Chemistry Lecture 4 PDF

Chemistry Dr Ramin Boroujerdi PhD | MSc | BSc | FHEA [email protected] Lecture 4 Chemical Equilibrium and Rate of Chemical Reactions  The Rate of a Chemical Reaction Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what mol...

Chemistry Dr Ramin Boroujerdi PhD | MSc | BSc | FHEA [email protected] Lecture 4 Chemical Equilibrium and Rate of Chemical Reactions  The Rate of a Chemical Reaction Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions, and what molecular events occur during the overall reaction. The rate of a chemical reaction—a measure of how fast the reaction proceeds—is defined as the amount of reactant that changes to product in a given period of time. A reaction with a fast rate proceeds quickly; a large amount of reactant is converted to product in a certain period of time. A reaction with a slow rate proceeds slowly; only a small amount of reactant is converted to product in the same period of time. Chemists seek to control reaction rates for many chemical reactions. For example, rockets can be propelled by the reaction of hydrogen and oxygen to form water. If the reaction proceeds too slowly, the rocket will not lift off the ground. If, however, the reaction proceeds too quickly, the rocket can explode. Reaction rates can be controlled if we understand the factors that influence them. The Rate of a Chemical Reaction (left side) In a reaction with a fast rate, the reactants react to form products in a short period of time. A+B→C (right side) In a reaction with a slow rate, the reactants react to form products over a long period of time. X+Y→Z *Reaction rates generally decrease as a reaction proceeds.  The Rate of a Chemical Reaction Collision theory of reaction rates is a theory that assumes that, for reaction to occur, reactant molecules must collide with an energy greater than some minimum value and with the proper orientation. The minimum energy of collision required for two molecules to react is called the activation energy, Ea. The value of Ea depends on the particular reaction. Simply:according to collision theory, chemical reactions occur through collisions between molecules or atoms. What Variables Affect Reaction Rates? 1. Concentrations of reactants Often the rate of reaction increases when the concentration of a reactant is increased. A piece of steel wool burns with some difficulty in air (20% O2) but bursts into a dazzling white flame in pure oxygen. The rate of burning increases with the concentration of O2. In some reactions, however, the rate is unaffected by the concentration of a particular reactant, as long as it is present at some concentration. What Variables Affect Reaction Rates? 2. Concentration of catalyst A catalyst is a substance that increases the rate of reaction without being consumed in the overall reaction. Because the catalyst is not consumed by the reaction, it does not appear in the balanced chemical equation (although its presence may be indicated by writing its formula over the arrow). A solution of pure hydrogen peroxide, H2O2, is stable, but when hydrobromic acid, HBr(aq), is added, H2O2 decomposes rapidly into H2O and O2. (Here HBr acts as a catalyst to speed decomposition) What Variables Affect Reaction Rates? 3. Temperature at which the reaction occurs Usually, reactions speed up when the temperature increases. It takes less time to boil an egg at sea level than on a mountaintop, where water boils at a lower temperature. Reactions during cooking go faster at higher temperature. What Variables Affect Reaction Rates? 4. Surface area of a solid reactant or catalyst If a reaction involves a solid with a gas or liquid, the surface area of the solid affects the reaction rate. Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area. A wood fire burns faster if the logs are chopped into smaller pieces. Similarly, the surface area of a solid catalyst is important to the rate of reaction. The greater the surface area per unit volume, the faster the reaction  Precise Definition of Reaction Rate The rate of a reaction is the amount of product formed or the amount of reactant used up per unit of time. So that a rate calculation does not depend on the total quantity of reaction mixture used, you express the rate for a unit volume of the mixture. Therefore, the reaction rate is the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of reactant per unit time. The usual unit of reaction rate is moles per liter per second, mol/L.s. *The reaction in picture is the formation of a product of formaldehyde with hydrogen sulfite ion. As the hydrogen sulfite ion is used up, the solution becomes less acidic and then changes to basic. Bromthymol- blue indicator marks the change from acidic to basic by changing from yellow to blue. The reaction shown here is complete in less than a minute.  