Spectrophotometry Lab November 29, 2018 Agenda - Nov 29, 2018

Final decision (thanks to Alaina) 60 multiple choice questions only. 90 mins. Exam 4 - Qu. 4 Classic mistake - RTQ Mrs. D-B! Introduction to spectroscopy Spectrophotometer lab. 4. s is the specific heat Consider four 100.0g samples of water, each in a separate beaker at 25.0℃. Into each beaker you drop 10.0g of a different metal that has been heated to 95.0℃. Assuming no heat loss to the surroundings, which water sample will have the lowest final temperature?. ℃ a. The water to which you have added iron (s = 0.45 J/ g) b. The water to which you have added aluminum (s =0.89 J/℃ g) c. The water to which you have added lead (s = 0.14 J/℃g) d. The water to which you have added copper (s = 0.20 J/℃g) 95oC

10g Fe 10 g Al 10 g Pb 10 g Cu o o o 0.45 J/ C 0.89 J/ C 0.14 J/ C 0.20 J/oC g g g g

25oC 25oC 25oC 25oC 10g Fe 10 g Al 10 g Pb 10 g Cu o o o 0.45 J/ C 0.89 J/ C 0.14 J/ C 0.20 J/oC g g g g Only used 0.14 J to raise 1g 1oC, so only that amount of energy is available to heat up the water, hence water will heat up the least with this one added; lowest temp.

25oC 25oC 25oC 25oC 4. s is the specific heat q = smΔT Consider four 100.0g samples of water, each in a separate beaker at 25.0℃. Into each beaker you drop 10.0g of a different metal that has been heated to 95.0℃. Assuming no heat loss to the surroundings, which water sample will have the lowest final temperature?. ℃ a. The water to which you have added iron (s = 0.45 J/ g) b. The water to which you have added aluminum (s =0.89 J/℃ g) c. The water to which you have added lead (s = 0.14 J/℃g) d. The water to which you have added copper (s = 0.20 J/℃g)

Lab 6: What is the relationship between the concentration of a solution and the amount of transmitted light through the solution?

Part 1: Looking for the relationship between transmittance and concentration of a solution.

Appropriate wavelength to take transmittance is 630 nm (do not adjust wavelength control during this lab). Stock solution Blue #1 Dye concentration is 7.3 microM (7.3 x 10-6 M)

Your groups will gather data as to the percent transmittance of the molecule in solution at various concentrations. Record results.

Each person then responsible for analysis (using Desmos.) Key is patience and organization. Groups of 3 or 4 - let me see everyone doing lab related activities safely, like college students. Not orming. Generally with the spectrometer

a) Cuvette should be more than half full and free of bubbles. b) Use the same cuvette for all your measurements. c) Make sure that the mark (fiducial line) on the cuvette aligns with the mark on the adapter toward the front of the instrument. d) Do not touch the cuvettes below the fiducial line. Use the lint free wipes to dry and clean the outside of your cuvette before inserting into the instrument. Using the spectrometer - be ready/don’t dither 1) Prepare your diluted solutions in a small beaker a) Measure xmL of Stock solution, pour into a clean empty beaker. Measure (10-x)mL of water (distilled) in the same measuring cylinder and pour into the dye and mix (swirl). Prepare all dilutions. Label. b) Check transmittance of water in your cuvette is 100% c) Empty water out, pour your diluted dye solution into the cuvette, find transmittance. Record. d) Empty dilution 1, rinse and repeat. Determine the relationship between transmittance and molarity

1) Graph transmittance as a decimal on the y-axis vs. [Dye] on x axis

2) 1/T (y-axis) vs. [dye] on x-axis

3) 1 x 10T vs. [Dye]

4) logT vs. [Dye]

5) - logT vs.[Dye] Per. 3 T vs. [Dye] 1/T vs. [Dye] 1x10T vs. [Dye] -LogT vs. Concentration microM The mathematical routine that results in a line that goes through zero and shows the linear relationship (with a positive slope) between transmittance and concentration. This is our Calibration line that can be used to obtain the concentration of the same absorbing species in a solution of unknown concentration. Now use the Calibration line to find the concentration of Blue Dye #1 in the Sports Drink.

