Need three graphs for my lab report

Anonymous
timer Asked: Nov 27th, 2018
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Hello there, this was a lab experiment. This lab runs from 75-79. Please go over the lab, go over my lab report. I have only one graph with three lines. You need to make it three separate graphs with plot points. That is all that's needed. , Please check to see if the link is working! http://www.ars-chemia.net/Classes/102/manual/102_manual.pdf

Syuzanna Abrahamyan Chem 102 Chemical Kinetics 11/15/18 Purpose: Chemical kinetics is the study of the speed at which chemical and physical processes take place. In a chemical reaction it is the amount of product that forms in a given interval of time or it can be defined as the amount of reactant that disappears in a given interval of time. Scientists that study rates at which processes occur are called kinetics. Reactions occur at many different rates. Some reactions in geological processes occur at imperceptibly slow speeds where product change takes place in intervals of years or even decades. (Patel) Q-13 Since the reaction takes place in a total volume of 50.50 mL, this volume must be taken into account in calculating the initial concentration of the two reactants. For example, In Experiment 1, since the 20.00 mL of 0.200 M KI added reacts in a total volume of 50.50 m L , the initial concentration of [I- ]0 can be calculated as follows: I  − = 0.2M KI o 20.00mL = 0.08 M KI 50.5 mL Similarly in experiment 1, the initial concentration of S O  = 0.1M (K S O ) 20.00mL = 0.04 M K S O 50.5 mL S O  is calculated as follows: 2− 2 8 0 2− 2 8 o 2 2 8 2 2 Similarly in experiment 1, the initial concentration of 8 S O  is calculated as follows: 2− 2 3 S O  = 0.005M (Na S O ) 10.00mL = 0.00099 M Na S O 50.5 mL 0 2− 2 3 o 2 2 3 2 2 3 Similarly for all the experiments it is calculated and tabulated below: S O  2− [I- ]0 Exp 1 Exp 2 Exp 3 Exp 4 0.2 M KI  2 10.00mL 50.5 mL 8 0 0.1M (K 2 S 2 O 8 ) = 0.04 M 20.00mL 0.2 M KI  50.5 mL = 0.08 M KI 20.00mL 0.2 M KI  50.5 mL = 0.08 M KI 20.00mL 0.2 M KI  50.5 mL = 0.08 M KI 20.00mL 50.5 mL = 0.04 M 0.1M (K 2 S 2 O 8 ) 20.00mL 50.5 mL = 0.04 M 0.1M (K 2 S 2 O 8 ) 10.00mL 50.5 mL = 0.02 M 0.1M (K 2 S 2 O 8 ) 20.00mL 50.5 mL = 0.04 M Q-14 Average Elapsed Time Run 1 Run 2 Run 3 T avg T1(secs) 60 62 T2(secs) 30 31 61 30.5 T3(secs) 78 71 67 72 T4(secs) 138.01 148.081 143.04 Q-15 Reaction Rates In order to determine the Reaction Rates, the following quantities must be known: Initial concentration of [S2O3 2- ] added and completely used up Concentration of [I2] produced, at the time the deep blue color appears. S O  added 2− 2 3 0 I 2  produced S O  = 2− 2 3 2 All runs 0.00099 = 0.00049M 2 10.00mL 50.5 mL = 0.00099 M Na 2S 2 O 3 0.005M (Na 2S 2 O 3 ) Q-16 Experiment no. 1 temp [I- ]0 S O  time Rate(Ms-1) 20.6 0.04 0.04 61 2 19.7 0.08 0.04 30.5 3 20.0 0.08 0.02 72 4 4.3 + 3.5 + 6 = 4.6 3 0.08 0.04 143.04 0.00049 = 8.03310 −6 61 0.00049 =16.0610 −6 30.5 0.00049 = 6.810 −6 70 0.00049 = 3.4310 −6 143.04 Reaction Order with respect to iodide: From experiment 1 and 2: Exp1 k 0.04 M  0.04 M  8.03310 −6 = = Exp2 k 0.08M m 0.04 M n 16.06610 −6 m n 0.5 = 2 − m log 0.5 = −m log 2 m =1 Reaction order with respect to thio-sulfate: From experiment 2 and 3: 2− 2 8 0 Exp2 k 0.08M  0.04M  16.0610 −6 = = Exp3 k 0.08M m 0.02M n 6.810 −6 m n 2.36 = 2 n log 2.36 = n log 2 n =1.23 1 Q-17   Rate Law equation: Rate = k I − S 2 O8 2−  Experiment 1 2− I − = 0.04, S 2O8 = 0.04 Rate 1=8.033X10-6 8.03310 −6 =5.02X10-3 M-1s-1 k= 0.040.04 Experiment 2 2− I − = 0.08, S 2O8 = 