Rates of reaction

Determining rate of reaction The rate of a chemical reaction is the speed at which reactants are used up or products are formed Rate of reaction is measured experimentally. The method is dependant on the type of chemical reaction

NaHCO3(s) + CH3COOH(aq) → CO2(g) + H2O(l) + Na+(aq) + CH3COO-(aq) The volume of CO2(g) produced vs time can be used as a measure of the rate of reaction

No more CO2 is produced after 90 seconds. Average rate = change in volume time 60/90 = 0.67 cm3 s-1 Average rate of reaction during the first 90 seconds = 0.67 cm3 s-1 Time (seconds) Volume of CO2 (cm3) 10 19.0 20 33.0 30 44.0 40 50.0 50 54.0 60 56.5 70 58.5 80 59.5 90 60.0 100 110

The rate at any particular time is calculated using the slope (gradient) and a tangent

Use the tangent to construct a triangle

Y Find the value of Y Y = 40 – 26 = 14

X Find the value of X Y = 42 – 17 = 25

To calculate the rate of reaction at that specific point (30 seconds) divide Y by X = 0.56 g/s

Practice

NaHCO3(s) + CH3COOH(aq) → CO2(g) + H2O(l) + Na+(aq) + CH3COO-(aq) The rate at which the mass decreases can be used as a measure of the rate of reaction

Using the data calculate the average rate of reaction Time (seconds) Mass (g) 196.270 10 196.235 20 196.210 30 196.189 40 196.178 50 196.171 60 196.166 70 196.163 80 196.161 90 196.160 100 110 Using the data calculate the average rate of reaction Average rate = change in mass 0.110 time 90 = 1.22x10-3 g/s

Rate of Reaction Defined Although the previous methods (volume and mass) are ways that the rate of a reaction can be measured, the change in concentration is the most common. Rate of reaction is the change in concentration of reactants or products per unit time Common units: mol dm-3 s-1 mol dm-3 min-1 Average rate = change in concentration / time As before a tangent can be used to calculate the rate at a given time

Relative rate Calculations A + B C D -1 a -∆[A] ∆t -1 b -∆[B] ∆t 1 c ∆[C] ∆t 1 d ∆[D] ∆t = = = 4 NH3 + 5 O2 4 NO + 6 H2O -1 4 -∆[A] ∆t -1 5 -∆[B] ∆t 1 4 ∆[C] ∆t 1 6 ∆[D] ∆t = = =

4 NH3 + 5 O2 4 NO + 6 H2O -1 -∆[A] -1 - 0.4 a ∆t 4 10 In the first 10 seconds the [NH3] decreased from 0.8M to 0.4M. What is the average reaction rate? -1 a -∆[A] ∆t -1 4 - 0.4 10 = 0.01 mol dm-3 s-1 At what rate is H2O being formed? Hint: look at the ratios 6 moles of H2O are formed for every 4 moles of NH3 that react with oxygen (6/4) x 0.02 = 0.015 mol dm-3 s-1 (d/a)

Practice Problems NaOH(aq) + CuSO4(aq) Na2SO4(aq) + Cu(OH)2(s) 2 In the first 15 seconds [NaOH] decreased from 2M to 1.4M. What is the average rate of reaction? At what rate is Cu(OH)2(s) formed? ( -1 / 2 ) x ( - 0.6 / 15) = 0.02 mol/dm-3/s For every 2 moles of NaOH used 1 mole of Cu(OH)2 is produced 0.02 / 2 = 0.01 mol/dm-3/s

Practice Problems 4 2 NO2(g) + O2(g) N2O5(g) At a particular moment during the reaction, molecular oxygen is reacting at a rate of 0.024 mol dm-3 s-1. At what rate is N2O5 being formed? At what rate is NO2 reacting 4 2

Na3PO4 + CaCl2 NaCl + Ca3(PO4)2 Practice Problems Na3PO4 + CaCl2 NaCl + Ca3(PO4)2 After 18 seconds the [CaCl2] decreases by 0.9M. At what rate is NaCl being formed? At what rate is Na3PO4 reacting? 2 3 6

Collision Theory Reaction rates are generally discussed in terms of collision theory For a reaction to occur particles must collide, and two conditions must be fulfilled: Molecules must collide with sufficient energy (activation energy) Molecules must collide with the correct orientation A collision can be successful or unsuccessful Results in a chemical reaction Does not result in a chemical reaction

The main factors that affect the rate of a chemical reaction are: 1) The concentration of reactants 2) Pressure (for reactions involving gases) 3) Surface area (solid reactants) 4) Temperature 5) Catalysts

Concentration The greater the concentration of reactants the greater the number of particles in a certain volume. Therefore, particles collide more often producing more successful collisions in a given amount of time.

