Physical Chemistry-I - Chemical Kinetics PDF

Physical Chemistry-I 1 – Chemical Kinetics Prepared By: Dr Abdul Rahim updated Sem1-24-25 by FRM  Chapter I : Outcomes You should be able to: Discuss the theories of reaction rates (Arrhenius theory, Lin...

Physical Chemistry-I 1 – Chemical Kinetics Prepared By: Dr Abdul Rahim updated Sem1-24-25 by FRM  Chapter I : Outcomes You should be able to: Discuss the theories of reaction rates (Arrhenius theory, Lindemann theory and theory of absolute reaction rates) and Enzyme catalysis Discuss chemical kinetics and the dependence of reaction rates on temperature Explain activation energy,Collision theory of reaction rates. Page 753 Text Physical Chemistry, Atkins, Peter & J de Paula, Oxford Book university Press, 11th / 10th edition 2 What are Chemical Kinetics? 3 Thermodynamics Does a reaction take place? Kinetics How fast does a reaction proceed? Reaction speed Measured by the change in concentration with time. Important factors which affect rates of reactions: Reactant concentration Temperature Action of catalysts Surface area Pressure of gaseous reactants 4 Kinetics Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). 5 Kinetics – In Brief Reaction Rates How we measure rates. Rate Laws How the rate depends on amounts of reactants. How to calculate amount left or time to reach a Integrated Rate Laws given amount. Half-life How long it takes to react 50% of reactants. Arrhenius Equation How rate constant changes with T. Mechanisms Link between rate and molecular scale processes. 6 Recall 7 A Pseudo first-order reaction can be defined as a second-order or bimolecular reaction that is made to behave like a first-order reaction. This reaction occurs when one reacting material is present in great excess or is maintained at a constant concentration compared with the other substance. A + B —-> C + D More details in practical 8 Collision Theory ❑ In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. i.e, Particles must ‘collide’ or hit against each other before they can react. ❑ Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. 16-09-2024 Prepared By: Dr Abdul Rahim 9 Collision Theory: As the average kinetic energy increases, the average molecular speed increases, and thus the collision rate increases. Activation Energy There is a minimum amount of energy required for reaction is known as activation energy, Ea. Ea Energy Reactants Products Reaction Progress 11 Reaction Coordinate Diagram To visualize energy changes throughout a process. Transition state. Activated complex Activation energy barrier Methyl isonitrile. 12 Temperature & Rate Generally, as temperature increases, so does the reaction rate. The rate constant k is temperature dependent. Why ? 13 Arrhenius Equation Activation energy J/mol Rate constant Frequency factor Temp K Gas constant J/mol.K 16-09-2024 Prepared By: Dr Abdul Rahim 14  The equation can be rewritten as: 𝐸𝑎 1 ln 𝑘 = − × + ln A 𝑅 𝑇 y = m × 𝑥+ c High activation energy signifies that the rate constant depends strongly on temperature. In cases that we have two rate constants at only two temperatures, activation energy can be obtained directly from the Arrhenius equation in the form 𝑘2 𝐸𝑎 1 1 ln = − 𝑘1 𝑅 𝑇1 𝑇2 15 Q1: For a reaction with an activation energy of 50 kJmol−1 , an increase in the temperature from 25 °C to 37 °C (body temperature) corresponds to ? Ans: k2 = 2.18k1 Q2: Find the activation energy for the reaction in which the rate constants were found to be 0.028 M–1s–1 at 327 oC and 23 M–1s–1 at 527 oC ? Ans: Ea = 130 kJ/mol 16 Lindemann Mechanism Unimolecular Reaction: A single molecule shakes itself apart or its atoms into a new arrangement. Eg: Isomerization reaction of cyclopropane to propene. 𝒅 [𝑨] Rate law of a Unimolecular reaction A P is = −𝒌𝟏 𝑨 𝒅𝒕 where P is products and k1 is the rate constant of this Unimolecular reaction. Bimolecular Reaction: A pair of molecules collide and exchange energy, atoms, or groups of atoms, or undergo some other kind of change. Eg: Dimerization of alkenes and dienes. 𝒅 [𝑨] Rate law of a Unimolecular reaction A + B P is = −𝒌𝟐 𝑨 [𝑩] 𝒅𝒕 where P is product and k2 is the rate constant of this Bimolecular reaction. 