0.04 Rate 1=16.06X10-6 16.0610 −6 =5.018X10-3 M-1sk= 0.080.04 Experiment 3 2− I − = 0.08, S 2O8 = 0.02 Rate 1=6.8X10-6 6.810 −6 =4.25X10-3 M-1s-1 k= 0.080.02 K avg= (5.02 + 5.018 + 4.25)10 −3 = 4.7610 −3 M −1 s −1 3         Q-18 Rate law   Rate = 4.7610 −3 I − S 2 O8 Q-19 Experiment 4(ice bath) I = 0.08, S O = 0.04 2− − 2 8 Rate 4=3.43X10-6 2−      k= 4.4310 −6 =1.38X10-3 M-1s0.080.04 Using Arrhenius Law k  E  1 1  ln  1  = a  −   k 2  R  T2 T 1   4.7610 −3  E a  1 1  = ln  −   −3   1.3810  8.314  273 + 4.6 273 + 20.1  E a = 4510 4 J / mol = 45kJ / mol Q 1(a) what is the overall order for the reaction? 2NO( g ) + CI 2 ( g ) → 2NOCI ( g ) Solution: The rate law of this reaction: Rate = K NO CI 2  2 The reaction is second order in NO and first order in Cl2.The overall order for the reaction is 03. Q1 (b) How does the reaction rate change when the nitrogen monoxide concentration is doubled and the chlorine concentration is halved? Solution: For experiment 1; Rate 1 = K NO 1 CI 1 2 For experiment 2; Rate 2 = K NO 2 2 CI 2 NO 2 NO 2 CL2 1 2CL2 2  1   2  1 Rate1 1 1 =  =  =  = 2 2 CI 2 2  2   1  2 Rate 2 NO CI 2 2 2 2 NO 2 2 1 Thus, rate of experiment-1 ( Rate1 ) is twice the rate of experiment-2 ( Rate 2 ). Question2: The initial rate of a reaction is found to increase by a factor of eight when the concentration of one reagent is doubled while all other reagent concentrations are held constant. What is the order of the reaction with respect to that one reagent? Define terms, set up the rate Exp2 law ratios and show cancellations for . Exp1 Solution 2: Rate = k A B  C  …………… (1) x y z When A is doubled, rate becomes 8 times. 8*Rate = k 2 A B  C  …………… (2) x y z From equation (1) and equation (2), we have A ; 2 x = 8 ; thus x=3 Rate = 8  Rate 2 Ax x Question 3: Given; At 593K, rate constant is 5.21  10 −4 and at 673K the rate constant is 7.42  10 −3 . At 593K, the reactant’s initial concentration was 0.2264 M, the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Thus, k 593 0.2264 = k 673 0.05999 x x x k 7.42  0.2264  = 14.2418   = 673 = k 593 0.521  0.05999  (3.774)x = 14.2418 x2 Thus, the order of decomposition is 2. Question 4: (a) The rate of reaction for a reactant, X − dx n = K X  …………………..(1) dt n is order of reaction is the rate of reaction dx n = K X  , when integrated noting dt that at t=0, X= Xo, initial concentration and at t=t, X= X, the equation becomes when n=0 is zero order reaction and when Eq.1 becomes − X =XO-Kt (2), For zero order reaction to obey, the plot of X vs. t need to be straight line with a slope of -K When n=1, the reaction is 1st order and Eq.1 becomes -dX/dt= K[X] The integrated equation is: ln(x)= ln(Xo)-Kt, so for 1st order reaction to obey, the plot of ln(x) vs. time ,t is straight line. When n=2, the reaction is 2nd order and eq.1 becomes -dX/dt= K[X]2, The integrated equation is 1/X= 1/XO+Kt, so the plot of 1/X vs t need to be straight line with a slope K. y = 0.005t + 1.109 R² = 0.997 14 12 y = -0.0004t+ 0.7220 R² = 0.6675 10 X,lnX,1/X 8 6 zero order 4 first order second order 2 0 -2 0 500 1000 -4 1500 t 2000 2500 y = -0.001t - 0.408 R² = 0.880 (b) Since the second order plot only gives a straight line, the decomposition reaction is second order with respect to A. (c) Second order equation is: 1/X= 1/XO+Kt Best fit line by graph is : y = 1.109 + 0.005t 1 / X = 1/X o + Kt K = 0.005 1/X o = 1.109 X o = 0.902 (d) Rate law equation -d X/d t= 0.005[X]2, Reference: Patel, N.A.; Swope, C. Chemical Kinetics. J. Phys. Chem. A [Online] 2001. 92, 1038-1050.

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