Pressure Increasing the pressure of gases decreases the distance between particles, therefore successful collisions occur more frequently.

Surface area Only the particles at the surface of a solid are able to collide with other particles. Breaking apart a solid into smaller pieces increases the number of particles that are available for successful collisions.

Increasing the temperature has a major effect on the rate of reaction. As temperature increases the average kinetic energy of the particles increases and collision occurs more frequently. However, the mass of the particles needs to be considered. Example – oxygen and helium are heated to 600 K. The mass of an O2 molecule is 8 times the mass of a helium atom. Therefore, the helium atoms will be travelling substantially faster at the same temperature. Increasing the temperature means that the particles collide more often and with greater energy, therefore there is a greater probability of a successful collision

Maxwell-Boltzmann distribution How has the number of particles with sufficient energy changed? Ea

Catalysts A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy, without being consumed. In other words a catalyst provides an alternative pathway that requires less energy There are two types of catalysts Heterogeneous catalysts: Different state of matter to the reactants Homogeneous catalysts: Reactants and catalysts share the same state of matter

The rate expression Rate of reaction is usually affected by a change in concentration of the reactants. Consider the following decomposition reaction: This reaction was experimentally measured and repeated with varying [A] The graph shows that the rate of reaction is proportional to the [A] rate ∝ [A] A constant of proportionality (rate constant (k)) can be added: rate = k[A] Units for rate = mol dm-3 s-1 A B

Rate = k[A]x [B]y A rate expression can be written for any reaction: aA + bB cC + dD Rate = k[A]x [B]y The x and y exponents must be determined experimentally: x = the order with respect to A y = the order with respect to B In order to know the value of x experiments have to be completed using a fixed amount of B and varying [A] In order to know the value of y experiments have to be completed using a fixed amount of A and varying [B] *** There is no connection between the chemical equation and the rate expression*** Note that the rate expression is only for reactants and that x and y are not the coefficients of A and B

Order of reactions Rate = k[A]0 Rate = k Units for k = mol dm-3 s-1 A reaction can be zero order, first order or second order with respect to the concentration of reactants Zero order : the rate of reaction does not vary with reactant concentration 2 NH3(g) N2(g) + 3H2(g) Rate = k[A]0 Rate = k Units for k = mol dm-3 s-1

Rate = k[A]1 Rate = k[A] Units for k = time-1 First order: The rate of reaction is directly proportional to the reactant concentration. Double the concentration = doubles the rate Triple the concentration = triple the rate Quadruple the concentration = quadruple the rate Rate = k[A]1 Rate = k[A] Units for k = time-1

Rate = k[A]2 (M/s) = K (M2) (M/s) = S-1 M-1 (M2) Second order: The rate of reaction is proportional to concentration squared Concentration x 2 = rate x 22 Concentration x 3 = rate x 32 Concentration x 4 = rate x 42 Rate = k[A]2 (M/s) = K (M2) (M/s) = S-1 M-1 (M2)

experimental data Reaction : 2 A B We want to determine: 1) The order of reaction with respect to A 2) The rate expression 3) The value of the rate constant (k), with untits 4) The rate of reaction when [A] = 1.3 mol/dm3

Experiment [A] (M) Rate (mol dm-3 s-1) 1 1.0 0.60 2 2.0 1.2 3 5.0 3.0 The experimental data shows that the rate of reaction is proportional to the [A]. = First order The rate expression is: rate = k[A] As it is first order [A] = [A]1 The find the value of k we can use the data from any of the experiments, wherein k = rate / [A] Experiment 1: k = 0.60 mol dm-3 s-1 1.0 mol dm-3 The rate when [A] = 1.3 mol dm-3 rate = k[A] rate = 0.60 x 1.3 = 0.78 mol dm-3 s-1

