20-10 Reaction Mechanisms

20-10 Reaction Mechanisms

• The symbol Z0 is used to represent the collision frequency.
• The fraction of sufficiently energetic collision is e-Ea>RT.
• Table 20.7 contains typical steric factors.
• As the complexity of the reactant molecule increases, the steric factor decreases.

• The fact that fewer collisions will occur with the proper orientation is reflected in this.

• The rate constant of a reaction can be expressed as the prod.

• For this reaction to occur in a single step.

• An unlikely event is a three-molecule collision.
• The reaction seems to follow a different path.
• Rate laws of chemical reactions are related to probable reaction mechanisms.

• In this section, we will explore the nature of elementary processes and then apply them to two simple types of reaction mechanisms.

• The process involves the collision of two substances.

• The rates of the forward and reverse processes may be equal in a condition of equilibrium.

• Some species are produced and consumed in different ways.
• The overall chemical equation and the overall rate law must not be used in a proposed reaction mechanism.

• In some cases, the rate of the overall reaction may be determined by one elementary process.

• We will apply these characteristics in our analysis of different mechanisms.

• Two or more elementary processes are involved in a multistep reaction.
• The reaction profile for a multistep reaction will include two or more transition states and one transition state for each step.

• The highest point along the reaction profile is the transition state of highest energy and corresponds to the rate-determining step.
• The rate-determining step is not necessarily the step with the highest activation energy.

• The conversion of A into B is slow, but step 1 is not the fastest.
• Why is this the case?

• The rate of conversion of B into C will be relatively slow because the concentration of B is kept relatively small.
• The rate of conversion of B into C is what determines the rate of reaction.

• The reaction between hydrogen and hydrogen monochloride produces hydrogen and hydrogen chloride.

• The following two-step mechanism seems plausible, and is the rate law for this reaction.

• Our mechanism suggests that step 1 occurs slowly but step 2 occurs rapidly.
• This shows that HI is consumed in the second step just as quickly as it was formed in the first.

• The rate at which HI is formed in this first step is the basis for the progress and rate of the overall reaction.
• The observed rate law for the net reaction is k3H243ICl4.
• If we have made a reasonable proposal, the proposed mechanism should give a rate law that is in agreement with the experiment.

• The species HI does not appear in the mental rate law.
• The intermediate species is a stable molecule.
• When postulating mechanisms, we have to invoke less well-known and less stable species, and we have to rely on the chemical reasonableness of the basic assumptions.
• A slightly more complicated reaction profile is caused by the presence of a reaction intermediate.

• There are two transition states and one reaction intermediate.
• The transition state for the first step is the highest point on the reaction profile.
• The difference between a transition state and a real species is not understood by atransition state.
• It is a reaction intermediate and only plex.
• The highest hypothetical is the transition state.

• Transition states can never be isolated, whereas reaction intermediates can sometimes be.

• Let's look at the following mechanism.

• There is a rapid equilibrium, but some of the N2O2 is slowly drawn off and eaten in the second step.

• We can assume that the first step of the mechanism progresses rapidly to equilibrium because we are told that it consists of a fast reversible reaction.
• If this is the case, the forward and reverse rates of reaction in the first step become equal.
• The ratio of rate constants can be replaced by a single constant.

• The Equilibrium constant numerical constant, K1, is an equilibrium constant.
• The expressions are rearranged to solve for the term 3N2O24.

• The significance of their 3N2O 24 is shown here.

• We obtained the thermody observed rate law by substituting this into equation.

• We have shown that the proposed mechanism is in line with the law.
• We can't say if this mechanism is the actual reaction path.

• After the relationships for the concentrations of any intermediates were established, the rate law of the reaction could be deduced from the rate of this step.
• More than one step can control the rate of a reaction.

• In this type of problem, we begin by identifying the slow step and using it to write the rate of reaction.
• We can assume that the equilibrium is established quickly because the fast step is given as an equilibrium.
• The common species between the two reactions, NO3 is an intermediate and can be eliminated by using the reaction equilibrium constant expression for the fast step.

• The rate equation for the rate rate of reaction is Eliminate 3NO34 if the rate of forward reaction is established quickly.

• The k1 observed rate law can be obtained if the value of 3NO34 rate is replaced with k23NO343NO4 in the rate equation.

• The final rate law is consistent with the experimental rate law.
• This alternative mechanism is plausible.
• It doesn't mean that this is the reaction mechanism.

• 2 NO2F1g2 is plausible.
• The rate law is k3NO243F24.

• This time, we will not make assumptions about the relative rates of the steps in the mechanism.

• The first, reversible reaction is written as two separate steps.

• One of the steps of the mechanism provides a convenient relationship to the observed rate of reaction.
• The last step involves the disappearance of O2.

• If the rate of change must be eliminated, keep in mind the intermediate N. The concentration of that remains constant throughout most of the reaction.
• The substance is constant.

• There are two rates for the steps that deplete the concentration of N2O2.
• The rate of disappearance of N2O2 is compared with the rate of appearance of N2O2, which is k13NO42.

• The proposed mechanism is based on the rate law.
• The rate law is more complex than the observed rate law.

• We did not make any assumptions about the relative rates of the three steps in the calculation.

• A complicated rate law is often the result of a steady-state analysis of any mechanism in which no ratedetermining step can be identified.
• The use of this type of rate law is shown in the section on enzyme catalysis later in the chapter.

• It seems that a relationship should exist between the equilibrium constant and the rate constants for the forward and reverse reactions.
• Elementary reactions can be shown that a relationship exists.

• The two rates become equal.

• The result is based on the assumption that the forward and reverse reactions are elementary reactions.

• Smog is a more familiar form of air pollution that occurs after a severe smog episode from the action of sunlight on the products of combustion.

• This type of smog is associated with high-temperature combustion processes.
• The oxides of nitrogen, principally NO1g2, are found in the exhaust from motor vehicles because the combustion of motor fuels takes place in air rather than pure oxygen.

• Unburned gasoline and partially oxidation hydrocarbons are some of the products found in the exhaust.
• These are the starting materials for photochemical smog.

• Many substances have been identified in the air, including NO, NO2, O3 and a variety of organic compounds derived from gasoline hydrocarbons.
• Ozone causes breathing difficulties for some people during smog episodes.

• PAN causes tear formation in the eyes.

• The hazy brown air that results in reduced visibility traffic congestion and heavy is the best-known symbol of features.

• Chemists who have been studying photochemical smog formation over the past several decades have determined that certain precursors are converted to conditions in Mexico City.

• Because the chemical reactions are very complex and still not fully understood, we will show how photochemical smog is formed.

• Smog formation begins with NO(g).

• NO1g2 is converted to NO21g2 to absorb ultraviolet radiation from the sun.

• A reaction forming ozone, O3.

• A large amount of ozone needs a plentiful source of NO2.
• This source was thought to be reaction at one point.

• The required levels of NO2 in photochemical smog are not achieved quickly enough by the reaction (20.31).