-logT = slope [dye]

-log (measured) = slope from graph [dye] Lab Report due Nov 17th. Create your own google document - MLA formatted paper. Date (1pt) Title (1pt) Beginning Questions (2pts) Safety (2pts) Procedure - paragraph (4pts) Data/Observations (use average values) (4pts)

Dilution Ratio Molar Average Measured concentration Measured Transmittance (µM) Transmittance as a decimal

10ml/0mL 6.8

0mL/10mL 0 100% 1.00

Transmittance for Sport’s Drink = ………... 11. At 25℃, the following heats of reaction are known.

2C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) ΔH = - 2,600.0 kJ/mol

C(graphite) + O2(g) → CO2(g) ΔH = - 394kJ/mol

2H2(g) + O2(g) → 2H2O(l) ΔH = - 574kJ/mol

At the same temperature, calculate ΔH for the reaction

C(graphite) + H2(g) → C2H2(g)

A. -2422 kJ B. -225 kJ C. 225 kJ D. 2422 kJ 11.

2C(graphite) + 2O2(g) → 2CO2(g) ΔH = 2( - 394kJ) and

H2(g) + ½ O2(g) → H2O(l) ΔH = ½ (- 574kJ) then

2CO2(g) + H2O(l)→C2H2(g) + 5/2 O2(g) ΔH = ½ (2,600.0 kJ)

Overall reaction

2C(graphite) + H2(g) → C2H2(g)

ΔH = -788 kJ) +(- 287kJ) + 1,300 kJ = 225 kJ C. 12 and 13. Aluminum reacts with hydrochloric acid according to the following equation: 0 2Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) ΔH rxn = -1052

kJ/molrxn 12. A sample of 27.0g of pure aluminum metal is added to 333 mL of 3.0 M hydrochloric acid. The volume of the hydrogen gas produced at standard temperature and pressure is a. 2.80 L

b. 5.60 L

c. 11.2 L

d. 22.4 L 12 and 13. Aluminum reacts with hydrochloric acid according to the following equation: 0 2Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g) ΔH rxn = -1052

kJ/molrxn 12. A sample of 27.0g of pure aluminum metal is added to 333 mL of 3.0 M hydrochloric acid. The volume of the hydrogen gas produced at standard temperature and pressure is a. 2.80 L

b. 5.60 L

c. 11.2 L

d. 22.4 L 12. 27 g x 1mol Al/27g = 1 mol Al 0.333L x 3.0 mol/L = 1 mol HCl so from the equation I can tell this is the limiting reagent (need 6 moles to react with 2 moles of Al, or 3 moles to react with 1 mol of Al) From equation 6 moles HCl produce 3 moles of hydrogen gas 1mole HCl produce 3/6 moles hydrogen gas = 0.5 mole 1 mole of gas at STP occupies 22.4 L So 0.5 mole occupies 22.4L x 0.5 = 11.2 L Answer C 13. What is the approximate density of the hydrogen gas produced at STP? a. 0.1 g/L b. 0.2 g/L c. 0.3 g/L d. 0.4 g/L 13. What is the approximate density of the hydrogen gas produced at STP? a. 0.1 g/L Density = mass/ volume = b. 0.2 g/L mass half a mole/ volume = c. 0.3 g/L 1g/11.2L = approx. 0.1 g/L d. 0.4 g/L 14. Which of the following reactions is not thermodynamically favored at low temperatures, but becomes favored as temperature increases?

o o Reaction ΔH (kJ/molrxn) ΔS (kJ/molrxn)

