Tuesday, April 5

Mr. Henderson began the class period having us take our lab notebooks out to copy down the title and purpose of the lab we would be working on at the end of the class period. Notes were given the previous day: Unit 11 Kinetics and Equilibrium chapter 11(reaction rates) consider 2 N2O5 --> 4 NO2 + O2 1 liter container Initially 10 moles/L 0 0 change -2 moles/L +4mol/L +1mol/L sometime later 8 moles/L 4mol/L 1mol/L (1hr later) Rate= (-change[N2O5])/(2 change t) Rate= (change[NO2])/(4 change t)= (change[O2])/(change t) In General: aA + bB --> cC + dD Rate= (-change [A])/(a change t)=(-change[B])/(b change t)=(change[C])/(c change t)=(change[D])/(d change t) We then looked back onto page one which we finished monday in class. 1) a. Rate= (- change[NH4])/(change t)= (-change[NO2-])/(change t)=(change[N2])/(change t)=(change[H2O])/2 change t b. Rate= (-change[N2])/(change t)= (-change[H2])/(3 change t)=(change[NH3])/(2 change t) 2) a. As the concentration of NO is doubled while keeping the Cl2 concentration constant, the rate of the reaction is increased by a factor of 4. This can be seen while comparing rows 1 and 2. b. As the concentration of Cl2 is doubled(while keeping the NO concentration constant), the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 2 and 3. c. Cl2: 1+ NO: 2 = Overall: 3 Mr. Henderson pointed out that in this specific question, the order of reactions for Cl2 and NO happens to be the coeficients. The overall reaction is simply determined by adding the order of reactions of both of the reactants which in this case is Cl2 and NO. So you then add Cl2's oreder of reaction,1, and add NO's order of reaction, 2, to get 3. d. Rate= k[NO2]^2 Rate=k[Cl2]-->Rate=k[NO]^2*[Cl2] k is the rate constant. e. consider row 3: 9.12=k(1)^2*(1) 9.12=k 3. a. As the concentration of I- is doubled while keeping the S2O8^2- concentration constant, the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 1 and 2. b. As the concentration of S2O8^2- is doubled(while keeping the I- concentration constant), the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 1 and 3. d. Rate=k[I-] Rate=k[S2O8^2-]--> Rate=k[I-]=[S2O8^2-]^2 We then were given time at the end of the class period to work on Lab KE1 Feeling Blue_No Longer.