Chem. 116 Quiz 2 spring 2009Name _________________Key_______ ___________________ Instructor: Martin Larter1. Nitric oxide can be reduced with hydrogen gas to give nitrogen and water vapor.2NO + 2H2  N2 + H2O (overall rxn)A proposed mechanism is: 2NO  N2O2 (fast)N2O2 + H2  N2O + H2O (slow)N2O + H2  N2 + H2O (fast)What rate law is predicted by this mechanism? (Show your work.)Rate = k[N2O2 ][H2] N2O2 = intermediate k1 [NO] = k-1 [N2O2][N2O2] = k1 [NO] 2k-1Rate = k k1 [NO]2 [H2] = k’[NO]2 [H2 k-1 2. A certain reaction has activation energy of 54.0 kJ/mol. As the temperature increased from 22oC to a higher value, the rate constant increases by a factor of 7.00. Calculate the higher temperature (R=8.314 J/mol*K).k2= 7 x k1Ln(k2/k1) = Ea/R(1/T1 -1/T2) = ln (7.00) = 54.0 kJ/mol x 1 - 1 8.314 x10-3 kJ/mol*K 295 K T21.9459 = 6.495 x103 K x 1 - 1 295 K T2T2 = 324 K = 51 oC3. Wilson’s disease is a hereditary problem in which the body is unable to excrete copper, which subsequently becomes stored in the liver. To study treatments for the disease, doctors inject patients with small doses of 64Cu, a radioactive isotope that decays via first-order kinetics with a half-life of 12.7 hr. If a solution of 64Cu(CH3CO2)2 is injected into a patient, what percentage of the 64Cu will remain in the body after 24 hours? k= ln 2/12.7 hr = 0.0545785 hr-1 ln [A]t = -0.0545785 hr-1 (24 hours) [A]o ln [A]t = -1.3098 [A]o [A]t = 0.2698 *100 =27.0 % remain in the body after 24 hours [A]o4. One proposed mechanism for the reaction of water with the organic compound ethyl acetate (CH3CO2C2H5) is:Step 1: CH3CO2C2H5 + OH- → CH3CO2H + C2H5O- SLOWStep 2: C2H5O- + H2O → C2H5OH + OH- FAST(a) Write the overall equation for this reaction, and identify any intermediates and/or catalysts.Overall reaction: CH3CO2C2H5 + H2O → CH3CO2H + C2H5OH Intermediate: C2H5O-Catalyst: OH-(b) Draw a fully labeled potential energy diagram for this reaction. Assume that the reactants are more stable that the products of the overall reaction.5. My dog, Baby, asked about a general assumption used in kinetics. The general assumption is that for every 10oC change the rate doubles. Is this a good assumption, explain in terms of the Arrhenius equation and collision theory. (Ea = 181.9 kJ/mol A=3.79 x1011)Each reaction rate coefficient k has a temperature dependency, which is usually given by the Arrhenius equation:To see what temperature rise is required to change the rate constant from k1 (at T1) to k2 (at T2), take the ratio of the Arrhenius equations for each of the two temperatures: k2 k1 = A exp(-Ea/RT2) A exp(-Ea/RT1) = exp ( - Ea R[1 T2 - 1 T1])By taking the ratio it can be seen the frequency factor cancels out and the activation energy is a constant that can be factored out along with the gas constant R. Thus there is a logarithmic relationship between k and 1/T. The increase in reaction rate (k) is not linear with temperatureLn(k2/k1) = Ea/R(1/T1 -1/T2)ln (2.0) = 181.9 kJ/mol/ 8.314 x10-3 kJ/mol*K (1/298K -1/T2)0.693 = 21878.75872 (0.0033557047-1/T2)0.693 = 73.41865342 - 21878.75872 K (1/T2)T2 =300.8 KIn this case it does not look like it is a good assumption that the reaction rate doubles with a ten degree change in temperature.If the activation energy has a particular value, and if the temperature change occurs in the right range, and if the reaction is an elementary one that obeys the Arrhenius equation, then a 10°C rise might double the reaction rate. It's a very iffy generalizationTo have a doubling of rate with a 10 oC change in the above example the activation energy would had to have been 52.8967