Energy!

 Energy: the capacity to do work or supplying heat Energy is detected only by its effects Energy can be stored within molecules etc as chemical potential  energy We will most often deal with energy changes with heat (q) For systems at constant pressure (no PV work) then  q = H (enthalpy)

Measuring Heat (q):units: Joule, calorie, Calorie (1 calorie = 4.184 J    1 Calorie = 1000 calories)Use these equivalencies to convert between heat units:

An 80 Cal apple has how many J? kJ?

80 Cal 1000 cal 4.184 J 1 X 1 Cal X 1 cal = 334720 J

334720 J 1 kJ 1 X 1000 J = 334.720 kJ

Heat is very often involved in chemical reactionsThere are two possibilities of how heat is involved: 1) heat is absorbed by the reaction ENDOTHERMIC2) heat is given off by the reaction EXOTHERMICHow much heat is given off or absorbed is measured with enthalpy (H)We are more interested in the change in enthalpy therefore HH = Hproducts - Hreactants

Exothermic reactions:combustion reactions are obvious examples of exothermic reactionsC6H12O6 + 6O2   6CO2 + 6H2O + 2803 kJ Heat is given off Heat is a product in a chemical reaction H of the products must be less than the H of the reactants Therefore H must be negative For this reaction H = - 2803 kJ We can also make a Enthalpy diagram for this reaction

Reactants

    H = -2803 kJ

 

Products

 

Reaction progress

Endothermic reactions:- Br2 + Cl2 + 29.4 kJ     2 BrCl Heat is absorbed by the reaction Heat is a reactant in a chemical reaction H of the products is more than the H of the reactants Therefore H must be positive For this reaction H = 29.4 kJ This reactions enthalpy diagram looks like:

        Products   H H = 29.4 kJ  reactants    reaction progress

Now we can use this information in stoichiometry type problems (aren’t you happy!)The heat term can be treated just like balanced coefficient on a substance in a mole ratio.

Ex: When 12.5 g of glucose (C6H12O6) is burned how many kJ of energy are released?C6H12O6 + 6O2   6CO2 + 6H2O + 2803 kJ

X X

Ex: When only 50.0 kJ of energy is available for the reaction of bromine and chlorine to form monobromine monochloride how many grams of the product are formed? Br2 + Cl2 + 29.4 kJ     2 BrCl

On Day 1 we looked at thermochemical equations like the one below:

(REVIEW For the equation below: the reaction is ____ thermic, the heat _______ (produced or absorbed) was ________, and the H of the reaction is ______)

C6H12O6 + 6O2   6CO2 + 6H2O + 2803 kJ

Some of you inquisitively wondered where I got the number for the heat term. The heat term can be found several ways. You could do the reaction in a calorimeter and determine the heat given off from the temperature rise of the water (we’ll learn that on day 3). Another way to find the heat term without the laboratory is to use Hess’s Law.

Hess’s Law:- Hess’s law states that if you add two or more equations to get a third equation (like adding equations in algebra) you can add the (s) of the original equations to find the H of the final equation- This is a great way to find the of an unknown reaction- To do this we need to know two things:

o If you reverse a reaction you change the sign of

o If you multiply or divide an equation by a whole number you do the same to

the EX: Find the of the following reaction: 2Al + Fe2O3   2Fe + Al2O3

 Given: 4Al + 3O2   2Al2O3   -3352.0 kJ

 

2Fe + 1 1/2O2     Fe2O3  = -822.2 kJ

822.2 kJ

-853.8 kJ

-1676.0 kJ

- We have found most of the fo (heat of formation) of many, many compounds (pg. 316)- We can use these to calculate the of most of our reactions using a more useful form of Hess's Law:

rxn = n(fo products) - m(fo reactants)

- 

 I know this looks bad but in use it is really quite elegant and usefulEX: What is the heat of the reaction for the decomposition of hydrogen peroxide?

- We have found most of the fo (heat of formation) of many, many compounds (pg. 316)- We can use these to calculate the of most of our reactions using a more useful form of Hess's Law:   rxn = n(fo products) - m(fo reactants)

- 

  I know this looks bad but in use it is really quite elegant and usefulEX: What is the heat of the reaction for the decomposition of hydrogen peroxide?

Summation(add up)

coefficients in front of the substance in the balanced equation

look up on a Hf table (316)

O2(g)

You try the next one!!!!

Ex: 15.0 kJ of heat is applied to 10.0 g of calcium carbonate. Heat causes calcium carbonate to decompose to calcium oxide and carbon dioxide gas. How many L of carbon dioxide gas at STP will be formed?

