Chemistry Lab 6o4j41

Chemical Kinetics Lab Background:​ ​This lab’s objectives include learning how to write rate law expressions,

determine orders of reaction by graphical analysis, and calculate rate constants as you study the oxidation of food dyes by sodium hypochlorite (NaOCL). Students will monitor the color change of food dye solutions over time and use the integrated rate law to determine the order of each reactant, the overall or the reaction and calculate the reaction’s rate constant.

Pre-Lab Questions: 1. A container of household bleach has 5.00% NaOCl by weight on its label. If the density of this bleach is 1.08 g/mL, calculate the molarity of this solution. a. Molarity = moles / L b. 1.08 g/mL = 1080 g / L c. Assuming 1 L, 1080 g * (5g / 100g) = 54 g NaOCl d. Molar Mass of NaOCl is 74 g/mol e. Moles of NaOCl = 54 g NaOCl / (74 g NaOCl / 1 mole NaOCl) = 0.73 mole NaOCl f. 0.73 mole / 1 L = 0.73 M 2. How many grams/L of NaCl are required in order to prepare a solution with a concentration equivalent to the solution in Question 1? a. Again assuming 1 L, 0.73 mol * (58.44 g NaCl / mol NaCl) = 42.66 g / L NaCl 3. What is the percent of NaOCl by weight for a liter of bleach that is 0.800 M NaOCl and has a density of 1.09 g/mL? a. 0.800 M = 0.800 mol / L NaOCl b. 74.8 g/mol * 0.800 mol/L = 59.2 g assuming 1 L. c. 1.09 g / mL = 1090 g / L d. 59.2 g / 1090 g * 100% = 5.4% NaOCl by mass

Data: ● Activity 1: - Blue #1: ​100 drops of NaOCl - Red #40: ​120 drops of NaOCl

● Activity 2: ○ Red Dye Absorbance Data: Time (s)

Absorbance (nm)

Time (s)

Absorbance (nm)

0

2.160 70

0.861

10

2.142 80

0.824

20

2.125 90

0.837

30

2.083 100

0.841

40

1.873 110

0.837

50

1.283 120

0.830

60

0.943 130

0.815

○ Red Dye Absorbance Graph:

○ Blue Dye Absorbance Data: Time (s) Absorba nce (nm)

Time (s) Absorba nce (nm)

Time (s) Absorba nce (nm)

Time (s) Absorba nce (nm)

0

2.270

100

2.153

200

1.402

300

0.303

10

2.252

110

2.128

210

1.257

310

0.255

20

2.239

120

2.118

220

1.110

320

0.214

30

2.243

130

2.090

230

0.949

330

0.181

40

2.231

140

2.050

240

0.821

340

0.147

50

2.215

150

2.010

250

0.706

350

0.123

60

2.212

160

1.946

260

0.605

360

0.100

70

2.200

170

1.845

270

0.500

80

2.187

180

1.726

280

0.426

90

2.165

190

1.546

290

0.359

○ Blue Dye Absorbance Graph:

● Activity 3:

○ Blue Dye NaCl Absorbance Data: Dye

m

n

Overall Order

Rate Expression

k (Rate Constant)

Blue #1

1

2

3

Rate = k[blue][NOCl]​2

4.43*10​-4 ​M​-1​s​-1

Red #40

1

2

3

Rate = k[red][NOCl]​2

4.43*10​-4 ​M​-1​s​-1

○ The germicidal bleach provided in this kit is 6.00% (m/v) NaOCl. The density of this solution is 1.09 g/mL. Convert this 6% to mol/L or molarity (M). a. M = mol/L b. 1.09 g/mL = 1090 g/L c. 1090 g/L * 0.06 = 65.4 g d. 65.4 g / 74.44 g = 0.879 mol e. 0.879 mol / 1 L = 0.879 M

Post Lab Questions: 1. The data in this experiment obeys the Beer-Lambert Law. Explain this law? a. The Beer-Lambert law describes the relationship between absorbance and concentration of an absorber of electromagnetic radiation. 2. Why is equimolar sodium chloride solution used to dilute the sodium hypochlorite in Trial 2 (Activity 2, Part 3) instead of distilled water?

a. Sodium chloride is used instead of distilled water because sodium chloride doesn’t react with sodium hypochlorite. It also is used to maintain the same ionic strength as the original NaOCl. 3. Sketch the graphs of concentration vs. time for a single-reactant reaction (aA → products) that would be zero order, first order, and second order.

4. Write a general equation and rate law for a reaction having two reactants. a. aA + bB -> C b. Rate = k[A]​x​[B]​y 5. Write the rate law for a reaction having two reactant in which one is in excess. a. aA + bB -> C , Assuming that reactant B is in excess k` = k[B]​y Rate = k`[A]​x 6. The orders with respect to reactants C and D of a reaction were determined experimentally to be 2 and 3. a. Write a general equation for this reaction, the rate law, and determine the overall order of the reaction. ● cC + dD -> E ● Rate = k[C]​2​[D]​3 ● Overall Order = 5 b. What would happen to the rate of the reaction if only the concentration of C were doubled? If only D were doubles? If both C and D were doubled? ● When C were doubled:- Rate of reaction increased by 4 times Rate’ =k[2C]​2​[D]​3​ = 4*k[C]​2​[D]​3​= 4*Rate ● When D were doubled:- Rate of reaction increased by 8 times Rate’ =k[C]​2​[2D]​3​ = 8*k[C]​2​[D]​3​= 8*Rate ● When both C and D were doubled:- Rate of reaction increased by 32 times

Rate’ =k[2C]​2​[2D]​3​ = 32*k[C]​2​[D]​3​= 32*Rate

Conclusion: ➢ Sources of error in this lab include: ○ Lack of proper materials. ○ Human error. ○ Instrumental Limitations (uncertainties in measuring instrument) ○ Errors due to limited space and time. ➢ This lab taught students how to determine the order or reactants experimentally and understand the concept of rate laws and their significance in chemical kinetics.