mod 2 lecture 4 chemical kinetics

• Temperature in Kelvin (K) equals temperature in Celsius (°C) + 273

Chemical Kinetics Part 1

• Chemical kinetics is module 2, lecture 4
• Learning objectives include understanding the rate of reaction, writing rate expressions, understanding rate laws, using initial rates, and using experimental data.
• Textbook for the material is Chapter 14

Thermodynamics and Kinetics

• Thermodynamics describes how far a reaction goes
• Kinetics describes how fast a reaction goes
• Thermodynamics and kinetics are not related.
• Reactions with a negative Gibbs Free Energy (△G < 0) may occur too slowly
• An example reaction is C(s) + O2(g) ⇌ CO2(g) with △G°= -394 kJ mol-1

Study of Reaction Rates

• There is a wide range of reaction speeds
• Studying reaction rates is important
• It helps us understand atmospheric processes, design new drugs, and create industrial processes

Chemical Kinetics

• The study of reaction speed has two parts
• Experimental: Measure reaction rates and study factors influencing rates
• Theoretical: Understand rates in molecular terms and use experimental data to probe reaction mechanisms

Rate Measurement

• Reaction rate is how the concentration changes with time
• For the reaction N2O4(g) ⇌ 2NO2(g):
  • Initially, only N2O4 is present
  • [N2O4] decreases rapidly at first, then slows to equilibrium
  • [NO2] increases rapidly at first, then slows to equilibrium

Understanding Rate

• Speed/velocity is the gradient of a plot of [N2O4] or [NO2] with time
• This gradient is called a rate
• The rate of N2O4 loss is half the rate of NO2 formation due to stoichiometry

Rate Expressions

• Rate = -(change in [N2O4] / change in time) = 1/2 (change in [NO2] / change in time)
• Calculus is used with very small change in time (dt) where: rate = -d[N2O4]/dt = 1/2 d[NO2]/dt
• A negative sign is placed in front of reactants to account for the decrease in concentration
• A rate has to be positive

Generalized Rate Expression

• For a general reaction aA + bB → cC + dD, the rate is:
  • Equal to -1/a (d[A]/dt) = -1/b (d[B]/dt) = 1/c (d[C]/dt) = 1/d (d[D]/dt)
• Definition of rate includes the stoichiometric coefficients a, b, c, and d
• [A] and [B] represent reactant concentrations; [C] and [D] represent product concentrations
• The rate can be expressed in terms of any reactant or product

Rate Units

• Units for rate are always concentration/time, typically mol L-1 s-1

Factors affecting reaction rate

• Chemical nature of the reaction
• Physical states of reactants and products
• Concentration of reactants
• Temperature (T)
• Catalysts increase rate (e.g., enzymes)
• Inhibitors decrease rate
• Catalysts and inhibitors undergo no net change

Reaction Rate

• Concentration does not always affect reactions the same way
• Increasing concentration increases rate for some reactions, such as steel wool oxidation

Oxidation of Ethanol

• For oxidation of ethanol in the liver, increased concentration has no effect on the rate:

• CH3CH2OH + NAD+ --(alcohol dehydrogenase)--> CH3COH + NADH + H+

• Reaction kinetics is based on how the rate depends on concentration.

Rate Law Equation

• Rate law describes how rate depends on concentration for the reaction aA + bB → cC + dD
• Rate = k[A]^x[B]^y
  • x is the order with respect to A
  • y is the order with respect to B
  • x + y is the overall reaction order
  • k is the rate constant
• Rate law must be determined experimentally
• Stoichiometric coefficients (a, b, c, d) are ignored when finding rate law

Example Rate Laws

• Rate = k [A]^x [B]^y
• 2N2O5(g) → 4NO2(g) + O2(g): rate = k [N2O5] (1st order)
• NO2(g) + CO(g) → CO2(g) + NO(g): rate = k [NO2]^2 (2nd order)
• C2H5Br + OH- → C2H5OH + Br-: rate = k [C2H5Br][OH-] (2nd order)
• (CH3)3CBr + OH- → (CH3)3COH + Br-: rate = k [(CH3)3CBr] (1st order)
• Rate laws are always determined experimentally
• x and y are 0, 1, or 2, but can be negative or fractions.
• x and y are not related to stoichiometric coefficients a, b, c, and d