Balance! CaCO3  CaO + CO2

Look up the Hf: -1207.0 --> -635.1  +  -393.5 (all in kJ/mol)

Do Hess' Law: Hrxn = n(Hfo products) - m(Hfo reactants)

  = [1(-635.1) + 1(-393.5)] - [1(-1207.0)]  = (-1029.0) - (-1207.0)  = +178.0 kJ/mol

Thermochemical equation is: CaCO3 + 178kJ CaO + CO2

Do the stoichiometry:

15.0 kJ  1 mol CO2  22.4L of CO2 gas 1   178 kJ  1 mol CO2 gas

10.0 g CaCO3 1 mol CaCO3 1 mol CO2 22.4 L of CO2 gas 1 100 g CaCO3 1 mol CaCO3 1 mol CO2

x xx

xx =

=

1.89 L

2.24 L

Answer is 1.89 L

Heat with no change of state:

- Heat Capacity

- Specific heat (C) q = m C T  (T = Tf - Ti)

- Calorimetry  o EX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C

o EX: 10.0 g of an unknown metal is heated to 100.0o C. The hot metal is added to 50 mL of water at 25.0o C. The final temperature of the solution is 35.6o C. What is the specific heat capacity of the unknown metal?

Heat with no change of state:

- Heat Capacity the amount of heat needed to increase the temperature 1 degree celcius - Specific heat (C) the amount of heat needed to raise 1 g of a substance 1 degree  q = m C T   (T = Tf - Ti) see chart pg. 296, therefore the C of water = 4.184 J

- Calorimetry measurement of heat change  o EX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C

q = m C T

q = (875)(4.184)(50)

q = 183050 J

o EX: 10.0 g of an unknown metal is heated to 100.0o C. The hot metal is added to 50 mL of water at 25.0o C. The final temperature of the solution is 35.6o C. What is the specific heat capacity of the unknown metal?

Heat with no change of state:

- Heat Capacity the amount of heat needed to increase the temperature 1 degree celcius

- Specific heat (C) the amount of heat needed to raise 1 g of a substance 1 degree  

q = m C T  (T = Tf - Ti) see chart pg. 296, therefore the C of water = 4.184 J

- Calorimetry measurement of heat change  o EX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C

q = m C T

q = (875)(4.184)(50)

q = 183050 J

o EX: 10.0 g of an unknown metal is heated to 100.0o C. The hot metal is added to 50 mL of water at 25.0o C. The final temperature of the solution is 35.6o C. What is the specific heat capacity of the unknown metal?

heat lost by metal = heat gained by water

-(m C T) = m C T

-(10)(x)(-64.4) = (50)(4.184)(10.6)

x = 3.44 J/ g(C)

q= Hfus (mol)

q= Hvap (mol)

Things to know: (where might one write them?)Hfus = 6.01 kJ/molHvap = 40.7 kJ/molCice = 2.1 J/g . oCCwater = 4.184 J/g . oCCsteam = 1.7 J/g . oC

q = m C T

q= Hfus (mol) q= Hvap (mol)

From -25 to 0:q = m C Tq= (10)(2.1)(25)q= 525 J

Melt the ice:q= Hfus (mol)q= (6.01)(.556)q= 3.34 kJ = 3340 J

Heat the water to boiling:q = m C Tq= (10)(4.184)(100)q= 4184 J

Boil the water:q= Hvap (mol)q= (40.7)(.556)q= 22.63 kJ = 22630 J

Heat the steam to 130:q = m C Tq= (10)(1.7)(30)q= 510 J

Add up the total heat:Total q= 525+3340+4184+22630+510 = 31189 J = 31.2 kJ

EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?

From -25 to 0:q = m C Tq= (10)(2.1)(25)q= 525 J

Melt the ice:q= Hfus (mol)q= (6.01)(.556)q= 3.34 kJ = 3340 J

Heat the water to boiling:q = m C Tq= (10)(4.184)(100)q= 4184 J

Boil the water:q= Hvap (mol)q= (40.7)(.556)q= 22.63 kJ = 22630 J

Heat the steam to 130:q = m C Tq= (10)(1.7)(30)q= 510 J

Add up the total heat:Total q= 525+3340+4184+22630+510 = 31189 J = 31.2 kJ

EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?

from ice at -25 to water at 0o525 + 3340 = 3865 J

from ice at -25 to water at 100o525 + 3340 + 4184 = 8049 J

from ice at -25 to steam at 100o525 + 3340 + 4184 + 22630 = 30679 J