Precise Definition of Reaction2NRate Consider the following gas-phase reaction: O (g) → 4NO (g) + O (g) 2 5 2 2 The rate for this reaction could be found by observing the increase in molar concentration of O2 produced. You denote the molar concentration of a substance by enclosing its formula in square brackets. Thus, [O2] is the molar concentration of O2. In a given time interval (Δt), the molar concentration of oxygen, [O2], in the reaction vessel increases by the amount Δ[O2]. The symbol Δ (capital Greek delta) means “change in”; you obtain the change by subtracting the initial value from the final value. The rate of the reaction is given by: This equation gives the average rate over the time interval Δt. If the time interval is very short, the equation gives the instantaneous rate—that is, the rate at a particular instant of time.  Precise Definition of Reaction Rate To understand the difference between average rate and instantaneous rate, it may help to think of the speed of an automobile. Speed can be defined as the rate of change of position, x; that is, speed equals Δx/Δt, where Δx is the distance traveled. If an automobile travels 84 miles in 2.0 hours, the average speed over this time interval is 84 mi/2.0 hr = 42 mi/hr. However, at any instant during this interval, the speedometer, which registers instantaneous speed, may read more or less than 42 mi/hr. At some moment on the highway, it may read 55 mi/hr, whereas on a congested city street it may read only 20 mi/hr. The quantity Δx/Δt becomes more nearly an instantaneous speed as the time interval Δt is made smaller.  Precise Definition of Reaction 5 2 Rate 2N O (g) → 4NO (g) + O (g) 2 2 Graph shows the increase in concentration of O2 during the decomposition of N2O5. It shows the calculation of average rates at two positions on the curve. For example, when the time changes from 600 s to 1200 s (Δt=600 s), the O2 concentration increases by 0.0015 mol/L (=Δ[O2]). Therefore, the average rate Δ[O2]/Δt=(0.0015 mol/L)/600 s = 2.5 × 10-6 mol/(L.s). Later, during the time interval from 4200 s to 4800 s, the average rate is 5 × 10-7 mol/(L.s). Note that the rate decreases as the reaction proceeds.  Precise Definition of Reaction Rate 2N O (g) → 4NO (g) + O (g) 2 5 2 Because the amounts of products and reactants 2 are related by stoichiometry, any substance in the reaction can be used to express the rate of reaction. In the case of the decomposition of N O to NO and O , we gave the rate in 2 5 2 2 terms of the rate of formation of oxygen, Δ[O2]/Δt. However, you can also express it in terms of the rate of decomposition of N2O5. Note the negative sign. It always occurs in a rate expression for a reactant in order to indicate a decrease in concentration and to give a positive value for the rate. Thus, because [N O ] decreases, Δ[N O ] is negative and -Δ[N O ]/ Δt is 2 5 2 5 2 5 positive.  Precise Definition of Reaction Rate 2N O (g) → 4NO (g) + O (g) 2 5 2 2 The rate of decomposition of N2O5 and the rate of formation of oxygen are easily related. Two moles of N2O5 decompose for each mole of oxygen formed, so the rate of decomposition of N2O5 is twice the rate of formation of oxygen. To equate the rates, you must divide the rate of decomposition of N 2O5 by 2 (its coefficient in the balanced chemical equation). Rate of formation of O2 = 1/2 (rate of decomposition of N2O5) Or Precise Definition of Reaction Rate Example Consider the reaction of nitrogen dioxide with fluorine to give nitryl fluoride, NO2F. How is the rate of formation of NO2F related to the rate of reaction of fluorine? 2NO2 (g) + F2 (g) → 2NO2F (g) Strategy We need to express the rate of this reaction in terms of concentration changes with time of the product, NO2F, and reactant, F2, and then relate these two rates. The rate of disappearance of reactants is expressed as a negative quantity of concentration change per some time interval. The rate of formation of products is expressed as a positive quantity of concentration change per some time interval. In order to equate rate expressions, we need to divide each by the coefficient of the corresponding substance in the chemical equation. Precise Definition of Reaction Rate Example Consider the reaction of nitrogen dioxide with fluorine to give nitryl fluoride, NO2F. How is the rate of formation of NO2F related to the rate of reaction of fluorine? 2NO2 (g) + F2 (g) → 2NO2F (g) Precise Definition of Reaction Rate Example Calculate the average rate of decomposition of N2O5, -Δ[N2O5]/Δt, by the following reaction during the time interval from t=600 s to t=1200 s (regard all time figures as significant). Use the data in the table. 2N2O5(g) → 4NO2(g) + O2(g) Strategy An average reaction rate is the change in concentration of a reactant or product over a time interval; in this case it’s the rate of decomposition of the reactant N2O5 (Δ[N2O5]/Δt). The Δ[N2O5] in the equation is the change in concentration of N2O5 (final value minus initial value). The Δt is the time interval (final minus initial) over which the concentration change occurred. Precise Definition of Reaction Rate Example Calculate the average rate of decomposition of N2O5, -Δ[N2O5]/Δt, by the following reaction during the time interval from t=600 s to t=1200 s (regard all time figures as significant). Use the data in the table. 