Tube 1 (Slope) = -0.0026 Abs/sec Tube 2 (Slope) = -0.0012 Abs/sec Tube 3 (Slope)

Question

Tube 1 (Slope) = -0.0026 Abs/sec

Tube 2 (Slope) = -0.0012 Abs/sec

Tube 3 (Slope) = -0.0014 Abs/sec

Molar abs of crystal violet is 53523 L/mol.cm

The specific reaction that used is : CV+ + OH---->CVOH

Kinetics Experiment: Crystal Violet: 3.04*10^-5 M NaOH: 0.1 M The molar absorptivity of crystal violet is 53523 L/mol-cm. Tube1: 5mL Crystal Violet + (2mL NaOh) Tube 2: 5mL Crystal Violet + (1mL NaOh + 1mL DlWater) Tube 3: 2.5mL Crystal Violet + (2mL NaOh + 2.5mL DlWater) Use the slope from each line to calculate the rate of reaction associated with that line? The Reaction order with respect to Crystal Violet and OH-. Use the calculated reaction orders and initial concentration and rate data and calculate the rate constant k. Thank you.

Explanation / Answer

A=cel(beer’s law)

Now [CV+]=A/el  e=Molar abs of crystal violet is 53523 L/mol.cm,l=1cm

Rate of the rxn=change of concentration/time elapsed

Absorbance data converted to concentration data

[CV+]=-0.0026/(53523  L/mol.cm* 1 cm)=4.8*10^-8 mol/L .sec

Rate of reaction=[CV+]final-[CV+]initial/time=[CV+]initial/t=4.8*10^-8 M/s

Cv+ has all reacted so [CV+]final=0

Tube 2 (Slope) = -0.0012 Abs/sec

Rate of reaction =-0.0012/53523=2.2*10^-8 M/s

Tube 3 (Slope) = -0.0014 Abs/sec

Rate=0.0014/53523=2.6*10^-8 M/s

Initial rate of reaction=k[CV+]initial^m [OH-]^n

K=rate constant

[OH-]=excess

So,Initial rate of reaction=k’[CV+]initial^m

  K’=k[OH-]

Tube1

[CV+]=3.04*10^-5 mol/L *0.005 L=15.2*10^-8 moles

Total volume=5.ml+2ml=7ml

[CV+]=15.2*10^-8 moles/0.007 L=2.17*10^-5 M

[OH-]=0.1 M*0.002L=0.0002moles

[OH-]=0.0002moles/0.007L=0.028M

Rate=4.8*10^-8 M/s=k (2.17*10^-5 M)^m (0.028M)^n.…………..(1)

Tube 2

[CV+]=2.17*10^-5 M

[OH-]=0.1 M*0.001L=0.0001moles

[OH-]=0.0001moles/0.007L=0.014M

Rate==2.2*10^-8 M/s=k(2.17*10^-5 M)^m (0.014M)^n.…………(2)

Tube 3

[CV+]=1.1*10^-5M

[OH-]=0.028M

Rate=2.6*10^-8 M/s=k(1.1*10^-5)^m (0.028)^n.…………(3)

Eqn(1)/eqn (2) gives,

2.2=(2)^n

Log 2.2=n log2

N=1

Eqn (1)/eqn(3)

1.8=(1.9)^m

Log 1.8=mlog 1.9

M=1

Order of reaction is 1 both with respect to CV+ and OH-

Now k can be easily calculated

Rate=4.8*10^-8 M/s=k (2.17*10^-5 M)(0.028M)  n=m=1

K=7.8*10^-2 S-1

Tube2

Rate==2.2*10^-8 M/s=k(2.17*10^-5 M) (0.014M)

K=7.1*10^-2 s-1

Tube 3

Rate=2.6*10^-8 M/s=k(1.1*10^-5) (0.028)

K=8.4*10^-2 S-1

K(average)=(7.8+7.1+8.4/3)*10^-2=7.8*10^-2 s-1

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