(please kindly type your work and use Excel to do graph ) thank you !!! --------
ID: 1059984 • Letter: #
Question
(please kindly type your work and use Excel to do graph ) thank you !!!
-------------------------/ this is what he submitted ( I don't know if it's right)
Chemistry Table 20.1: Data 450 0357 450 0361 900 150 0.298 30 600 0305 600 299 1200 0,472 200 0.227 400 0.311 7500 0271 750 253 1800 0413 250 0.175 509 026 950 0214 950 0.201 2400 356 300 0.149 60 0.221 1250 0.1511250 0.147 3000 0.333 350 0.118 163 1550 fo 103 1550 0.104 3600 0283 400 0.101 80 0.119 Table 20.2: Conditions 7.5 x 105 0.50 40.0 050 30.0 7.5 x 105 5 7.5 x 10-5 0.50 Focus Questions: 1. What is the rate law for the dissociation of ferroin? 2. What might be the reaction mechanism? 3. What is the activation energy for this reaction? Part 1: Determination of the Rate Law In this part you will analyze the data from runs 1 & 2 (table 12.1). In run 1 the initial concentration of Ferroin, [Fe(phen)32 lo 7.5 x 105 M and the initial concentration of sulfuric acid, [H2S04]o 0.50 M. In run 2, the concentration of ferroin was the same, but the concentration of H2SO4 was reduced by a factor of 10. Both runs were conducted in a thermostatic bath at 40.0°C (table 12.2). Copy the data into a spreadsheet. Add a new calculated column with time in minutes. Experiment 20
Explanation / Answer
The test run data is plotted and the following are the plots
( Test run-1)
this is a 1st order reaction ln CA= -0.0011*t-0.5075
rate constant K=0.0011 and half life =0.693/K= 0.693/0.0011=630 sec, CAO= initial concentration = e*(-0.5075)=0.602
Ca= e(-0.0011t-0.5075)
dcA/dt = -e((-0.0011t-0.5075)* 0.0011
at t=0 dCa/dt = initial rate = -e(-0.5075)*0.0011 =-0.000662 M/sec
for tesr run2, the rate constant is same as for run-2
for run-3 also the equaition
K= 0.0005, half life =0.693/0.0005=1386 sec and initial rate = -e(-0.5521)*0.0005=-0.000288 M/sec
for run-4, the plot is shown below
K= 0.0041, hlaf life = 0.693/0.0041=169 sec and initial rate = e(-0.427)*0.0041=-0.002675 M/sec
for 5th run
K =0.0206, Half life =0.693/0.0206=33.64 sec and initial rate = exp(-0.427)*0.0206=-0.013441 M/sec
T K (/sec
30+273.15=303.15 0.0005
40 +273.15= 313.15 0.0011
50+273.15 =323.15 0.0041
60+273.15= 333.15 0.020
So a plot of lnK Vs 1/T is shown below
but from Arhenius theory lnK= -Ea/R+lnK0
Ea/R = 12447 and Ea= 12447*8.315=103235 joules=103.235 Kj/mole
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