2N2O5(g) → 4NO2(g) + O2(g)  *Both beakers contain the same amounts of reactants, However, the beaker on the right contains more water and thus lower concentrations of reactants. Dependence of Rate on Concentration Experimentally, it has been found that a reaction rate depends on the concentrations of certain reactants as well as the concentration of catalyst, if there is one. A rate law is an equation that relates the rate of a reaction to the concentrations of reactants (and catalyst) raised to various powers. Consider the reaction of substances A and B to give D and E, according to the balanced equation: You could write the rate law in the form: Here k, called the rate constant, is a proportionality constant in the relationship between rate and concentrations. The exponents m, n, and p are frequently, but not always, integers. They must be determined experimentally, and they cannot be obtained simply by looking at the balanced equation. Reaction Order You can classify a reaction by its orders. The reaction order with respect to a given reactant species equals the exponent of the concentration of that species in the rate law, as determined experimentally. The overall order of a reaction equals the sum of the orders of the reactant species in the rate law. Reaction Order Example Bromide ion is oxidized by bromate ion in acidic solution. The experimentally determined rate law is Rate=k[Br-][BrO3-][H+]2. What is the order of reaction with respect to each reactant species? What is the overall order of the reaction? The exponent of each of the reactants given in the experimentally determined rate law is the reaction order of that reactant. The overall order of the reaction is the sum of the orders of the reactant species in the rate law. The reaction is first order with respect to Br- and first order with respect to BrO3- ; it is second order with respect to H+. The reaction is fourth order overall (=1 + 1 + 2). Change of Concentration with Time A rate law tells you how the rate of a reaction depends on reactant concentrations at a particular moment. But often you would like to have a mathematical relationship showing how a reactant concentration changes over a period of time. Such an equation would be directly comparable to the experimental data, which are usually obtained as concentrations at various times. In addition to summarizing the experimental data, this equation would predict concentrations for all times. Using it, you could answer questions such as: How long does it take for this reaction to be 50% complete? to be 90% complete? Using calculus, we can transform a rate law into a mathematical relationship between concentration and time called an integrated rate law. We have zero-order, first-order and second-order reactions. Change of Concentration with Time Zero-Order Rate Law aA → products Keeping in mind that any real number raised to the zero power is 1, the expression for zero-order rate law is usually written as: This rate law indicates that the rate of a zero-order reaction does not change with concentration. The relationship between concentration and time for a zero-order reaction is: (zero-order integrated rate law) Here [A]t is the concentration of reactant A at time t, and [A] 0 is the initial concentration. k is rate constant. Change of Concentration with Time First-Order Rate Law aA → products Using calculus, you get the following integrated rate law equation: (first-order integrated rate law) Here [A]t is the concentration of reactant A at time t, and [A] 0 is the initial concentration. The ratio [A]t/[A]0 is the fraction of reactant remaining at time t. The symbol “ln” denotes the natural logarithm (base e=2.718). k is rate constant. Change of Concentration with Time Second-Order Rate Law aA → products Using calculus, you can obtain the following relationship between the concentration of A and the time: (second-order integrated rate law) Here [A]t is the concentration of reactant A at time t, and [A] 0 is the initial concentration. k is rate constant. Change of Concentration with Time Example The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.80 × 10-4/s at 45 oC. a) If the initial concentration is 1.65 × 10 -2 mol/L, what is the concentration after 825 s? b) How long would it take for the concentration of N 2O5 to decrease to 1.00 × 10-2 mol/L from its initial value, given in a? Strategy Part a of this problem wants us to determine the concentration of a reactant (N 2O5) after some period of time has elapsed. Part b of the problem asks us to determine the amount of time that has elapsed for a certain change in concentration of the reactant. An integrated rate law (first-order here), which relates concentration changes to time, will enable us to complete both