First-Order Reactions

• Processes that are first order overall (rate = k [A]^x [B]^y):
  • Only one reactant A: aA → products (1st order in A)
    • Example: 2N2O5(g) → 4NO2(g) + O2(g); rate = k [N2O5]
  • Two or more reactants: aA + bB + ... → products
    • 1st order in A and zero order in all other reactants
    • Example: (CH3)3CBr + OH- → (CH3)3COH + Br-; rate = k [(CH3)3CBr]

Pseudo-First Order Reactions

• The rate law is: rate = k [A][B]
• Concentration of B is much larger than A: [B] >> [A]
• Concentration of [B] does not change much
• [B] is approximately constant
• rate = k’[A]
• k’ (= k[B]) is called the pseudo first-order rate constant

Rate Order

• The affect on doubling concentration
• If the order is equal to 0, there is no Change
• If the order is equal to 1, the rate of reaction doubles
• If the order is equal to 2, the rate of reaction quadruples

Rate Constant

• 'k' is independent of concentration
• 'k' increases with temperature
• Units for 'k' depend on overall order x + y
  • If the overall order is:
    • 0, the units of k are mol L-1 s-1
    • 1, the units of k are s-1
    • 2, the units of k are L mol-1 s-1
    • 3, the units of k are L2 mol-2 s-1

Rate vs Time

• Rate and time are inversely related
• Dunedin and Christchurch are ~400 km, so if:
  • It takes 4 hours to get there, that is one speed
  • If it takes 8 hours to get there, that is another speed
• Reaction takes twice as long means the rate is half as fast

Rate Constant Example

• Second order reaction rate = (change in concentration / change in time) = k [A] [B]
• The units are mol L-1 / s
• The unit for rate is always mol L-1 s-1
• k = rate / ([A] [B]) = (mol L-1 s-1) / (mol2 L-2 ) = L mol -1s -1

Measuring Rate

• Units of 'k' change depending on the order of reaction
• Based on initial rates of reaction
• Vary concentration of one reactant, while all others are held constant
• aA + bB → cC + dD rate = k[A]x [B]y

Initial Rates

• Initial rate = k[A]0x [B]0y
• [A]0 and [B]0 are initial concentrations at time = 0
• Orders x and y are unknown
• To find x: measure initial rates for different values of [A]0 while holding [B]0 constant
• To find y: measure initial rates for different values of [B]0 while holding [A]0 constant

Initial Rate Importance

• As reaction proceeds [A] and [B] decrease
• The rate will decrease, thus examine reaction rate over a short time scale
• [A] and [B] remain close to [A]0 and [B]0: Studying initial rates for which the change in [A] and [B] with time is linear

Calculating k

• initial rate = k[A]0x [B]0y
• Obtain values of x and y from experiment
• Calculate k from any set of rate measurements:k= initial rate / ([A]0x [B]0y )
• Remember to work out the correct units for k

Experiment Example

• Experiment used to calculate the experimental rate law for the reaction: S2O8^2- + 3I- → 2SO4^2- + I3-
• Initial rate = k [S2O8^2-]0^x [I-]0^y
• Compare Expts 1 and 2 where [I-]0 is constant in both Expts
• [S2O8^2-]0 is doubled and initial rate doubles
• This makes x = 1
• Compare Expts 1 and 3 where [S2O8^2-]0 is constant in both Expts
• [I-]0 is doubled and initial rate doubles
• This makes y = 1
• Ensure that one reactant is held constant

Calculated Rate Law

• If x and y are determined, the rate law= k[S2O8^2-][I-]
• The overall order the is reaction of S2O8^2- + I- is x + y = 2
• This makes it a 2nd order reaction
• Using data from the experiment, k can be calculated
• This makes the rate constant at 298 K to be 36 L mol -1 s-1

POP Quiz

• The overall homework is to complete Chemistry – the central science 15th Ed by Brown et al. chapter problems 14.7 14.8a 14.12 14.13 14.14
• The answers are provided on the Blackboard