3 A + B C + D Experiment [A] (M) [B] (M) Rate (mol dm-3 h-1) 1 0.1 0.10 0.50 2 0.30 4.5 3 0.20 Given the experimental data above, calculate each of the following for the reaction: The order with respect to A The order with respect to B The overall order of reaction The rate expression The value of k The rate when [A] = 1.60 M and [B] = 0.30 M

Order with respect to A: 0.30 / 0.10 = 3 4.50 / 0.50 = 9 Order with respect to B: 0.20 / 0.10 = 2 4.5 / 4.5 = 0 Second order with respect to A Experiment [A] (M) [B] (M) Rate (mol dm-3 h-1) 1 0.10 0.50 2 0.30 4.5 3 0.20 Zero order with respect to B

d) The rate expression: rate = k[A]2[B]0 rate = k[A]2 c) The overall order of reaction is the sum of the orders with respect to A and B 2 + 0 = 2 Therefore, the overall order is 2 d) The rate expression: rate = k[A]2[B]0 rate = k[A]2 e) The value of the rate constant (k) Substitute in the values for rate and [A] from any of the experiments Rate = k[A]2 0.50 = k 0.102 k = 0.50 / (0.102) = 50 4.50 = k 0.302 k = 4.50/ (0.302) = 50 Experiment [A] (M) Rate (mol dm-3 h-1) 1 0.10 0.50 2 0.30 4.5 3 Units = mol dm-3 h-1 (mol dm-3)2

f) The rate when [A] = 1.60 M and [B] = 0.30 M Rate = k[A]2 Rate = 50 x 1.602 = 128 mol dm-3 h-1 Recall that [B] is not required to calculate the rate as it is zero order with respect to B

Practice Problems Determine the following: The order with respect to A Experiment [A] (M) [B] (M) Rate (mol dm-3 s-1) 1 1.20 2.00 5.00 x10-3 2 2.40 1.00 x10-2 3 8.00 0.16 Determine the following: The order with respect to A The order with respect to B The overall order of reaction The rate expression The value of K (with units) The rate of reaction when [A] = 3.2 M and [B] = 2.4 M

The Arrhenius equation The Arrhenius equation shows the variation of the rate constant with temperature As the temperature increases the rate constant increases exponentially A is related to the frequency of collisions and the orientation of collisions 𝑒 −𝐸𝑎/𝑅𝑇 represents the fraction of collisions that have ≥Ea

Variables Ea – KJ/mol R – 8.31 J/mol/K T – K The units of A and k are always the same (vary with order of reaction)

k, A and R are all constants This equation can be simplified by taking the natural log of both sides of the equation (natural log is opposite to e) and rearranging to a linear equation. ln k = -Ea x 1 + ln A R T y = m x + b k, A and R are all constants This equation is often used to calculate the activation energy of a reaction Ea = (ln A – ln k) x RT

CALCULATING ACTIVTION ENERGY There are two methods for calculating Ea Equation method (without a plot). This method is used to estimate Ea from experiments at only 2 temperatures. Graphical Method (ln k vs 1/T). This method is used to calculate Ea from experiments at varying temperatures ln k1 = ln A - −𝐸𝑎 𝑅𝑇1 Subtract equation 1 from equation 2 and rearrange ln 𝑘2 𝑘1 = −𝐸𝑎 𝑅 1 1 𝑇2 𝑇1 - ln k2 = ln A - −𝐸𝑎 𝑅𝑇2 ln k = -Ea x 1 + ln A R T

ln 𝑘2 𝑘1 = −𝐸𝑎 𝑅 1 1 𝑇2 𝑇1 - Step 3 – Rewrite with calculated values Temperature K (s-1) 35°C 1.4 x10-4 45°C 5.0 x10-4 Step 4 – Solve for Ea Step 1 – convert temperatures to kelvin T1 = 35 + 273.15 = 308 K T2 = 45 + 273.15 = 318 K Step 2 – Put values in the equation Step 5 – Units

Graphical method

Graphical method −𝐸𝑎 𝑅 = ∆𝑦 ∆𝑥 ln k = -Ea x 1 + ln A R T Temperature K (s-1) 35°C 1.4 x10-4 45°C 5.0 x10-4 Steeper slope = larger Ea X = 1/Temperature y = ln k −𝐸𝑎 𝑅 = ∆𝑦 ∆𝑥

