ID: 0003259 5. From your knowledge of x and y in the equation (as well as the ra

Question

ID: 0003259

5. From your knowledge of x and y in the equation (as well as the rate in each experiment from your graph), calculate k from your data for each solution. Rate = k[S2O8-2]x[I-]y

Rate of reaction, [S2O8-2]/t, as calculated from graphs (that is from slopes of lines)

Solution 1 = 6.4 x 10-6 mol/L-s

Solution 2 = 1.8 x 10-5 mol/L-s

Solution 3 = 2.8 x 10-5 mol/L-s

REPORT SHEET Rates of Chemical Reactions I: A Clock Reaction A. Preliminary Experiments 1. What are the colors of the following ions? K Color I sc Color less 2. The color of the starch I2 complex is Blut B. Kinetics Experiment Solution 1. Initial [S20s 0.05 M; initial [I] 0.05 M. Time experiment started Aliquot Time (s) between no. appearances of color Cumulative times (s) Total moles of S208 consumed 2.0x10 4.0x10 6.0x10 4 8.0x10 10.0x10 12.0x10 y20 s 14.0x10 Solution 2. Initial [s,os 0.10 M; initial [I] 0.05 M. Time experiment started Total moles of S208 Time (s) between Cumulative times (s) Aliquot no. appearances of color consumed 2.0x10 4.0x10 6.0x10 4 8.0x10 4 10.0x10 12.0x10 002 14.0x10 4 Copyright c 2012 Pearson Education, Inc. drawn mus

Explanation / Answer

the rate law ,

rate=k [S2O82-]^x [I-]^y                                  

solution 1-

initial rate =6.4 *10-6 M/s

[S2O82-]o=initial concentration=0.05M

[I-]o=initial conc=0.05M

rate=k [S2O82-]^x [I-]^y      

6.4 *10-6 M/s=k(0.05M)^x *(0.05M)^y....................(1)   

solution 2-

initial rate =1.8 *10-5 mol/L-s

[S2O82-]o=initial concentration=0.1M

[I-]o=initial conc=0.05M

rate=k [S2O82-]^x [I-]^y     

1.8 *10-5 mol/L-s=k(0.1M)^x (0.05M)^y...................................(2)

solution 3-

initial rate =2.8 *10-5 mol/L-s

[S2O82-]o=initial concentration=0.05M

[I-]o=initial conc=0.1M

rate=k [S2O82-]^x [I-]^y     

2.8 *10-5 mol/L-s=k(0.05M)^x (0.1M)^y...................................(3)

eqn (1)/eqn(2),

6.4 *10-6 M/s/1.8 *10-5 M/s=k(0.05M)^x *(0.05M)^y/k(0.1M)^x (0.05M)^y

0.355=(0.5)^x *(1)^y

or,0.355=(0.5)^x

log 0.355=xlog 0.5

or,-0.450=x(-0.301)

x=0.45/0.301=1.4 (approx 1)

eqn (2)/eqn(3),

1.8 *10-5 mol/L-s/2.8 *10-5 mol/L-s=k(0.1M)^x (0.05M)^y/k(0.05M)^x (0.1M)^y

0.643=(2)^x (0.5)^y

or,0.643=(2)^1 (0.5)^y

or 0.321=(0.5)^y

log 0.321=y log(0.5)

0.4=y(0.30)

y=0.4)/0.3=1.3 (approx 1)

y=1

putting value of x and y in eqn (1),

6.4 *10-6 M/s=k(0.05M)*(0.05M)

k=0.00256=2.6*10^-3 M-1 s-1

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---