of these problems. Change of Concentration with Time Example The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.80 × 10-4/s at 45 oC. a) If the initial concentration is 1.65 × 10 -2 mol/L, what is the concentration after 825 s? b) How long would it take for the concentration of N 2O5 to decrease to 1.00 × 10-2 mol/L from its initial value, given in a? Part a. In this case, you need to use the equation relating concentration to time for a first-order reaction, which is: Substituting the appropriate values, you get: Change of Concentration with Time Example The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.80 × 10-4/s at 45 oC. a) If the initial concentration is 1.65 × 10 -2 mol/L, what is the concentration after 825 s? b) How long would it take for the concentration of N 2O5 to decrease to 1.00 × 10-2 mol/L from its initial value, given in a? To solve for [N2O5]t, you take the antilogarithm (antiln) of both sides. This removes the ln from the left and gives antiln(-0.396), or e-0.396, on the right, which equals 0.673. Hence, Change of Concentration with Time Example The decomposition of N2O5 to NO2 and O2 is first order, with a rate constant of 4.80 × 10-4/s at 45 oC. a) If the initial concentration is 1.65 × 10 -2 mol/L, what is the concentration after 825 s? b) How long would it take for the concentration of N 2O5 to decrease to 1.00 × 10-2 mol/L from its initial value, given in a? Part b. You substitute into the same first-order equation relating concentration to time. The left side equals -0.501; the right side equals -4.80 × 10-4/s × t. Hence, or Half-Life of a Reaction As a reaction proceeds, the concentration of a reactant decreases, because it is being consumed. The half-life, t1/2, of a reaction is the time it takes for the reactant concentration to decrease to one- half of its initial value. Half-Life of a Reaction Example Sulfuryl chloride, SO2Cl2, is a colorless, corrosive liquid whose vapor decomposes in a first-order reaction to sulfur dioxide and chlorine. At 320 oC, the rate constant is 2.20 × 10-5/s. What is the half-life of SO2Cl2 vapor at this temperature? We are asked about the half-life of a first-order reaction, so we can solve this using the appropriate half-life equation. So, the time required for one-half (50.0%) of the SO 2Cl2 to decompose is 3.15 × 104 s, or 8.75 hr.  Chemical Equilibrium Chemical equilibrium is the condition of a system where the concentrations of the reactants and products do not change with time. Also, the system does not show any change in its properties. When a reaction starts, eventually, the rate of the reverse reaction (which is increasing) equals the rate of the forward reaction (which is decreasing). At that point, dynamic equilibrium is reached. This condition is not static—it is dynamic because the forward and reverse reactions are still occurring but at the same constant rate. Dynamic equilibrium—In a chemical reaction, the condition in which the rate of the forward reaction equals the rate of the reverse reaction. Chemical Equilibrium Chemical reactions often seem to stop before they are complete. Such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants. Actually, two reactions are occurring, and the eventual result is a mixture of reactants and products, rather than simply a mixture of products.  The Equilibrium Constant A Measure of How Far a Reaction Goes The equilibrium-constant expression for a reaction is an expression obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. The equilibrium constant, Kc (also showed as Keq), is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. * A and B are reactants, C and D are products, and a, b, c, and d are the respective stoichiometric coefficients in the chemical equation.  The Equilibrium Constant A Measure of How Far a Reaction Goes To write an equilibrium constant expression for a chemical reaction, we examine the chemical equation and follow the definition for the equilibrium constant. For example, suppose we want to write an equilibrium expression for this reaction: The equilibrium constant is [NO2] raised to the fourth power multiplied by [O2] raised to the first power divided by [N2O5] raised to the second power. Notice that the coefficients in the chemical equation become the exponents in the equilibrium expression: The Significance of the Equilibrium Constant What does an equilibrium constant tell us? What does a large equilibrium constant (Keq >> 1) imply about a reaction? It indicates that the forward reaction is largely favored and that there will be more products than reactants when equilibrium is reached. Consider the reaction: The equilibrium constant is large, meaning that at equilibrium the reaction lies far to the right—high concentrations of products, tiny concentrations of reactants *The symbol >> means much greater than. The Significance of the Equilibrium Constant What does an equilibrium constant tell us? What does a small equilibrium constant (Keq