Lab 20 Experiment 20

Gary Bagdasarian 5/19/20 Experiment 20 Focus Questions: The rate law for the dissociation of ferroin is k[Fe( phen )_3^2+]. The reaction mechanism is: Fe(phen)_3^2+(aq) <-> Fe(phen)_2^2+ + phen(aq) Fe(phen)_2^2+(aq) <-> Fe(phen)^2+ + phen(aq) Fe(phen)^2+(aq) <-> Fe^2+(aq) + phen(aq) phen(aq) + H3O+(aq) <-> phenH+ + H2O(l) The activation energy for this reaction is roughly 126.588964 kJ/mol*K. Post-Lab Questions: In an elementary step, the coefficients equal the reaction order its rate law because it’s just one step, therefore, all the reactions collide adeptly. If a reaction were to occur in one step, then all the reactants will contribute to the reaction to the products. However, in a multi-step reaction, only the slowest step is important for the rate law and that is not representative of all the reactants in the reaction. A reaction can be thermodynamically favo rable when it is spontaneous and gravitates towards a lower energy state. A kinetically favorable reaction has low activation energy. When looking at a reaction thermodynamically, it should have negative free energy for the reaction to proceed. Kinetically unfavorable refers to the reaction having a higher activation energy. It tells us how fast a reaction will occur, and a high activation energy means a slower reaction time. a) A+B+D -> F b) The catalyst is C and the intermediate is E. c) Step 1: bimolecular, k[A][B] Step 2: bimolecular, k[C][D] Step 3: unimolecular, k[E] d) The mechanism is not consistent with the actual rate law because the slowest step is C+D=E. The rate law for that is k[C][D] and doesn’t match with the rate law of k[A][B][D]. Data: Part 1: Run 1 Time(sec) [A] ln[A] 1/A Time(min) 0 0.607 -0.49923 1.647446 0 150 0.504 -0.68518 1.984127 2.5 300 0.427 -0.85097 2.34192 5 450 0.357 -1.03002 2.80112 7.5 600 0.305 -1.18744 3.278689 10 750 0.271 -1.30564 3.690037 12.5 950 0.214 -1.54178 4.672897 15.833 1250 0.151 -1.89048 6.622517 20.833 1550 0.103 -2.27303 9.708738 25.833 Calculations: Half-life, graph [A] vs. time 0.607/2 = 0.3035 at 10 minutes Half-life = 10 minutes Rate constant, graph ln[A] vs. time y=-0.0671x-0.5057 ln[A]=-kt+ln[A]0 k=0.0671 Initial rate, graph [A] vs. time y=0.6031 x 10^-0.067x y’=(0.6031 x (-0.067)) x 10^-0.067x y’=(0.6031 x (-0.067)) x 10^-0.067(0) = -0.0404077 Initial rate=0.0404077 Run 2 Time(sec) [A] ln[A] 1/A Time(min) 0 0.607 -0.49923 1.647446 0 150 0.524 -0.64626 1.908397 2.5 300 0.412 -0.88673 2.427184 5 450 0.361 -1.01888 2.770083 7.5 600 0.299 -1.20731 3.344482 10 750 0.253 -1.37437 3.952569 12.5 950 0.201 -1.60445 4.975124 15.833 1250 0.147 -1.91732 6.802721 20.833 1550 0.104 -2.26336 9.615385 25.833 Calculations: Half-life 10 min. Rate constant =0.0682 Initial Rate: y=0.6002 x 10^-0.068x y’=(0.6002 x (-0.0682)) x 10^-0.068x y’=(0.6002 x (-0.0682)) x 10^-0.068(0) = -0.0408136 Initial rate=-0.0408136 The rate order is first order. This is because the ln[A] vs. time graph is linear and there isn’t much change in half-life. Fe(phen)_3^2+(aq) Fe(phen)_2^2+(aq) + phen(aq) slow Fe(phen)_2^2+(aq) <-> Fe(phen)^2+(aq) + phen(aq) fast Fe(phen)^2+(aq) <-> Fe^2+(aq) + phen(aq) fast phen(aq) + H3O+(aq) <-> phenH+(aq) + H2O(l) fast Part 2: Run 3 Time(s) [A] ln[A] 1/A Time(min) 0 0.601 -0.50916 1.663894 0 300 0.565 -0.57093 1.769912 5 600 0.529 -0.63677 1.890359 10 900 0.498 -0.69716 2.008032 15 1200 0.472 -0.75078 2.118644 20 1800 0.413 -0.88431 2.421308 30 2400 0.356 -1.03282 2.808989 40 3000 0.333 -1.09961 3.003003 50 3600 0.283 -1.26231 3.533569 60 Calculations: Half-life, use graph [A] vs. time 0.601/2 = 0.3005 at 55 minutes Rate constant, use graph ln[A] vs. time ln[A]=-kt+[A]0 y=-0.0124x-0.5104 k=0.0124 Initial rate, use graph [A] vs. time y=0.6002 x 10^-0.012x y’=(0.6002 x (-0.012)) x 10^-0.012x y’=(0.6002 x (-0.012)) x 10^-0.012(0) = -0.0072024 Initial rate=-0.0072024 Run 4 Time(s) [A] ln[A] 1/A Time(min) 0 0.611 -0.49266 1.63661 0 50 0.513 -0.66748 1.949318 0.833333 100 0.425 -0.85567 2.352941 1.666667 150 0.298 -1.21066 3.355705 2.5 200 0.227 -1.48281 4.405286 3.333333 250 0.175 -1.74297 5.714286 4.166667 300 0.149 -1.90381 6.711409 5 350 0.118 -2.13707 8.474576 5.833333 400 0.101 -2.29263 9.90099 6.666667 Half-life, use graph [A] vs. time 0.611/2 = 0.3055 at 2.5 minutes Rate constant, use graph ln[A] vs. time y=-0.2847x-0.4715 k=0.2847 Initial rate, use graph [A] vs. time y=0.6241 x 10^-0.285x y’=(0.6241 x (-0.285)) x 10^-0.285x y’=(0.6241 x (-0.285)) x 10^-0.285(0)=-0.1778685 Initial rate=-0.1778685 Run 5 Time(s) [A] ln[A] 1/A Time(min) 0 0.606 -0.50088 1.650165 0 10 0.515 -0.66359 1.941748 0.166667 20 0.426 -0.85332 2.347418 0.333333 30 0.343 -1.07002 2.915452 0.5 40 0.311 -1.16796 3.215434 0.666667 50 0.276 -1.28735 3.623188 0.833333 60 0.221 -1.50959 4.524887 1 70 0.163 -1.81401 6.134969 1.166667 80 0.119 -2.12863 8.403361 1.333333 Half-life, use graph [A] vs. time 0.606/2=0.303 at 7 minutes Rate constant, use graph ln[A] vs. time y=-1.1492x-0.4556 k=1.1492 Initial rate, use graph [A] vs. time y=0.6341 x 10^-1.149x y’=(0.6341 x (-1.149)) x 10^-1.149x y’=(0.6341 x (-1.149)) x 10^-1.149(0)=-0.7285809 y’=-0.7285809 Initial rate=-0.7285809 Run Temperature(C) Temperature(K) 1/T k ln[k] 1 40 313 0.003195 0.0671 -2.70157 2 30 303 0.0033 0.0124 -4.39006 3 50 323 0.003096 0.2847 -1.25632 4 60 333 0.003003 1.1492 0.139066 ln[k]=-(Ea/R)(1/T) + ln[A] y=-15191x+45.777 Ea/R=15191 Ea=15191 x R =15177 x 0.008314 =126.29 kJ/mol