• Some Demonstrations of Heats of Reactions! • Heats of Formation and Heats of Reactions: Hess's Law and Calculating Heat

Today

• Some demonstrations of heats of reactions!

• Heats of Formation and Heats of Reactions: Hess’s law and calculating heats of reactions from values you can look up

• DPV is small enthalpy correction– use enthalpy of bond and energy interchangeably--- • Extensive vs intensive – Your Questions/ Announcements - Where is the placement exam retake? 320 Havemeyer at 9 am Friday (tomorrow)

- When will students get the results back? Will it be before program change date? Friday is also the change date. You will receive the scores later on in the day on Friday.

- Can students who got an A (scored 13 and higher) still retake to try and get a little higher? No

- Is the previous score kept if the retake yields a lower grade? Yes

- Will the solutions to Homework 0 be posted? Yes Your Questions/ Announcements

- Are there points associated with OWL? Is it required? No - Are there points associated our answers (right vs wrong) on Poll Everywhere? Is it required? No

- Will we post scores for the rest of the placement exam Yes, but it does NOT contribute to your course grade.

- What is happening next week? We start Chapter 10. Homework for next week is slightly shortened compared to the distributed list. Read Ch 10 and do Problems 12-22. You will have quiz on chapter 9 in recitation. We’ll do all the remaining problems for Ch 10 the following week, catching up to where we were. - Meet your Prof? Many of you have signed up but many have not and there are plenty of slots available. No points, no grade consequences, but come meet us anyhow. - Suggest an experiment? You will have a couple of changes to propose an experiment (no points or bonus points). Right now the google sheet is open and editable for proposing where to measure lead. Heat Capacity: The heat required to change the temperature

• When reactions generate heat, we normally don’t measure the heat directly, rather we often measure the change in the temperature of the system, and infer from that how much heat was added or removed from the system. In the problems you are assigned you may be asked to calculate how much heat (or how much energy) is involved in changing the temperature of a substance.

• Section 9.4 Heat Capacity: The heat required to change the temperature

For many substances there is a linear relationship relating the heat required , q, to achieve a temperature change, DT. The proportionality constant is the heat capacity, C, q= C DT.

If an ideal gas heated at constant volume Cv=3/2Rs. If it is heated at constant pressure Cp= 5/2R.

We will treat the heat capacities of solids and liquids as empirical values, which are highly variable and more difficult to understand theoretically. See table 9.3 p 314. Question 14: calculations using heat capacity

The enthalpy of fusion of ice is 6.020 kJ/mol. The heat capacity of liquid water is 75.4 J/moloC. What is the smallest number of ice cubes at 0oC, each containing 1 mol of water, necessary to cool 500. g ( a pint) of liquid water initially at 23oC to below 1oC?

• a. 1 • b. 8 • c. 14 • d. 15 • e. 126 Question 14: answer

• The enthalpy of fusion of ice is 6.020 kJ/mol. The heat capacity of liquid water is 75.4 J/moloC. What is the smallest number of ice cubes at 0oC, each containing 1 mol of water, necessary to cool 500 g ( a pint) of liquid water initially at 23oC to below 1oC? n mol*6.020 kJ/mol = 75.4 J/moloC* 22oC*(500g/ 18 g/mol) (1kJ/1000J) = 46.1kJ n =7.7 • B. Note: units first, estimate next, calculate next, then consider the error/sig fig and likely systematic error. Heat Capacity

• The molar specific heat capacity C, is the amount of heat required to change the temperature of 1 mole of the substance by 1 K. q = n C DT, where n is the number of moles. • The total heat capacity for a specific object C’ depends on the number of moles n -- so C’ = nC. • In these equations we have used a molar specific heat capacity; it is an extensive property, and you need to multiply by the number of moles. • If a phase change occurs, heat is added or removed for the phase change, and is calculated in a separate term. What about water associated with the ice?

• If the final temperature is above 0C, there would be an additional heat term due to warming the water associated with the ice cubes. How would you add this to the equation? Results of 14.

• Each individual should report the minimum number of ice cubes. Question 15: Systematic errors in calorimetry

A) ΔHcalc is less positive because the reaction absorbs heat from the calorimeter.

B) ΔHcalc equals the actual value because the calorimeter does not absorb heat.

C) ΔHcalc is more negative because the calorimeter always absorbs heat from the reaction.

D) ΔHcalc is less negative because the calorimeter absorbs heat from the reaction.

