CHEM 122 11th EditionExam # 3 Study Guide Lectures: 17-23Lecture 17 (February 27th)Collision Theory model: Rate of a reaction is proportional to the number of effective collisions per second- Effective collision: A collision that results in the formation of productso only a very small number of collisions actually result in an effective collision.- What is required for a reaction to occur?1. Molecules must collide2. The energy of the collision must exceed some minimum energy3. They must collide in the proper orientation (head on, from the side etc.)Effect of Temperature on Rate:- Temperature affects the rate constant (k) as temperature increases.- General Rule: the rate constant for a reaction increases by about a factor of 2 to 3 for each 10 degrees Celsius increase in temperature.- Why is there this strong temperature dependence on the rate constant of a reaction?o As temperature increases, molecules move faster, therefore, there are more collisions per second.o But the increased collision frequency doesn’t account for the dramatic temperature effect on rate constants.- Activation energy: minimum energyLecture 18 (March 2nd)Activation Energy- The relationship between Ea (activation energy) forward and reverse reactions: Ea(forward)−∆ H=Ea(reverse) - Why does temperature have a significant effect on rate constants and reaction rates?o Only collisions whose kinetic energy exceeds activation energy can react, and only a very small fraction of collisions actually exceed the activation energy.Svante Arrhenius: Chemist in the late 1800’s who studied the effect of temperature on the reaction rates.- Arrhenius equation: k =A e−EaRT o K: rate constantoEa: Activation energyoR: Gas constant (8.314J/mol K)oT : Absolute temperature (kelvin)oA : Frequency factor  The frequency factor takes into account both the collision rate as well as the probability that the collisions have the proper orientation.- As Ea increases, k increases- As T increases, k increases- The Arrhenius equation can be rearranged to the equation of a straight line (y=mx+b)olnk=(−EaR)1T+lnAo A plot of ln k and 1/T yields a straight line- Fraction of collisions whose collision energy exceeds activation energy:oF=e−EaRTLecture 19 (March 4th)Activation Energy- Measuring the rate constant, k and 2 temperatures can determine the activation energy.olnk1k2=EaR(1T2−1T1)Reaction Mechanisms: the sequence of events leading from reactants to products.- Elementary Reactions: collisions of 2 molecules at a time, requires at least 2 steps.- Reaction Intermediate: species that appears in a reaction mechanism but neither is a reactant or a product.- Molecularity: the number of reactant molecules that participate in a single elementary reaction.Molecularity of Elementary Reactions:- Unimolecular: a single molecule is involved in the mechanismo A products Rate=k[A]- Bimolecular: A collision between two reacting moleculeso most commono A + B Products Rate= k[A][B]- Termolecular: the simultaneous collision between 3 reacting moleculeso very rareo A + B + C Products Rate= k[A][B][C]- Coefficients of the equation determine the exponents in the Rate law.Determining the rate Law From a Multistep Reaction Mechanism:- the rate law for the overall reaction is determined by the slowest elementary step.- Equilibrium: Rate (forward)= Rate(reverse)Lecture 20 (March 6th)Catalyst: a substance that increased the rate of a reaction without undergoing a permanent chemical change itself.- Heterogeneous catalyst: catalyst in a different phase than the reactants.o Most common in catalysts.- Homogenous catalysts: Catalyst is in the same phase as the reactants.- How does a catalyst speed up a reaction?o By lowering the activation energy for the reaction.- Since the process of creating hydosulfuric acid is so slow a catalyst is needed to speed up the process. This process is known as the Lead chamber process.- Haber Process: the process of using a heterogeneous catalyst, a solid metal surface to speed up the process of creating ammonia with diatomic nitrogen and hydrogen.Lecture 21 (March 9th)Chemical Equilibrium: when the rates of a reaction forward and backward are equal.- There is no net change in the concentration of components in the reaction- Dynamic Equilibrium: reactants are still being converted to products and products are still being converted to reactants.- Equilibrium constant Expression: aA + bB pP + qQ Kc=[ P]p[Q]q[ A ]a[B ]b=productsreactants- Reaction Quotient: numerical value for the mass action expression- Equilibrium Constant K(¿¿c)¿: at equilibrium the reaction quotient’s constant for a particular reaction- Reverse Reaction: KC'=1Kc- When the equation coefficients are multiplied by a factor the equilibrium constant can by found by: Kc' '=(Kc)xCoal Gasification: the conversion of coal into gaseous or liquid fuelsWater Gas: CO + H2Lecture 22 (March 11th)Equilibrium Law Expressions: Kp=Pconcentrationsof productsPconcentrations of reactants-Kp: Concentration in partial pressure-P: Partial pressure of concentration ex.) [A]xA general equationrelating the ∂ pressure K∧the concentration K : -Kp=Kc(RT )∆ n-Kp:Concentration in partial pressure-Kc: Concentration expression-R : Ideal gas constant (.0821 L atm/ mol K)-T : Temperature K-∆ n: Change in mol (mole products-mole reactants)Reaction Quotients:- For finding the initial concentration we calculate the reaction quotient (Q)- Comparison of Q with the equilibrium constant (K) allows us to predict the reaction direction of reaction. o Q=K: the reaction is at equilibriumo Q<K: the reaction proceeds to generate productso Q>K: the reaction proceeds to generate more reactantsEquilibrium Calculation:When given the initial concentrations we can use those to find the equilibrium concentrations.1. Create an ICE chart. Initial .01 M 0 M 0 MChange -x +x +xEquilibrium .01-x x x2. Fill in the initial molarity, which is given.3. With the chemical equation given multiply x by the moles of each reactant and product that is in a gaseous state. In this example the moles of each molecule is 1. The products and reactants have to be opposite signs. In this example the reactants are negative and the products are positive.4. Then add the initial concentration with the change 5. Plug it into the Equilibrium Expression equation: (x )(x).01−xKc in this example K is given to us as 10.6. By simple algebra the equation is then turned into a