A. CO(g) + 2H2(g) → CH3OH(l) - 566 -173

B. 2H2O(g) → 2H2(g) + O2(g) 484 90.0

C. 2N2O(g) → 2N2(g) + O2(g) -164 149 2+ - D. PbCl2(s) → Pb (aq) + 2Cl (aq) 23.4 -12.5 14. Which of the following reactions is not thermodynamically favored at low temperatures, but becomes favored as temperature increases? ΔG = ΔH - TΔS

o o Reaction ΔH (kJ/molrxn) ΔS (kJ/molrxn) A. Always -ΔG - 566 -173

B. 2H2O(g) → 2H2(g) + O2(g) 484 90.0

C. 2N2O(g) → 2N2(g) + O2(g) -164 149 2+ - D. PbCl2(s) → Pb (aq) + 2Cl (aq) 23.4 -12.5 Never - ΔG 14. Which of the following reactions is not thermodynamically favored at low temperatures, but becomes favored as temperature increases? ΔG = ΔH - TΔS

o o Reaction ΔH (kJ/molrxn) ΔS (kJ/molrxn) A. Always -ΔG - 566 -173

B. 2H2O(g) → 2H2(g) + O2(g) 484 90.0

C. -ΔG at low T -164 149

2+ - D. PbCl2(s) → Pb (aq) + 2Cl (aq) 23.4 -12.5 Never - ΔG 15.Given the following data: 0 0 CO2(g) ΔH f = -393.5 kJ/mol H2O(g) ΔH f = -285.8 kJ/mol 0 C4H10(g) ΔH f = -124.7 kJ/mol

0 Calculate the ΔH rxn for the following reaction:

2 C4H10(g) + 13 O2(g) →8 CO2(g) + 10 H2(g)

a. - 6255.4. kJ b. - 5756.6 kJ c. -40. 6 kJ d. 539.4 kJ 15. Given the following data: 0 0 CO2(g) ΔH f = -393.5 kJ/mol H2O(g) ΔH f = -285.8 kJ/mol 0 C4H10(g) ΔH f = -124.7 kJ/mol

0 Calculate the ΔH rxn for the following reaction:

2 C4H10(g) + 13 O2(g) →8 CO2(g) + 10 H2(g)

o o o a. - 6255.4. kJ ΔH reaction = sumΔH f (products) - sumΔH f b. - 5756.6 kJ (reactants) c. -40. 6 kJ

d. 539.4 kJ See page 249 15. Given the following data: 0 0 CO2(g) ΔH f = -393.5 kJ/mol H2O(g) ΔH f = -285.8 kJ/mol 0 C4H10(g) ΔH f = -124.7 kJ/mol

0 Calculate the ΔH rxn for the following reaction:

2 C4H10(g) + 13 O2(g) →8 CO2(g) + 10 H2(g)

o o o a. - 6255.4. kJ ΔH reaction = sumΔH f (products) - sumΔH f b. - 5756.6 kJ (reactants) c. -40. 6 kJ

d. 539.4 kJ See page 249 16.

0 ΔH rxn = 2(201.kJ) + 2(-393.5 kJ) + 4 (-285.8 kJ) = Split up methanol form 2 CO2 form 4 water

Overall = -1528.2 kJ B. 17. Which of the following graphs describes a pathway of reaction that is endothermic? Looking for products at higher PE than reactants. 17. Which of the following graphs describes a pathway of reaction that is endothermic? Looking for products at higher PE than reactants. 18. (seen before) - important general principles

The dissolution of an ionic solute in a polar solvent can be imagined as occurring in three steps, as shown in the figure above. In step 1, the separation between ions in the solute is greatly increased, just as will occur when the solute dissolves in the polar solvent. In step 2, the polar solvent is expanded to make spaces that the ions will occupy. In the last step, the ions are inserted into the spaces in the polar solvent. Which of o the following best describes the enthalpy change, ΔH , for each step? a. All three steps are exothermic b. All three steps are endothermic c. Steps 1 and 2 are exothermic, and the final step is endothermic. d. Steps 1 and 2 are endothermic, and the final step is exothermic. 18. (seen before) - important general principles The dissolution of an ionic solute in a polar solvent can be imagined as occurring in three steps, as shown in the figure above. In step 1, the separation between ions in the solute is greatly increased, just as will occur when the solute dissolves in the polar solvent. In step 2, the polar solvent is expanded to make spaces that the ions will occupy. In the last step, the ions are inserted into the spaces in the polar solvent. Which of the o following best describes the enthalpy change, ΔH , for each step? D) Steps 1 and 2 are endothermic, and the final step is exothermic. 19. Using Hess’s Law and the equations below find ΔHo at 25℃

