Reverse Reactions and Chemical Equilibrium In all the preceding work on chemical kinetics, reaction rates and mechanisms, attention has been focused upon chemical reactions which are proceeding in one direction only (forward). You should realize, however that many reactions are reversible, i.e. they can occur in either direction. In particular, for a given forward reaction aA + bB ---> lL + mM one can imagine of the reverse process lL + mM ---> aA + bB

Reverse Reactions and Chemical Equilibrium Now if no product molecules L & M are present at the beginning of a reaction, the reverse reaction may well be forgotten about until an appreciable concentration of products builds up, since the rate of the reverse reaction will be proportional to some product of the concentrations of L and M. In general, however, the reverse reaction becomes important eventually, since the forward reaction slows down with time (reactant concentrations being reduced by conversion to products) whereas the reverse reaction rate increases with time (as product concentrations build up). Eventually a point in time is reached when the Rate Forward Reaction = Rate Reverse Reaction Rate Forward Reaction = Rate Reverse Reaction

Reverse Reactions and Chemical Equilibrium At this point CHEMICAL EQUILIBRIUM is achieved; unless the system is disturbed by a temperature change or by adding excess reactant or product molecules, the equality of rates is maintained as are the "equilibrium" concentrations of all chemical species. The variation of reaction rates with time is illustrated graphically as follows: At this point CHEMICAL EQUILIBRIUM is achieved; unless the system is disturbed by a temperature change or by adding excess reactant or product molecules, the equality of rates is maintained as are the "equilibrium" concentrations of all chemical species. The variation of reaction rates with time is illustrated graphically as follows:

Reverse Reactions and Chemical Equilibrium In any chemical system, which a state of equilibrium is reached the reaction seems to be stopped. The macroscopic visible properties that we can see are constant. In reality the reactions are still taking place. H2(g) + I2(g) 2 HI(g) Both the forward and reverse reactions are proceeding at the same rate when a reaction is in a state of equilibrium. The observable properties and concentrations of all participants (species) become constant when a chemical system reaches a states of equilibrium. The observable properties and concentrations of all participants (species) become constant when a chemical system reaches a states of equilibrium.

Conditions of Equilibrium A chemical system is said to be in a state of equilibrium if it meets the following criteria: 1. The system is closed. 2. The observable macroscopic properties are constant. 3. The reaction is sufficiently reversible so that observable properties change and then return to the original rate when a factor that affects the rate of the reaction is varied and then restored to it's original value.

Equilibrium Expression Let's consider in detail the forward and reverse processes for some general "atom exchange" reaction AB + CD ---> AC + BD Assuming that the forward reaction occurs by a one-step bimolecular mechanism, then Rate Forward Reaction = Rf = kf [AB][CD] Rate Forward Reaction = Rf = kf [AB][CD] Similarly, assuming that the reverse reaction also occurs by a one-step mechanism Rate Reverse Reaction = Rr = kr [AC][BD]

Activated Complex in Revesible Reactions Now according to the "Principle of Microscopic reversibility", the activated complex which must be achieved in the reverse reaction is identical with activated complex for the forward reaction; thus the energy profile for the forward and reverse reactions can be illustrated by the same plot: Now according to the "Principle of Microscopic reversibility", the activated complex which must be achieved in the reverse reaction is identical with activated complex for the forward reaction; thus the energy profile for the forward and reverse reactions can be illustrated by the same plot:

Activated Complex in Revesible Reactions In the graph, Ef represents the activation energy for the forward reaction and Er that for the reverse; since the activated complex is identical for both directions it follows that ΔH = Er - Er Thus, if the reaction is exothermic (as shown), the activation energy for the reverse reaction must be greater that for the forward reaction. If the reaction is endothermic, the converse is true. If ΔH = 0 then Ef = Er.

Equilibrium Expression Returning to the rate law constant expression, we know that at equilibrium Rf = Rr therefore k f [AB] e [CD] e = k r [AD] e [CB] e where the subscript e denotes equilibrium concentrations. By rearrangement we can obtain the expression [AD] e [CB] e = K f = a constant K eq [AB] e [CD] e K r Returning to the rate law constant expression, we know that at equilibrium Rf = Rr therefore k f [AB] e [CD] e = k r [AD] e [CB] e where the subscript e denotes equilibrium concentrations. By rearrangement we can obtain the expression [AD] e [CB] e = K f = a constant K eq [AB] e [CD] e K r