E) ΔHcalc is more positive because the reaction absorbs heat from the calorimeter. • Melting the ice : is it endothermic? Yes. It cools something else (here to water) or pulls heat out of something else. • Is D H positive or negative? Positive. • If it cools the vessel as well it won’t cool the water as much. The total enthalpy , total heat at constant pressure is the sum of those two terms and is a defined value.

• The apparent DH will be less positive The Sign of the Enthalpy Change: Endothermic Conversely, if the reaction makes weaker bonds and breaks stronger bonds, DH is positive and the reaction is sometimes said to be “uphill” in potential energy. When the enthalpy change of a reaction is positive, heat is absorbed by the system from the surroundings during the reaction, and the reaction is referred to as endothermic. Question 15: answer

When a student performs an endothermic reaction in a calorimeter, how (if any) does the calculated value of ΔH differ from the actual value if the heat exchanged with the calorimeter is not taken into account? The endothermic reaction has a positive calculated DH . The system cools down. However it will apparently cool down less if the calorimeter can equilibrate bit with it (also cool down).

A) ΔHcalc is less positive because the reaction absorbs heat from the calorimeter.

B) ΔHcalc equals the actual value because the calorimeter does not absorb heat.

C) ΔHcalc is more negative because the calorimeter always absorbs heat from the reaction.

D) ΔHcalc is less negative because the calorimeter absorbs heat from the reaction.

E) ΔHcalc is more positive because the reaction absorbs heat from the calorimeter. Question 16: A calculation involving heat capacity and hot aluminum. A 26.9 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter containing 200 g of water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. What is the final temperature of the water?

a. 86.4oC b. 80.1oC c. 54oC d. 22.8oC e. 21.1oC Question 16: solution A 26.9 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter containing 200 g of water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. What is the final temperature of the water?

The heat absorbed by the ice cubes is equal and opposite to the heat released by the water; their sum is zero. 200*4.18 J/oC *(T - 21.1 oC) + 24 J/mol oC *1mol*(T -86.4oC) = 0

836 (T) + 24*T = 836*21 + 24*86 T=10842/442= 22.8 oC

D The overall sign for the term associated with water is positive– heat is added to the water. The overall sign of the term for the aluminum is negative, heat is removed from the aluminum. Question 17: Cold Packs, Ammonium Nitrate

• A. -4oC • B. 7.4°C • C. 21.4oC • D. 45.9oC • E. 54.4°C Formula Weight?

• Formula mass or formula weight of a molecule is the sum of the atomic weights of the atoms in its empirical formula. • The molecular mass or molecular weight of a molecule is calculated by adding together the atomic weights of the atoms in the molecular formula. • The molecular formula indicates the type and number of atoms in a molecule. The empirical formula (or simplest formula) is based on the mole ratio of elements present in a compound. • These two can differ in some cases. For example for glucose the molecular formula is C6H12O6 while its empirical formula is CH2O. For this problems they are identical however. Question 17: solution

A calorimeter contains 100 g of water at 22.5°C. A 20 g sample of NH4 NO3is added to the water in the calorimeter. After the solid has dissolved, what is the temperature of the water? The enthalpy of solution for dissolving ammonium nitrate is 25.41 kJ/mol. And the FW is 80.04 g. Assume that no heat is lost to the calorimeter or the surroundings and that the heat capacity is 4.18 J/g°C for solution (the same as that of pure water). 4.18 J/g oC*100g *(22.5-T oC) (1kJ/1000J)= 25.41 kJ/mol *(20g/80 g/mol) (22.5-T)= 6350/418 = 15.2 T = 7.3 oC • Someone asked, what about cooling the 20 g of ammonium nitrate? Excellent question. We don’t have enough information to add that correction term, so it was left out. Results of Experiment 17

• Each individual in the group should report the final temperature. Question 18: Hot Packs Magnesium Sulfate

A calorimeter contains 100 g of water at 22.5°C. A 20 g sample of MgSO4 added to the water in the calorimeter. After the solid has dissolved, what is the temperature of the water? The enthalpy of solution for dissolving ammonium nitrate is -80 kJ/mol. And the FW is 120.4 g/mol . Assume that no heat is lost to the calorimeter or the surroundings and that the heat capacity is 4.18 J/g°C for solution (the same as that of pure water).

A. 7.4°C B. 21.4oC C. 45.9oC D. 54.4°C E. 120oC • Here is where we ended. Try this problem on your own.