C2H5OH(l) + 3 O2(g) → 2CO2(g) + 3H2O(l)

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) ΔH = W kJ/mol

2C(graphite) +3H2(g) + ½ O2(g) → C2HgOH(l) ΔH = X kJ/mol

C2H4(g) +H2O(l) → C2H5OH(l) ΔH = Y kJ/mol

A. W - Y B. X - 2Y C. X + 2W + Y D. 2X - W + Y 19. Using Hess’s Law and the equations below find ΔHo at 25℃

C2H5OH(l) + 3 O2(g) → 2CO2(g) + 3H2O(l)

C H (g) + 3 O (g) → 2 CO (g) + 2 H O(l) ΔH = W kJ/mol 2 4 2 First 2 2 2C(graphite) +3H2(g) + ½ O2(g) → C2HgOH(l) ΔH = X kJ/mol C2H5OH(l)→ C2H4(g) +H2O(l) ΔH = - Y C2H4(g) +H2O(l) → C2H5OH(l) ΔH = Y kJ/mol

Then C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)

A. W - Y ΔH = W kJ/mol B. X - 2Y Overall C H OH(l) + 3 O (g) → 2CO (g) + 3H O(l) C. X + 2W + Y 2 5 2 2 2 ΔH = W - Y D. 2X - W + Y 19. Using Hess’s Law and the equations below find ΔHo at 25℃

C2H5OH(l) + 3 O2(g) → 2CO2(g) + 3H2O(l)

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) ΔH = W kJ/mol

2C(graphite) +3H2(g) + ½ O2(g) → C2HgOH(l) ΔH = X kJ/mol

C2H4(g) +H2O(l) → C2H5OH(l) ΔH = Y kJ/mol

A. W - Y B. X - 2Y C. X + 2W + Y D. 2X - W + Y 20. Seen before

10 pairs ions 5 sets 5 sets ions in double VOL

20. Answer D is best choice.

10 pairs ions 5 sets 5 sets ions in double VOL 2M 1M 0.5 M

20. Answer D is best choice.

10 pairs ions 5 sets 5 sets ions in double VOL 2M 1M 0.5 M

21. Solutions A and B are both clear and colorless. When Solution A is mixed with solution B, the temperature of the mixture increases and a yellow precipitate is observed. What can be concluded from these observations? a. The reaction is thermodynamically favored (spontaneous) at all temperatures. b. The reaction is thermodynamically favored (spontaneous) only at high temperatures. c. The reaction is thermodynamically favored (spontaneous) only at low temperatures. d. The reaction is not thermodynamically favored (spontaneous) at any temperatures. 21. Solutions A and B are both clear and colorless. When Solution A is mixed with solution B, the temperature of the mixture increases and a yellow precipitate is observed. What can be concluded from these observations? a. The reaction is thermodynamically favored (spontaneous) at all temperatures. b. The reaction is thermodynamically favored (spontaneous) only at high temperatures. c. The reaction is thermodynamically favored (spontaneous) only at low temperatures. d. The reaction is not thermodynamically favored (spontaneous) at any temperatures. ΔG = ΔH - TΔS - ΔH (exothermic) BUT ΔS negative (more order in a solid than a liquid) 21. Solutions A and B are both clear and colorless. When Solution A is mixed with solution B, the temperature of the mixture increases and a yellow precipitate is observed. What can be concluded from these observations? a. The reaction is thermodynamically favored (spontaneous) at all temperatures. b. The reaction is thermodynamically favored (spontaneous) only at high temperatures. c. The reaction is thermodynamically favored (spontaneous) only at low temperatures. d. The reaction is not thermodynamically favored (spontaneous) at any temperatures. ΔG = ΔH - TΔS - ΔH (exothermic) BUT ΔS negative (more order in a solid than a liquid) 22. A student mixes dilute AgNO3(aq) with excess NaCl(aq) to form AgCl(s), as represented by the net ionic equation above. Which of the diagrams below best represents the ions that are present in significant concentrations in the solution? (Ksp for AgCl is 1.8 x 10-10).