Equilibrium Expression The relationships derived above for a particular reaction type are, in fact, completely general for any reaction regardless of its complexity or mechanism. That is, for any reaction in which the forward and reverse reaction rates have achieved equality (equilibrium as denoted by a double arrow), aA + bB lL + mM aA + bB lL + mM then the product of the equilibrium concentration of the products, each raised to its coefficients in the balanced equation, divided by the product of the equilibrium concentrations of the reactants, each raised to its coefficients in the balanced equation, is a characteristic constant for the system. then the product of the equilibrium concentration of the products, each raised to its coefficients in the balanced equation, divided by the product of the equilibrium concentrations of the reactants, each raised to its coefficients in the balanced equation, is a characteristic constant for the system.

paradox Before leaving this topic, one paradox" should be cleared up - that is, the appearance of coefficients in the balanced equation as the concentration exponents in K, whereas as I have emphasized earlier in the rate unit, the exponents in the rate laws for the forward and reverse reactions need not equal these coefficients unless the reaction proceeds by a single-step mechanism. Before leaving this topic, one paradox" should be cleared up - that is, the appearance of coefficients in the balanced equation as the concentration exponents in K, whereas as I have emphasized earlier in the rate unit, the exponents in the rate laws for the forward and reverse reactions need not equal these coefficients unless the reaction proceeds by a single-step mechanism.

Resolving of the paradox The paradox is resolved by considering the particular reaction AB + C -----> AC + B Where the reaction mechanism is deduced from the initial rate data is AB ---> A + B (slow) A + C ---> AC (fast)

Resolving of the paradox to yield a rate law for the forward direction to yield a rate law for the forward direction Rf = k[AB] Now if true chemical equilibrium has been achieved in the system, it must follow that EQUILIBRIUM HAS ALSO BEEN ESTABLISHED IN EVERY STEP OF THE REACTION. That is, at equilibrium the reverse of each step is also important, so the mechanism is step 1 AB ---> A + B R 1 = k 1 [AB] step 2 A + B ---> AB R 2 = k 2 [A][B] step 3 A + C ---> AC R 3 = k 3 [A][C] step 4 AC ---> A + C R 4 = k 4 [AC] Rf = k[AB] Now if true chemical equilibrium has been achieved in the system, it must follow that EQUILIBRIUM HAS ALSO BEEN ESTABLISHED IN EVERY STEP OF THE REACTION. That is, at equilibrium the reverse of each step is also important, so the mechanism is step 1 AB ---> A + B R 1 = k 1 [AB] step 2 A + B ---> AB R 2 = k 2 [A][B] step 3 A + C ---> AC R 3 = k 3 [A][C] step 4 AC ---> A + C R 4 = k 4 [AC]

Resolving of the paradox Now since R 1 = R 2 at equilibrium, k 1 [AB] = k 2 [A][B] and since R 4 = R 3 at equilibrium, k 4 [AC] = k 3 [A][C] and since R 4 = R 3 at equilibrium, k 4 [AC] = k 3 [A][C] By equating the ratio of the left-hand sides of the two equations to the ratio of the right-hand sides, we obtain By equating the ratio of the left-hand sides of the two equations to the ratio of the right-hand sides, we obtain or upon rearrangement or upon rearrangement (Note that theconcentrations of the intermediate species cancel out.)

Resolving of the paradox Therefore, even though the reaction proceeds by a multi-step mechanism, we have proved in this case (and can prove it for any other particular case) that the ratio of product to reactant concentrations, each raised to the corresponding coefficient in the balanced overall equation, is still a constant Keq characteristic of the reaction if complete equilibrium has been established. Thus there is no "conflict" between the use of reaction orders which are not equal to balanced equation coefficients and the rule that in the equilibrium constant one always obtains exponents which equal the coefficients. The properties and concentrations of an equilibrium system are constant because the rates of the forward and reverse reactions are equal so that it appears that no reactions are occurring. Therefore, even though the reaction proceeds by a multi-step mechanism, we have proved in this case (and can prove it for any other particular case) that the ratio of product to reactant concentrations, each raised to the corresponding coefficient in the balanced overall equation, is still a constant Keq characteristic of the reaction if complete equilibrium has been established. Thus there is no "conflict" between the use of reaction orders which are not equal to balanced equation coefficients and the rule that in the equilibrium constant one always obtains exponents which equal the coefficients. The properties and concentrations of an equilibrium system are constant because the rates of the forward and reverse reactions are equal so that it appears that no reactions are occurring.

Reference: SCH4U Equilibrium:introduction