A) The Nature of the Reactants B) The Concentration of the Reactants C)Temperature D)Presence of a Catalyst E)Surface area in Heterogeneous Systems

In homogeneous reaction systems the rate is affected by: a) The Nature of Reactants Consider this reaction: CaCO 3 (s) + HCl(aq) --> CO 2 (g) + H 2 O(l) +CaCl 2 (aq) This reaction is quite fast. Why? HCl is completely ionized. This means there are no HCl particles present in a solution of HCl. Rather there are only H 1+ ions and Cl 1- ions present. The H1 + ions are able to easily collide with the CaCO 3 (s) particles and a successful reaction occurs. This reaction however is quite slow. CaCO 3 (s) + 2HC 2 H 3 O 2 (aq) -->CO 2 (g) + 2H 2 O(l) + Ca(C 2 H 3 O 2 ) 2 because HC 2 H 3 O 2 (aq) particles are only partially ionized.

The bonds within Acetic Acid are stronger so a solution of this acid contains mostly HC 2 H 3 O 2 particles and very few H 1+ ions. This is a stronger bond than the bond between H 1+ and Cl 1- H-Cl H 1+ + Cl H | H - C - C = O | | H O - H 100 H 1+ + C 2 H 3 O

The Concentration of Reactants If a reaction is to occur the reacting particles must first collide. The more collisions the faster the reaction proceeds. Consider this reaction H 2 (g) + I 2 (g) > 2HI(g) It has been determined, experimentally, that if [H 2 ] doubles while [I 2 ] remains constant the rate of reaction doubles. This can be expressed as rate  [H 2 ]

Also if [I 2 ] doubles while [H 2 ] is constant, the reaction rate (r) also doubles. So r  [I 2 ] So if r  [I 2 ] and [H 2 ], Reaction Rate (r)= k [I 2 ] [H 2 ] where k is the reaction rate constant which is specific to the reaction and the conditions under which it is carried out. This reaction rate expression implies that the reaction in question must proceed directly from reactants to products with no intermediate steps. This kind of reaction is called elementary. In general for an elementary reaction

aA + bB -----> cC r = k[A] a [B] b The Reaction Order is a + b Consider this elementary reaction (a reaction which proceeds directly from reactants to products with no intermediate step) 2A + B ---> A 2 B the rate expression is r = k[A] 2 [B] 1 the reaction order is = 3, or 2nd order for A, and 1st order for B, Reaction Orders > 3 are usually not elementary (there are intermediate steps). Now consider this reaction

4 HBr(g) + O 2 (g) ----> 2 H 2 O(g) + 2Br 2 (g) if this is an elementary reaction then the rate expression would be r = k [HBr] 4 [O 2 ] 1, the reaction order is = 5 However experimentation shows that for this reaction if [HBr] doubles while [O 2 ] is constant the rate, r, doubles, so r  [HBr]; also if [O 2 ] is doubled and [HBr] is constant the rate, r, doubles This means the rate expression is r = k [HBr][O 2 ] and the reaction order is 2. Why are the rate expressions different? This reaction must proceed in steps. The steps are

1.HBr + O 2 ---> HOOBrslow 2HOOBr + HBr ---> 2 HOBrfast 32HOBr + 2 HBr --> 2H 2 O + 2Br 2 fast adding steps gives 4 HBr(g) + O 2 (g) ----> 2 H 2 O(g) + 2Br 2 (g) A reaction can only go as fast as its slowest step. The rate of step 1 is the same as the overall rate. Most reactions have reaction mechanisms which limit overall rates. Data is collected in an attempt to determine the rate limiting step. Here is an example of how data is used….

For a reaction where A + B ----> C the following data was collected. This problem is solved in 2 steps. 1st to see how [A] affects the rate consider 2 trials where [B] does not change. Trials 2 and 3.

When [A] doubles, the Rate is unchanged. This means [A] has no effect on the rate. To see how [B] affects the rate, two trials are considered where [A] is unchanged. Trials 1 and 2 fulfill this requirement.

When [B] doubles the Reaction Rate increases by a factor of 8. What does this mean? Remember This is the equation we are trying to solve reaction rate = [B] x

= Reaction Rate from trial 2 Reaction Rate from trial 1 [B] from trial 2 [B] from trial 1 x 8 = 2 x X = 3 r =k[B] 3

Now try this one yourself. Determine the rate expression using this data To see how the [A] affects the rate consider 2 trials where [B] does not change. Trials 2 and 3

= Reaction Rate from trial 3 Reaction Rate from trial 2 [A] from trial 3 [A] from trial 2 x 4 = 2 x X = 2 r =k[A] 2

To see how the [B] affects the rate consider 2 trials where [A] does not change. Trials 1 and 2

= Reaction Rate from trial 2 Reaction Rate from trial 1 [B] from trial 2 [B] from trial 1 x 2 = 2 x X = 1 r =k[B] 1

The overall rate expression is r =k[A] 2 [B] 1 Let’s try one more using this data

To see how the [A] affects the rate consider 2 trials where [B] does not change. Trials 1and 2 = Reaction Rate from trial 2 Reaction Rate from trial 1 [A] from trial 2 [A] from trial 1 x 2 = 2 x X = 1 r =k[A]

To see how the [B] affects the rate consider 2 trials where [A] does not change. Trials 2 and 3 = Reaction Rate from trial 3 Reaction Rate from trial 2 [B] from trial 3 [B] from trial 2 x 8 = 2 x X = 3 r =k[B] 3

The overall rate expression is r =k[A][B] 3

As temperature increases Ek increases, molecular collisions increase so the rate of reaction, r, increases. In general a 10 K increase in temperature doubles r. eg. 290 K, r = 4.0 molL -1 min -1 what is the rate at 310 K? 16 molL -1 min -1

If a successful reaction is going to occur the reacting molecules must reach the activated complex. To do this they must come together with sufficient velocity to overcome electron-electron repulsion. If the energy of all reacting particles could be found the following distribution curve would be obtained.

# of particles EkEk T1T1 T2T2 EaEa Only particles to the right of Ea have sufficient kinetic energy to collide and reach the activated complex

Catalysts are substances which speed up chemical reactions without themselves being consumed. This is why a catalyst is written over the arrow between reactants and products. catalase H 2 O > H 2 O + 1/2O 2 In general what a catalyst does is provide a new reaction path containing a different, lower energy, activated complex. Because less energy is needed to reach the activated complex the reaction tends to go faster. This can be illustrated by the following graph

Uncatalyzed reaction catalyzed reaction Potential energy Reaction Coordinate

If the reaction involves a heterogeneous system (more than one phase) a chemical reaction can only occur at the surface between them. A pile of starch burns very slowly, but a fine particle spray of the same starch burns very rapidly since the oxygen required for the reaction can interact over a much larger surface.

O2O2 O2O2 O2O2 O2O2 Limited Surface Area O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 O2O2 Much larger Surface Area