AgCl will precipitate out; (very small Ksp shown) All Ag+ ions will be one. EXCESS NaCl so will still be some chloride ions and sodium ions in final beaker. Nitrate ions are spectators - so want some of them also.

23. A 100-mL sample of water is placed in a coffee-cup calorimeter. Solid NaCl is then dissolved in the water. The temperature of the water decreases from 20.5 ℃ to 19.7℃ and is then allowed to return to room temperature (20.5 ℃ ). Determine the signs for ΔH, and ΔS for the process of dissolving NaCl and ΔG for the entire process at constant temperature. ΔH positive (endothermic process) ΔS positive (increasing disorder as solid dissolves and dissociates into more particles (ions) ΔG -ve at this temp. (process was spontaneous) 23. A 100-mL sample of water is placed in a coffee-cup calorimeter. Solid NaCl is then dissolved in the water. The temperature of the water decreases from 20.5 ℃ to 19.7℃ and is then allowed to return to room temperature (20.5 ℃ ). Determine the signs for ΔH, and ΔS for the process of dissolving NaCl and ΔG for the entire process at constant temperature. ΔH positive (endothermic process) ΔS positive (increasing disorder as solid dissolves and dissociates into more particles (ions) ΔG -ve at this temp. (process was spontaneous) Ans. C 24. Which reaction would have the most positive ΔSo?

a. CO(g) + 2H2(g) → CH3OH(l)

b. 2CH3OH(g) + 3O2(g) → 2CO2 (g)+ 4H2O(g)

c. HCl(g) + NH3(g) → NH4Cl(s)

d. Ba(OH)28H2O(s) + 2NH4NO3(s) →

Ba(NOs)2(s) +2 NH3(g ) + 10 H2O(l)

Looking for more motion/particles in products than reactants. 24. Which reaction would have the most positive ΔSo?

a. CO(g) + 2H2(g) → CH3OH(l)

b. 2CH3OH(g) + 3O2(g) → 2CO2 (g)+ 4H2O(g) 5 to 6

c. HCl(g) + NH3(g) → NH4Cl(s)

d. Ba(OH)28H2O(s) + 2NH4NO3(s) →

Ba(NOs)2(s) +2 NH3(g ) + 10 H2O(l) 0 to 2

Looking for more motion/particles in products than reactants. 25. Consider the freezing of liquid water at - 10℃ and 1 atm. For this process, what are the signs for ΔH, ΔS, and ΔG?

Liquid to solid disorder to more order ΔS is negative

Freezing is exothermic (see slide) 15 ΔH is negative

At -10oC freezing would be spontaneous ΔG is negative

Answer: 25. Consider the freezing of liquid water at - 10℃ and 1 atm. For this process, what are the signs for ΔH, ΔS, and ΔG?

Liquid to solid disorder to more order ΔS is negative

Freezing is exothermic (see slide) 15 ΔH is negative

At -10oC freezing would be spontaneous ΔG is negative

Answer: D Qu. 26 - 30

o K(s) + ½ Cl2(g) → KCl(s) ΔH = -437kJ/molrxn The elements K and Cl react directly to form the compound KCl according to the equation above. Refer to the information above and the table below to answer the questions that follow.

26. How much heat is released or absorbed when 0.050 mol of Cl2(g) is formed from KCl(s)? a. 87.4 kJ is released b. 43.7 kJ is released c. 43.7 kJ is absorbed d. 87.4 kJ is absorbed

Qu. 26 - 30

o K(s) + ½ Cl2(g) → KCl(s) ΔH = -437kJ/molrxn The elements K and Cl react directly to form the compound KCl according to the equation above. Refer to the information above and the table below to answer the questions that follow.

26. How much heat is released or absorbed when 0.050 mol of Cl2(g) is formed from KCl(s)? a. 87.4 kJ is released b. 43.7 kJ is released c. 43.7 kJ is absorbed d. 87.4 kJ is absorbed

What remains in the reaction vessel after equal masses 27. of K(s) and Cl2(g) have reacted until either one or both of the reactants have been completely consumed? a. KCl only b. KCl and K only

c. KCl and Cl2 only

d. KCl, K, and Cl2 only K at mass 40g/mol 1: 0.5 in equation Chlorine = 2 x 35.5 = 71 g/mol need 35.5g for 1 molK If use 40g of K and 40 g of chlorine, will have some chlorine left over. What remains in the reaction vessel after equal masses 27. of K(s) and Cl2(g) have reacted until either one or both of the reactants have been completely consumed? a. KCl only b. KCl and K only

c. KCl and Cl2 only

d. KCl, K, and Cl2 only K at mass 40g/mol 1: 0.5 in equation Chlorine = 2 x 35.5 = 71 g/mol need 35.5g for 1 molK If use 40g of K and 40 g of chlorine, will have some chlorine left over as well as the product KCl. o Which of the values of ΔH for a process in the table is 28. (are) less than zero (i.e. indicate(s) an exothermic process)? Looking for forming bonds/attractions not breaking them. a. z only b. y and z only c. x, y, and z only d. w, x, y, and z o Which of the values of ΔH for a process in the table is 28. (are) less than zero (i.e. indicate(s) an exothermic process)? Looking for forming bonds/attractions not breaking them. a. z only b. y and z only c. x, y, and z only d. w, x, y, and z It is observed that the reaction producing KCl from its 29. elements goes essentially to completion. Which of the following is a true statement about the thermodynamic favorability of the reaction? a)The reaction is favorable and driven by an enthalpy change only. b)The reaction is unfavorable and driven by an entropy change only. c)The reaction is favorable and driven by both enthalpy and entropy changes. d) The reaction is unfavorable due to both enthalpy and entropy changes. It is observed that the reaction producing KCl from its 29. elements goes essentially to completion. Which of the following is a true statement about the thermodynamic favorability of the reaction? a)The reaction is favorable and driven by an enthalpy change only. (entropy is unfavorable) b)The reaction is unfavorable and driven by an entropy change only. c)The reaction is favorable and driven by both enthalpy and entropy changes. d) The reaction is unfavorable due to both enthalpy and entropy changes. Cl (g) + 2 e- → 2 Cl- (g) 30. 2

o Which of the following expressions is equivalent to ΔH for the reaction represented above? a. x + y b. (x/2) - y c. x + 2y d. x - y

Cl2(g) → 2 Cl (g) x 2Cl(g) + 2 e- → 2 Cl- (g) 2y - - Overall: Cl2(g) + 2 e → 2 Cl (g) x + 2y Cl (g) + 2 e- → 2 Cl- (g) 30. 2

o Which of the following expressions is equivalent to ΔH for the reaction represented above? a. x + y b. (x/2) - y c. x + 2y d. x - y

Cl2(g) → 2 Cl (g) x 2Cl(g) + 2 e- → 2 Cl- (g) 2y - - Overall: Cl2(g) + 2 e → 2 Cl (g) x + 2y Homework Read and make your own notes on the following:

10.1 Intermolecular Forces pages 426 - 429

Dipole-Dipole

Hydrogen bonding

London Dispersion forces (or just London forces, or just dispersion forces)

Changes of state p.463 - 66

Vapor pressure 459-63

Phase diagrams 467- 71 including info in Chemical Impact section.