Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: 181

Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: The exponent n is determined by experiment. k is called the rate constant. Note that this constant does not depend on the concentration of acetyl chloride. 182

For a more general reaction such as: a A + b B c C + d D 183

For a more general reaction such as: a A + b B c C + d D The forward rate (left right) is given by: 184

For a more general reaction such as: a A + b B c C + d D The forward rate (left right) is given by: Note that x and y are NOT necessarily related to the stoichiometric coefficients a and b in the above equation. They are numbers which are constants for a given reaction. 185

Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers. 186

Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers. Reaction Order: The power to which the concentration of a reactant needs to be raised in the rate law expression. For a reaction with more than one species present, it is the sum of the reaction orders of the individual species in the rate law. 187

Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B. 188

Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B. The reaction would be described as third-order (sum of the reaction orders of the individual species reacting). 189

Important Point The exponents x and y that determine the order of the reaction are not found from the stoichiometric coefficients in the overall balanced equation. 190

Important Point The exponents x and y that determine the order of the reaction are not found from the stoichiometric coefficients in the overall balanced equation. These exponents must be determined experimentally. 191

Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) 192

Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) The rate law is given by: 193

Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) The rate law is given by: Note that the rate is NOT given by: 194

For the reaction H 2(g) + I 2(g) 2 HI (g) 195

For the reaction H 2(g) + I 2(g) 2 HI (g) The rate of reaction is: 196

For the reaction H 2(g) + I 2(g) 2 HI (g) The rate of reaction is: It is just a coincidence that the exponents on the concentration terms in this rate law match the stoichiometric coefficients in the overall balanced equation for this example. 197

For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) 198

For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) The rate of reaction is: 199

For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) The rate of reaction is: The exponents are not required to be integers, though this is often the case. 200

Integrated rate laws: Concentration changes over time 201

Analysis of some simple classes of reaction by Order 202

Analysis of some simple classes of reaction by Order Zero-order reactions 203

Analysis of some simple classes of reaction by Order Zero-order reactions The simplest class of reactions to treat are zero- order. In practical applications, this case does not arise very frequently, but it is simple to treat. 204

For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. 205

For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, 206

For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, then 207

For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, then So the rate is a fixed constant k and does not change with time. 208

Notation: The concentration of reactant A at time t is denoted as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by, which is the initial concentration of A. 209

Notation: The concentration of reactant A at time t is denoted as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by, which is the initial concentration of A. From the expression we can write 210

Recall that the equation of a straight line is 211

Recall that the equation of a straight line is So comparison with, indicates that a plot of versus t will yield a straight line with a slope = - k. 212

Recall that the equation of a straight line is So comparison with, indicates that a plot of versus t will yield a straight line with a slope = - k. 213

A plot of the rate versus time would look like: 214

First-order reactions From a practical standpoint, this is an important case, and occurs commonly. 215

First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then 216

First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then 217

First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then The equation can be solved (with a slight modification and using calculus). 218

The result is: 219

The result is: Because of the following property of logs: 220

The result is: Because of the following property of logs: then the above equation can be written as 221

The preceding equation can be put in the form (by taking antilogs of both sides of the equation): 222

The preceding equation can be put in the form (by taking antilogs of both sides of the equation): that is: 223

The preceding equation can be put in the form (by taking antilogs of both sides of the equation): that is: You can convert the preceding equation back into by taking ln of both sides of the equation. 224

Note that the equation is that of a straight line if we identify y with the term and x with the time t (the constant b would be zero). 225

Note that the equation is that of a straight line if we identify y with the term and x with the time t (the constant b would be zero). 226

An alternative plot can be made: Rewriting as so the following plot can be made: 227

An alternative plot can be made: Rewriting as so the following plot can be made: 228

A plot of versus t will show the following appearance: 229

A plot of versus t will show the following appearance: This is an exponential fall-off of the concentration with time. 230

Second-order reactions 231

Second-order reactions We will examine two cases: Both cases arise fairly commonly. 232

Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. 233

Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, 234

Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, then 235

Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, then This equation can be solved (using calculus) for to yield the following result. 236

237

This is of the form of a straight line if you identify y with and x with the time t. 238

This is of the form of a straight line if you identify y with and x with the time t. The constant b would be and the slope = k. 239

240