17 Explanation of Unimolecular reactions was provided by Frederick Lindemann and then elaborated by Cyril Hinshelwood. Reactant molecule A becomes energetically excited by 𝐴 + 𝐴 → 𝐴∗ + 𝐴 𝒅 [𝑨∗ ] 2 collision with another A molecule in a bimolecular step = 𝒌𝟏 𝑨 𝒅𝒕 (Activation Process) 𝐴∗ + 𝐴 → 𝐴 + 𝐴 𝒅 [𝑨∗ ] = −𝒌𝟐 𝑨∗ [𝑨] 𝒅𝒕 The energized molecule (A*) might lose its excess energy by collision with another molecule 𝐴∗ → 𝑃 𝒅 [𝑨∗ ] (Deactivation Processes) = −𝒌𝟑 𝑨∗ 𝒅𝒕 18 If the Unimolecular step is the rate-determining step, the overall reaction will have first-order kinetics Applying the steady-state approximation to the net rate of formation of A* 𝒅 [𝑨∗ ] = 0 = (Formation – decomposition) 𝒅𝒕 0=𝒌𝟏 𝑨 2−(𝒌𝟐 𝑨∗ 𝑨 + 𝒌𝟑 𝑨∗ ) 𝒌𝟏 𝑨 2 𝑨∗ = 𝒌𝟑 +𝒌𝟐 𝑨 𝒅 [𝑨∗ ] 𝒌𝟏 𝒌𝟑 𝑨 2 The rate law for the formation of P is = 𝒌𝟑 𝑨∗ = 𝒅𝒕 𝒌𝟑 +𝒌𝟐 𝑨 19  𝒌𝟏 𝒌𝟑 𝑨𝟐 𝑲 𝑨 𝟐 𝒌𝟐 Divide the equation with k2. We will get = ′ 𝒌𝟑 𝒌𝟐 𝑲 + [𝑨 ] + [𝑨] 𝒌𝟐 𝒌𝟐 Case 1: 𝑲 𝑨 𝟐 = = 𝑲𝑨 At High Pressure K’ > [A], 𝑲′ 𝑺𝒆𝒄𝒐𝒏𝒅 𝒐𝒓𝒅𝒆𝒓 𝒌𝒊𝒏𝒆𝒕𝒊𝒄𝒔 As the pressure is reduced the rate of the bimolecular process in which A* loses its excess energy becomes negligible compared to the rate at which A* goes on to form products. 16-09-2024 Prepared By: Dr Abdul Rahim 20 Catalysts Homogeneous and Heterogeneous Catalysts Homogeneous Catalyst: A catalyst that exists in the same phase as the reactants Heterogeneous Catalyst: A catalyst that exists in a different phase from that of the reactants Catalysts ❖ Catalyst is a substance that increases the rate of the reaction without being substantially consumed in the process. ❖ Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. ❖ Catalysts change the mechanism by which the process occurs. ❖ It neither changes the thermodynamics of the reaction nor the equilibrium composition but only affects the Kinetics 22 ❖ Catalyst lowers the activation energy for both forward and reverse reactions. ❖ This means , the catalyst changes the reaction path by lowering its activation energy and consequently the catalyst increases the rate of reaction. 23  How a Heterogeneous Catalyst works ? One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. Substrate has to be adsorbed on the active sites of the catalyst 24 Enzymes Enzymes are biological catalysts. The substrate fits into the active site of the enzyme much like a key fits into a lock. Enzymes follow zero order kinetics when substrate concentrations are high. It means, no increase in rate of the reaction when substrate is added more. 25  For a given initial concentration of substrate, [S]0, the initial rate of product formation is proportional to the total concentration of enzyme, [E]0. For a given [E]0 and low values of [S]0, the rate of product formation is proportional to [S]0. For a given [E]0 and high values of [S]0, the rate of product formation becomes independent of [S]0, reaching a maximum value known as the maximum velocity, νmax. 26  k1 E + S  ES ⎯⎯→ E + P k2 k -1 E = Enzyme S = Substrate P = Product ES = Enzyme-Substrate complex k1 rate constant for the forward reaction k-1 = rate constant for the breakdown of the ES to substrate k2 = rate constant for the formation of the products 27  d P  The rate of product formation ν= = k 2 ES …………………………………1 dt Apply the steady-state approximation d ES = k1 E S − k −1 ES − k 2 ES ≈ 0 dt 𝑘1 𝐸 𝑆 𝐸𝑆 = …………………………………2 𝑘−1 + 𝑘2 𝑘−1 + 𝑘2 = 𝐾𝑀 → Michaelis constant 𝑘1 28  Total concentration of the enzyme is [E]0 = [E] + [ES] [E] = [E]0 − [ES] [[E]0 − [ES]] 𝑆 Apply in equation 2 𝐸𝑆 = 𝐾𝑀 Substrate is typically in large excess relative to the enzyme. There fore [S] ≈ [S]0. [E]0 [S]0−[S]0 𝐸𝑆 =𝐾𝑀 𝐸𝑆 [E]0 [S]0 =[S]0 𝐸𝑆 + 𝐾𝑀 𝐸𝑆 [E]0 [S]0 = 𝐸𝑆 {[S]0 + 𝐾𝑀 } 29  𝑆 𝐸0 0 𝐸𝑆 = …………………………………3 𝐾𝑀 + 𝑆 0 𝐸0 𝐸𝑆 = 𝐾𝑀 1+ ൗ𝑆 0 Apply in equation 1 𝑑𝑃 𝐸0 …………………………………4 = 𝑘 2 𝐾𝑀 =ν 𝑑𝑡 1+ ൗ𝑆 0 Michaelis–Menten equation 30 When [S]0 > KM, the rate reaches its maximum value & independent of [S]0 𝜈𝑚𝑎𝑥 = 𝑘2 𝐸 0 Substitution of this definition of 𝜈𝑚𝑎𝑥 into equation 4 𝜈𝑚𝑎𝑥 ν= 𝐾𝑀 1+ ൗ𝑆 0 31 An enzyme hydrolyzed a substrate concentration of 0.03 mmol/L, the initial velocity was 1.5x10-3 mmol/L.min-1 and the maximum velocity was 4.5x10-3 mmol/L.min-1. Calculate the KM value? Ans: KM = 0.06 Urease enzyme hydrolyzed urea at [S]= 0.03 mmol/L with a KM value of around 0.06 mmol/L. The initial velocity observed was 1.5x10-3 mmol/L.min-1. Calculate the maximum velocity of the enzymatic reaction? Ans: νmax = 1.5 ×10-3 × 3 =4.5 × 10 -3 32