Calculations using the Arrhenius equation Calculate the frequency factor (A): This requires a plot of ln k vs 1/T. The y-intersect = ln A elnA = A Calculate the rate constant at 300°C ln 𝑘2 𝑘1 = −𝐸𝑎 𝑅 1 1 𝑇2 𝑇1 ln k2 – ln k1 = −𝐸𝑎 𝑅 1 1 𝑇2 𝑇1 - - ln k2 = −𝐸𝑎 𝑅 1 1 𝑇2 𝑇1 + ln k1 -

Calculating the fraction of molecules with energy > activation energy F = e-Ea/RT Calculate the temperature at which k has a specific value: ln k – ln A = -Ea/RT T = -Ea / ((ln k – ln A) R)

Practice problems 1) Determine the value of Ea given the following values of k at the temperatures indicated: 600 K: k = 2.75 x 10-8 L mol-1 s-1 800 K: k = 1.95 x 10-7 L mol-1 s-1 2) Using the data from the following table, determine the activation energy of the reaction: Temperature (K) Rate Constant, k(s-1) 375 1.68 x 10-5 400 3.5 x 10-5 500 4.2 x 10-4 600 2.11 x 10-3

Mechanisms of reactions Why can’t the rate equation be derived directly from the chemical equation for a reaction? Example: 2NO2(g) + F2(g) 2NO2F(g) For this reaction to occur in a single step all 3 molecules must collide at exactly the same time. Doubling the [reactant] would double the collision rate, and therefore double the rate of reaction. In this case the rate expression would be: rate = k[NO2]2[F2] The rate expression obtained experimentally is: rate = k[NO2][F2] Why are the expressions different? [NO2]2 because there are 2 NO2 molecules Because the reaction does not occur in just one step

Rate expressions can be derived from elementary steps 2NO2(g) + F2(g) 2NO2F(g) Step 1 : NO2 + F2 NO2F + F Step 2 : NO2 + F NO2F Sum : 2 NO2 + F2 2 NO2F This reaction consists of elementary steps F does not appear in the overall equation. F is a reaction intermediate Rate expressions can be derived from elementary steps Step 1: rate = k1[NO2][F2] Step 2: rate = k2[NO2][F] Step 1 expression = experimental rate expression Therefore, step 2 has no effect on the rate

Rate determining step Step 1 : NO2 + F2 NO2F + F Step 2 : NO2 + F NO2F Step 1 occurs significantly more slowly than step 2 The slowest step in a reaction mechanism is called the rate determining step Step 2 is much faster and has, effectively, no influence on the overall rate of reaction. Therefore, the concentration of these species do not occur in the rate equation.

Rate equation When writing the rate equation based on reaction mechanisms the rate equation contains concentrations of reactants involved up to and including the rate determine step, not including any intermediates. A reaction between NO and H2 occurs in the following three-step process: Step 1: NO+NO→N2O2 (fast) Step 2: N2O2+H2→N2O+H2O (slow) Step 3: N2O+H2→N2+H2O (fast) 1) What is the rate determining step? 2) Write the balanced equation for the overall reaction. 3) Are there any intermediates? 4) Write the rate equation.

Identifying Intermediates and catalysts A + B C + D D + E B + F Catalysts appear on the reactant side first and reappear on the product side later (not consumed) Intermediates appear on the product side first and reappear on the reactant side later (consumed) Overall reaction: Intermediate(s): Catalyst:

A + B C + D C + E F + G F + A B + E Overall reaction: Intermediate(s): Catalyst(s):

H2O2 + I- H2O + IO- slow H2O2 + IO- H2O + O2 + I- fast Overall reaction: Intermediate(s): Catalyst(s): Rate equation:

Br2 2Br Fast Br + H2 HBr + H Slow Br2 + H HBr + Br Fast 2Br + H2 2 HBr Fast Overall reaction: Intermediate(s): Rate equation:

Cl2 2Cl Fast Cl + CO COCl Fast COCl + Cl2 COCl2 + Cl Slow 2Cl Cl2 Fast Overall reaction: Intermediate(s): Rate equation: