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• Topics to be covered this semester include Chemical Kinetics, Chemical Equilibrium, Electrochemistry.
• Dr. Nelson will be teaching this course

Chemical Kinetics

• Simple rate equations, orders of reactions, and rate constants will be examined.
• How temperature affects rate constants and activation energy.
• Catalysis as an introduction.
• Reaction rates of chemical reactions differ greatly; some reactions finish when compounds are mixed, while others take hours even with refluxing.
• Reaction rates can be described by specifying how quickly reactant or product concentration changes over time.
• Reaction rate = change in concentration / change in time
• A + B produces C; one may track the decay of A or B or the formation of C.
• A concentration-time profile would show a "negative" change in reactant concentration and a "+ve" change in product concentration over time.
• Monitoring the same reaction should yield a measurable rate value, regardless of whether a product or reactant is monitored.
• rate = -d[A]/dt = -d[B]/dt = d[C]/dt
• Stoichiometry affects the units for the reaction rate

Reaction Rate

• Units are affected by stoichiometry mol dm-3 s-1
• Rate is divided by the stoichiometric coefficient, normalized for each species.
• The rate law depends on reactant concentrations: rate is proportional to [reactants]^x.
• Rate law is expressed as rate = k[A]^m[B]^n
• k represents the rate constant.
• Rate laws/equations are determined by measuring the rate, and the orders m & n bear no relation to stoichiometric coefficients.
• Indices or orders m + n determine the overall reaction order.
• m &/or n can be positive or negative, zero, fractions, or integers (whole numbers).
• If rate expression -d[A]/dt = k[A]^m[B]^n, m=1 and n=0, then -d[A]/dt = k[A], indicating a first-order reaction.
• First order reaction mol dm-3 s-1 = k x mol dm-3, rate = k[A]

Rate Constant

• Is equal to: k = mol dm-3 s-1 / mol dm-3 = s-1
• Zero order reaction: rate = k if m & n = 0
• Zero order reaction: k => mol dm-3 s-1
• The units for a second-order reaction rate = k [A][B] or rate = k [A]^2
• k = dm3 mol-1 s-1

Additional Points

• Rate can be expressed by reactants or products; the stoichiometric coefficient influences the differential rate expression.
• Rate constant is not influenced by stoichiometric coefficients or concentrations.
• The rate equation is written only with respect to reactants.
• Rate constant units depend on the overall order of the rate expression.
• Rate is always in mol dm-3 s-¹, due to its differential nature.
• The reaction Rate relies on reactant concentration, except for zero-order reactions.
• Rate constants do not depend on reactant concentration, only on temperature.

Zero Order Reactions

• In a zero order reaction rate = -d[A]/dt = k
• Rearranging would result in the following expression: -d[A] = kdt or d[A] = -kdt
• The integrated form of a zero order reaction is: [A]t = [A]0 - kt
• Plotting [A] over time results in a straight line with slope = -k.
• Plotting d[A]/dt (rate) vs [A] yields zero slope, because the rate is independent of [A] (zero order).

First Order Reactions

• In a first order reaction -d[A]/dt = k[A]
• The expression is in linear form, and a d[A]/dt plot against [A] should yield a straight line with slope = -k in a 'difficult plot.'
• Multiplying the expression by -dt yields d[A] = -k[A]dt:
• A graph of [A] vs. time will show an exponential decrease in A over time.
• A plot of ln[A] vs. t gives a straight line.
• Integrated form d[A] = -k[A]dt.
• The integrated form of a 1st order reaction is: ln[A]t= ln[A]0 – kt
• Which is in the form y= a - mx, with a plot of ln [A]t vs t resulting in a straight line.
• Half-life is the time it takes for the reactants to reach half of their initial concentration.
• Half-life for a 1st order reaction does not rely on the initial concentration.
• The 1st order reaction's half-life is calculated by: t1/2 = ln2/k = 0.693/k
• k represents the rate constant.
• The concentration after a number of half-lives may be found via: [A]t=[A]0(1/2)^x, where x = # of half lives.

Second Order Reactions

• Begins rate = k [A]^m[B]^n
• If m=2 and n=0, or if [A]=[B] and m=n=1 then results in: rate = k[A]^2, ie a second order rate expression
• -d[A]/ dt = k[A]^2
• Solution results in 1/[A]t = 1/[A]0 + kt
• Certain texts may show the solution as: 1/[A]t = 1/[A]0 + 2kt although 1/[A]t vs t, yields a linear relationship for these second-order reactions and a value for the rate constant via the slope.
• Half life of is t1/2 = 1/ k[A]0

Reaction Order Rate Laws

• Summarized as:
  • Zero order reaction rate law: r=k Integrated Rate Law: [A]t = [A]0 - kt Plot:[A] vs t Half life: t1/2 = 1/2[A]0 k
  • First order reaction rate law:r = k[A] Integrated Rate Law: ln[A]t = ln[A]0 - kt Plot: ln[A] vs t Half life: t1/2 = ln 2/k
  • Second order reaction rate law: r = k [A]2 Integrated Rate Law: 1/[A]t = 1/[A]0 + kt Plot; 1/[A] vs t Halflife: t1/2 = 1/(k[A])

Calculating the Order of Reactions

• Initial rate method: (A + B -> C)
  • Measurement of the slope at the beginning of the reaction provides with an initial rate
  • Change the concentration of one species while holding the others constant to see how the rate is affected by each reactant.
• Recall for the reaction: aA + bB → cC,
  • a rate expression may be used to determine the data set where [A] and [B] were systematically varied and the initial rate determined: Rate = k[A]^m[B]^n where m & n are the reaction orders.

Calculating Rate Constant

• Since rate = k[A]^m[B]^n, with fixed temp
  • [A] / mol dm-3 [B] / mol dm-3 Initial rate / mol dm-3 s-1 - 0.001 0.005 2.0 x 10-4 (r1) - 0.001 0.010 2.0 x 10-4 (r2) - 0.002 0.010 4.0 x 10-4 (r3)
• r1 = k[A1]^m[B1]^n and r2 = k[A2]^m[B2]^n etc.
• And r1/r2 = k[A1]^m[B1]^n / k[A2]^m[B2]^n .......etc. k is independent of how reactants are concetrated and will cancel out in the ratio equation since the experiments are performed at same temp.
• r1/r2 = ([A1]/[A2])^m ([B1]/[B2])^n
• If interested in n then the coefficient to m must be l in order to simplify the calculations.
• The equation can be simplified by making [A1] = [A2]
• For ex. lets plug in some #s - r1/r2 =2.0×10-4 / 2.0×10-4 = (0.001)^n (0.005)^n / (0.001)^m (0.010)^n 2.0×10-4 / 2.0×10-4 = (0.005/0.010)^n
• Thus 1 = (1/2)^n, Taking the log of the expression and log |= n log (1/2)
• Thus n = 0

Determining Values.

• r2/r3 = 2.0×10-4/4.0×10-4 = (0.001/0.002)^m,
• Thus: log (1/2) = m log(1/2) and m = 1
• Following can be said about: - k = rate/[A]
• i.e it is a 1st order reaction k = (2.0 x 10-4 / 0.001) = 0.2 s -1
• The 1/2 life for the (1st order) reaction: can also be determined
• t1/2 = ln 2/k = 0.693/ 0.2 s-1 =3.5 s

Example

• Given the initial rate data at 25 °C will determine this reaction: NH4+(aq) + NO2-(aq) -> N2(g) + 2 H2O(1)
• Experiment # Initial [NH4+] / M Initial [NO2] / M Initial rate of consumption of NH4+ / M s -1 - 1 0.24 0.10 7.2 × 10-6
• 2 0.12 0.10 3.6 × 10-6
• 3 0.12 0.15 1.22 × 10-5 Can determine the rate law.

Solution

• Begins w general laws - k[NH4+]^x[NO2-]^y, find order. and the order 1st wrtNH4 + x comparison:
• r2/r1 is given by the equation: (3.6×10-6)/7.2×10-6
• Rearranges to (0.12/0.24)^x ( 0.10/0.10)^y
• And the number solution becomes 0.5 = 0.5; therefore x = I The order wrt NO2 can be found by comparing the rates wrt reactors 2 and3 and the equation:
• r3/r3 = (1.22×10-5)/(3.6×10-6)
• rearranges to (0.12/0.12)^x (0.15/0.10)^y
• Equals 3.39 = 1.5y;
• With y log 1.5 = log 3.39
• And y = (log3.39/log1.5) = 3 Thus x = I and y = 3 The order w values inserted is give b: k[NH4+]x[NO2-]y
• The rate law: rate = k[NH4+][NO2-]3 Rate Constant: begins k[NH4+][NO2-]3 and equation:
• k = rate /(NH4+)(NO₂)³ and the answer is (7.2 × 10-6 Ms-1/(0.24M) (0.10M)3
• Gives the rate constan k = 0.030M-3 s-1

Reaction Mechanisms

• Chemical kinetics focuses on reaction rates and rate constants, reaction mechanisms are also significant.
• A reaction mechanism is defined as the series of molecular events/reaction steps defining pathway from reactants to products.
• Each step is an elementary step with its own rate and constant.
• The slowest step is the rate determining step (r.d.s).
• If k2E is a unimolecular reaction.
• The rate is k[NO₂]*[CO], but rate ≠ k[NO₂][CO]
• From the mechanism the rate : k [NO2]^2 in step I
• The rate law relies on molecularity the r.d.s.
• The overall rate in the slowest step of the reaction.

Criteria

• The elementary steps sum up to the overall reaction.
  • Can consistent the observations with law rate.
• The discussion on chemical kinetics centered around reaction rates and rate constant. The reaction mechanism. Is the sequence of molecular events or reaction steps that defines the pathway from reactants to products.
• A -> E each Step has its own rate and rate constant for each elementary step. Is called the rate determining step (r.d.s). which if k2

Molecularity

• Each step has its own molecularity.
  • Is the # of reacting species in a step.

Reaction Mechanisms

• The elementary have to must an overall reaction is proposed.
  • The elementary sums to give the overall reaction. Species are formed in the first step.

Collision Theory

• Gases are in constant random motion
• Energy a T (K)
• Aspects of the KMT can be applied to liquids (with some modifications)
• What are all molecules made of?
• What are the constituents of atoms, and in chemical relations what are the atoms interacting? What is the electrical nature of the part of atom interacting in a chemical reaction. Is hypothetical AABB-2 AB Upon collision electron clouds repel each other.The collisions must be forceful enough to overcome the repulsive force. Like wise collision enough to all demanding process. There energy barrier must be surpassed in order for the process.

Reactions

• The conditions are as follows.
• Includes Collision Correct orientations Collision Theory.
• The Sufficient Collision energy.

  • Potential looks at conversion of all products to determine hypothetically.

  • What is energy what goes to products from reaction Enthalapy is thermodynamic in its properties The kinetic must also be.

  • Energy upon collision must the reaction which to proceed to correct. A potential configuration EP curve atoms maximum will call the transition state and activated completes.

Collisions

• Rate is affected by: px fx collision

  • Where p is Fraction describes correct orientation upon collision is a steric factor.
  • Fraction collision efficient the reaction.

• A+BA + The B is the rate in collision

• The alpha [A] and alpha [BC]

• The there and cons involved pxfxz[A][Bc]

  • Rate a is a Z proportionality collision.

• Law that give: is give K[a][b]c]

• This meant k=piz or zEa/RT A = pr ex factor, Ea R = universal Constant and Temperature-absolute temperature.

• The parameter z is called for the pre exponential factor collision .

• Temperature to rates

  • Affect Temperature. Most Collisions occur are low energy.

• That large Energy is a little fraction.

Molecules

Only E> Ea will reach .

• They Ea back each. recall-we pre factor depends in energy.
• The the gets of as well.
• What are the conditions like collision increases.

Applications which and equations k = Ae Ea/RT

• This is and be equation linearized ink = lna (E/R) I(/)

Catalysis

Reaction that not by react concentration temperture from catalyze. A and substance the the a reaction used Catalyst is gas liquids and solid.Catalyst are very important driving force behind the efficiency

• Bodies like catalysts, and enzymes, and forming products and yields . and more efficient the conversation products Catalyst are more catalyst reactants and products

Pathways

kAe Ea/RT alpha Rate .

• pathway catalyst is better pathway to pathway that factor an equation reaction . a function a lower energy.

E = Catalyze a rate.

• Reactions and equation smaller factor .

• This an and is. Heterogeneous. -Same equation. Overall.

• NO catalyst up and regenerated homogenous what advantages is the .

• the can and is more with from .Different the .

• Hydrogen on from

• And the an reaction their and reaction what result from .