Ch6 lec 3
Revision as of 20:15, 14 April 2020 by Bes (talk | contribs) (→Ch 6, Sec 6.4 and 6.5...additional problem)
(4/14/20, bes)
Greetings Gen Chem folks!
- This week we are going to continue our discussions of thermochemistry (Ch 6). In previous lecture notes we have covered the critical content associated with Sections 6.1-6.5. There is a lot more in these section then i have discussed, so focus on these lecture notes and not so much on the extended textbook material. As noted, all chemistry and biochemistry majors will be required to take Chem 312: Thermodynamics in their junior and senior year where we will get into more of the details.
Review of Section 6.4: Enthalpy
- - Nearly all chemical reactions generate (exothermic) or consume (endothermic) energy/heat.
- - We tabulate the amount of heat involved in an rxn using the enthalpy (ΔH) which most often has units of J or kJ per mol, kJ/mol of something.
- Remember, some of this can be confusing because the ΔH units can be displayed as just kJ or kJ/rxn <--as written. Depending on the stoichiometric coefficients, this is per X mole of reactant or product.
- - The types of problems that are asked involve:
- Q1: Given the ΔH for a reaction, determine how much heat is generated (implying exothermic reaction) when X moles of a reactant is converted to products.
- Q:2 Given the ΔH for a reaction, determine how much heat is generated (implying exothermic reaction) when Y grams of a reactant is converted to products.
- Q3: Given the amount of heat from a rxn and the amount (grams or moles) of a reactant or product, calculate the ΔH for the reaction.
See WA Ch6.1...
Review of Section 6.5: Calorimetry
- - When an aqueous chemical reaction occurs, the heat (q) generated or consumed goes into the water or comes out of the water.
- If you add heat to water the temperature of the water increases (+ΔT),
- If you remove heat from water the temperature of the water decreases (-ΔT).
- - The equation used to relate the heat to ΔT is:
- q = mass of water * spec heat * ΔT
- - where the spec heat for water is 4.184 J/(goC).
- q = mass of water * spec heat * ΔT
- - Unfortunately, we introduced "calorimetry" in the first lab period...where we introduced heat (q), ΔH, specific heat, ΔT...all at once...sorry. Now that you know a little bit more about how to calculate the heat involved, we can now understand better how this heat generated or consumed can result in water heating up or cooling down.
The types of calorimetry problems that are asked involve:
- Q1: Given the ΔH for a reaction and/or how much heat is generated when X moles/grams of a reactant is converted to products, what is the change in temperature of water in which the rxn occurred?
Ch 6, Sec 6.4 and 6.5...additional problem
- (see Chemistry in Action, page 250)
Now this enthalpy stuff sounds a little confusing, but you are much more familiar with this science topic then you are currently aware. I am making reference to the "calorie" content of food.
So what kind of foods do you eat and how many calories do you get from these food items?
- - first off, we eat a lot of different foods, but nearly all foods are made up of the following:
- carbohydrates (i.e. sugars like glucose)
- fats (i.e. fatty acids, triglycerides)
- protein (i.e. alanine)
- Glucose (a simple carbohydrate) --> C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (g)
- ΔHrxn = -2,801 kJ/mol of glucose, molecular weight = 180.16 g/mol, therefore -15.5 kJ/gram.
- Stearic Acid (a fat) --> C18H36O2 (s) + 24 O2 (g) --> 18 CO2 + 18 H2O (g)
- ΔHrxn = -11,407 kJ/mol of stearic acid, molecular weight = 284.48 g/mol, therefore -40.1 kJ/gram.
- Alanine (a protein) --> These are a bit more complicated since amino acids contain the element nitrogen...
- 4 C3H7NO2(s) + 15 O2 (g) --> 12 CO2 (g) + 14 H2O (g) + 2 N2 (g)
- ΔHrxn = -1,577 kJ/mol of alanine,molecular weight = 89.09 g/mol, therefore -17.7 kJ/gram.
- 4 C3H7NO2(s) + 15 O2 (g) --> 12 CO2 (g) + 14 H2O (g) + 2 N2 (g)
| Type | Example | kJ/mol | kJ/g | Ave value Calories/g |
| Carbohydrate | Glucose | -2,801 | -15.5 | 4 |
| Fat | Stearic Acid | -11,407 | -40.1 | 9 |
| Protein | Alanine | -1,577 | -17.7 | 4 |
- Note that the negative sign is not shown; it is assumed that food gives off heat/energy and so the sign is dropped.
As you can see in the table above, there is a column with average Calorie per gram. How did we calculate this...unit conversion of 1 Calorie = 4.184 kJ:
- (Note: there is also a "small c" calorie; 1 calorie = 4.184 J, all food is measured in the "big C" Calorie.)
- - for glucose --> -15.5 kJ/g * 1 Calorie/4.186 kJ = 3.7 Calories/gram ~ 4 Calories/gram
- - for stearic acid --> -40.1 kJ/g * 1 Calorie/4.184 kJ = 9.6 Calories/gram ~ 9 Calories/gram
- - for alanine --> -17.7 kJ/gram * 1 Calorie/4.184 kJ = 4.2 Calories/gram ~4 Calories/gram
Example 1 - Carbohydrates
- 1) I have a bag of granulated sugar for cooking.
- 2) the packaging says, 30 Calories/serving
- 3) the packaging says, 8 grams per serving
- 4) Therefore, 30 Calories/8 grams = 3.8 Calories/gram' ~ 4 Calories/gram
- Conclusion: this is nearly a pure carbohydrates, it makes sense that the calorie content is ~4 Cal/g.
Example 2 - Fatty Food
- 1) In my kitchen i have canola oil for cooking.
- 2) the packaging says, 120 Calories/serving
- 3) the packaging says, 14 grams per serving
- 4) Therefore, 120 Calories/14 grams = 8.6 Calories/gram ~ 9 Calories/gram.
- Conclusion: this is nearly a pure fat, it make sense that the calorie content is ~ 9 Cal/g.
Example 3 - Protein
- 1) Someone left "Nitro-tech Whey Gold" protein supplement in my kitchen.
- 2) the packaging says, 120 Calories/serving
- 3) the packaging says, 33 grams per serving
- 4) Therefore, 120 Calories/33 grams = 3.6 Calories/gram' ~ 4 Calories/gram.
- Conclusion: As noted above, protein is a bit more confusing due to nitrogen in the chemical formula, but the calorie content is ~ 4 Cal/g.
Example 4 - Bark Thins with Almonds
- 1) I have some "Bark Thins with Almonds" (chocolate i got for Easter).
- 2) the packaging says - 220 Calories/serving
- 3) the packaging says - 40 grams per serving
- 4) Therefore, 220 Calories/40 grams = 5.5 Calories/gram
- Conclusion: the packaging states 13 g total fat, 22 g total carbohydrates, and 4 g protein:
- 13 g * 9 Cal/g = 117 Cal
- 22 g * 4 Cal/g = 88 Cal
- 4 g * 4 Cal/g = 16 Cal
- ----> 221 Cal total with 117/221 * 100 = 52.9% calories from fat, 88/221 *100 = 39.8 calories from carbohydrates, and 4/221 * 100 = 1.8 % calories from protein.
Now go find a three food item and record/calculate the following:
- 1) What is the food?
- 2) How many Calories (notice the "big C" on the label) are in this food item per serving?
- 3) How many grams per serving?
- 4) Calculate the Calories per gram...and % calories/fat, % calories/carbohydrate, and % calories/protein,
Email me the results...by thurs, April 16, 5 pm.