sequence of steps leading the reaction mechanism, then you are to determine the to product formation. rate law expression Relative rates of elementary steps—which one is faster EXAMPLE 1 or slower The experimental rate law for the reaction between 𝑁𝑂2 and Presence of catalyst—there is a catalyst if (a) In step 𝐶𝑂 to produce 𝑁𝑂 and 𝐶𝑂2 is 𝑟𝑎𝑡𝑒 = 𝑘[𝑁𝑂2 ]2. The reaction 1, it is a reactant and in the final step, it is a product is believed to occur via two steps: and; (2) in the overall reaction, it is deleted. Step 1: 𝑁𝑂2 + 𝑁𝑂2 → 𝑁𝑂 + 𝑁𝑂3 Step 2: 𝑁𝑂3 + 𝐶𝑂 → 𝑁𝑂2 + 𝐶𝑂2 CATALYSIS It is the process known of having the presence of Answer the following: catalyst in a reaction. 1. What is the equation for the overall reaction? Cancel the intermediates: HETEROGENOUS CATALYSIS o Step 1: 𝑁𝑂2 + 𝑁𝑂2 → 𝑁𝑂 + 𝑁𝑂3 A catalysis is heterogenous if the reactants and the o Step 2: 𝑁𝑂3 + 𝐶𝑂 → 𝑁𝑂2 + 𝐶𝑂2 catalyst are in different phases. Overall reaction: 𝑁𝑂2 + 𝐶𝑂→ 𝑁𝑂 + 𝐶𝑂2 Normally, the catalysts are solids while the reactants are liquids or gases. 2. What is the intermediate? Common examples of heterogenous catalysis: Intermediate product: 𝑁𝑂3 We cannot consider 𝑁𝑂2 as an intermediate 1.) HABER PROCESS because it appeared in the overall reaction. 3. What can you say about the relative rates of steps 1 and 2? 𝑅𝑎𝑡𝑒 = 𝑘[𝑁𝑂2 ]2 is the rate law for step 1, so step 1 is slower than step 2. The rate law expression is derived from step 1. It is the rate-determining step. Figure 6. 𝐾2 > 𝐾1 Ammonia production Ammonia is usually used in the production of EXAMPLE 2 fertilizers, explosives, and they are also used in 2𝑁𝑂 (𝑔) + 𝑂2 (𝑔)→ 2𝑁𝑂2 (𝑔) laboratory experiments In the illustration, the white molecules are the Step 1: 𝑁𝑂 + 𝑁𝑂→ 𝑁2 𝑂2 hydrogen molecules, the purple molecules are the Step 2: 𝑁2 𝑂2 + 𝑂2 → 2𝑁𝑂2 nitrogen molecules, and the brown molecules are Overall reaction: 2𝑁𝑂 + 𝑂2 → 2𝑁𝑂2 the catalysts. Catalysts are all in solid form: Iron (Fe), Aluminum Answer the following: oxide (Al2O3), or Potassium oxide (K2O). 1. Which is the rate determining (slowest) step? BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS Reactant molecules and products are all gases. ENZYMES You can see in the third illustration (right side) the catalyst is not consumed. 2.) OSTWALD PROCESS Figure 7. Nitric acid production Nitric acid is useful in the manufacture of fertilizers, dyes, drugs, and explosives. They are also very Figure 9. useful in laboratories for experiments. It is considered to be a biological catalyst, usually Catalyst is platinum which is a solid and the reacting protein in nature. substances are gases. They are selective because they will only act on These can be considered reaction mechanisms. certain substrate or reactants. There are 2 elementary steps that leads to production of nitric acid. HOMOGENOUS CATALYSIS ADVANTAGE OVER HETEROGENOUS CATALYSIS 3.) CATALYCTIC CONVERTERS Homogenous catalysis can be carried out under atmospheric conditions. o Hindi siya choosy sa condition or wala siyang pinipiling atmospheric condition. It can function selectively for a particular reaction type. The cost for the process is lesser because this kind of catalysis makes use of less precious metals like platinum and gold. SUMMARY OF FACTS AND CONCEPTS Figure 8. 1. The rate of a chemical reaction is the change in the Present in automobiles concentration of reactants or products over time. The In a catalytic converter, we convert the poisonous rate is not constant but varies continuously as gases into good gases. concentrations change. Example 1: 𝐶𝑂 + 𝑢𝑛𝑏𝑢𝑟𝑛𝑒𝑑 𝐻𝐶→ 𝐶𝑂2 + 2. The rate law expresses the relationship of the rate of 𝐻2𝑂 (𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛) reaction to the rate constant and the concentrations of o The reaction between carbon monoxide the reactants raised to appropriate powers. The rate with the unburned hydrocarbon is constant k for a given reaction changes only with converted into carbon dioxide and water temperature. through oxidation. 3. Reaction order is the power to which the Example 2: 𝑁𝑂 + 𝑁𝑂2 → 𝑁2 + 𝑂2 (𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛) concentration of a given reactant is raised in the rate o Nitrogen monoxide and nitrogen dioxide law. Overall reaction order is the sum of the powers are all poisonous gases. o With the help of the catalytic converter, to which reactant concentrations are raised on the rate they are converted into good gases, which law. The rate law and the reaction order cannot be are nitrogen and oxygen gases. determined from the stoichiometry of the overall equation for a reaction; they must be determined by HOMOGENOUS CATALYSIS experiment. For a zero-order reaction, the reaction Reactants and catalysts are dispersed in a single rate is equal to the rate constant. phase, usually in liquids. 4. The half-life of a reaction (the time it takes for the Common examples of homogenous catalysis: concentration of a reactant to decrease by one-half) 1.) Acid catalysis: Sulfuric acid production or the Lead can be used to determine the rate constant of a first- Chamber Process (𝐻2 𝑆𝑂4 ) order reaction. 2.) Base catalysis 5. In terms of collision theory, a reaction occurs when 3.) Enzyme catalysis: The reactants are the substrate molecules collide with sufficient energy, called the while the catalyst is the enzyme. The products, activation energy, to break the bonds and initiate the substrate, and enzyme are all in aqueous solution. reaction. The rate constant and the activation energy are related by the Arrhenius equation. 6. The overall balanced equation for a reaction may be the sum of a series of simple reactions, called elementary steps. The complete series of elementary steps for a reaction is the reaction mechanism. 7. If one step in a reaction mechanism is much slower than all other steps, it is the rate-determining step. 8. A catalyst speeds up a reaction usually by lowering the value of Ea. A catalyst can be recovered unchanged at the end of a reaction. BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN PHA612 LEC: CHEMICAL KINETICS 9. In heterogeneous catalysis, which is of great industrial importance, the catalyst is a solid and the reactants are gases and liquids. In homogeneous catalysis, the catalyst and reactants are in the same place. Enzymes are catalysts in living systems. REFERENCES Chemical kinetics video lecture part I discussed by Mr. Adrien Kyle M. Jacinto, MSc Chemical kinetics video lecture part II discussed by Prof. Ma. Cristina C. Doria, PhD --END-- BAMBA, GO, LEDESMA, MANALO, OLIVA, SALALIMA, TAN