Heat in Chemical Reactions

10.1 Chemical Reactions that Involve Heat

1. Heat: EnergyEnergy (symbol - qq) that is transferred from one object to another due to a difference in temperaturetemperature. Measured in Joules (symbol - JJ)

2. Thermochemistry: The study of heat heat changeschanges in a chemical reaction.

3. Types of Chemical Reactions

a. Exothermic Reactions: releaserelease heat into their surroundings. HHeat is a productproduct of the reaction and temperature increasesincreases. This occurs during bond formationformation.

Combustion reactions are exothermic: burning propaneC3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ

Exothermic Reaction

surroundings

ΔT = +ΔH =

-

Exothermic Reaction

b. Endothermic Reactions: Heat is absorbedabsorbed by the reactants and stored in the chemical bondsbonds of the products. Heat acts as a reactantreactant and temperature decreasesdecreases. This occurs during bond breakingbreaking.

Electrolysis of water requires electricalelectrical energy.

2H2O + 572kJ → 2H2 + O2Endothermic Reaction

surroundings

ΔT= -ΔH=+

Exothermic/Endothermic

Reactions

10.2 Heat and Enthalpy Changes

1. Enthalpy: The heatheat content of a system at constant pressure (symbol is HH ).

2. Enthalpy Change: The heat absorbed or released during a reaction (symbol is ΔΔHH ).

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4. When reactions take place at standard temperature and pressure, qq = HH.

5. Standard Enthalpy Change (H): Enthalpy change that occurs when reactants in their standard states (mostmost stablestable formform) change to products in their standard states. STP Standard Temperature and Pressure are: 00 C and 11 atm.The H is listed after the equation. If the H is positive the reaction is endothermicendothermic and heat was absorbedabsorbed. If the H is negative, the reaction is exothermicexothermic and heat was releasedreleased.

6. The amount of heat absorbed or released in a reaction depends upon the number of molesmoles of reactants.

7. Enthalpy Changes in Stoichiometry Problems:

Ex) How much heat will be released if 5.0 g of H2O2 decomposes?

2H2O2 → 2H2O + O2 ΔH = -190 kJ

2 22 2

2 2 2 2

1 mol H O -190 kJ5.0 g of H O = -14 kJ or 14 kJ released

34 g of H O 2 mol H O

Ex) How much heat is transferred when you eat a 10. g Jolly rancher which is made of glucose (C6H12O6)? It reacts in your body with oxygen according to the following equation. If 4.184kJ = 1 Cal, how many Calories are in the Jolly Rancher?

C6H12O6 + 6O2 → 6CO2 + 6H2O ΔH = -2803 kJ

6 12 66 12 6

6 12 6 6 12 6

1 mol C H O -2803 kJ 1 Cal10. g of C H O = -160 kJ = 38 Cal

180 g of C H O 1 mol C H O 4.184 kJ

C6H12O6 + 6O2 → 6CO2 + 6H2O + 2803 kJ

3. Enthalpy Diagrams: #1 #2

#1 #2

a. Which has a higher enthalpy? Products or Reactants R P

b. Was heat absorbed or released? R A

c. Is this an endothermic or exothermic reaction?

ExoEnd

o

d. Is ΔH for this reaction positive or negative?

- +

e. Would the ΔH be on the left or right side of the yield sign? R L

f. Is the reverse reaction exothermic or endothermic?

Endo

Exo

g. Rewrite each equation with the heat term

in the reaction as a reactant or product:

#1) C3H8 + 5O2 → 3CO2 + 4H2O + 2043 kJ

#2) C + H2O + 113kJ → CO + H2

= “change in” H = Hproducts ─ Hreactants

H Hproducts Hreactants

Exothermic Reaction - low

high

Endothermic Reaction + hig

hlow

10.3 Hess’s Law - (1802-1850)

1. The enthalpy change for a reaction is the sumsum of the enthalpy changes for a series of reactions that addsadds upup to the overall reaction.

2. This is also called the Law of Heat of Law of Heat of Summation (Summation (ΣΣ))

3. This allows you to determine the enthalpy change for a reaction by indirectindirect means when a direct method cannot be done.

4. Steps for using Hess’s Law

1. Identify the compounds2. Locate the compounds on the Heat of Reaction

Table.3. Write the reaction from the table so the

compound is a reactant or product.4. Write appropriate ΔH for each ‘sub equation.”

• If needed, multiply equation and enthalpy change value. (coefficients)

• If you reverse the equation, change the sign of the enthalpy change.

5. Add the equations to arrive at the desired net (original) equation.

6. Add ΔH (enthalpy changes) of each “sub equation.”

10.3 Calorimetry:1. The Kinetic Theory states that heat

results from the motionmotion & vibrationvibration of particles.

2. Heat: The transfer of kinetickinetic energy from a hotterhotter object to a coldercolder object. Heat is dependentdependent on composition and amountamount.

3. Temperature is a measure of how hot or cold something is; specifically it is the measure of the average kinetic energy average kinetic energy (speed)(speed) of the particles in an object. It is independentindependent of amount.

Heat• Temperature is not the same as heat.

• Temperature is a measure of the average kinetic energy of the particles in an object.

• A temperature change is a result of a energy transfer.

• Julius Sumner Miller - Physics - Heat & Temperature

• Heat vs. Temperature Animation

4. Calorimetry is the study of heatheat flowflow and measurementmeasurement.

5. Calorimetry experiments determine the heats of reactions (enthalpy changes) by making accurate measurements of temperature changes produced in a calorimetercalorimeter.

6. A Calorimeter is an insulatedinsulated device used to measure heat absorbed or released in a chemical or physical change.

7. Specific Heat (Cp): The amount of heat needed to raise 1 g1 g of a substance by 1C.

Formula for specific heat:

m=mass (substance) T=change in temperature of the substance

(Tf─Ti)

Specific Heat of Water = 4.184 J/g ºC = 1 calorie or .001food

Calorie

p

qC =

mΔT

8. Measuring Specific Heat of a Metal:Ex #1) What is the specific heat of a nickel if

the temperature of a 32.2 g sample of nickel is increased by 3.5ºC when 50. J of heat is added.

p

qC =

mΔT +

+50.J

32.2g 3.5 C

0.44 J/g C

Ex #2) How much heat is absorbed to be able to increase the temperature of a 26.2 g sample of aluminum (Cp = 0.897 J/gºC) from 25.3 ºC to 65.9 ºC?

pq = mC ΔT

o + oq = .897 J/g C 26.2g 40.6 C

q = +954 J

9. Measuring Heat (q) of a Substance Dissolved in Water:

You can rearrange this formula to determine the heat released or absorbed by the surroundings (solution) as the substance dissolves based on this assumption:

q reaction = -q surroundings

q = 1) Calculate qfor the surroundings (solution) and

determine qrxn.2) Calculate the moles of solute dissolved in the water.3) Calculate H =

heat (q)

mol

pmC ΔT ΔT = Tf - Ti

Ex) When a 4.25 g sample of solid NH4NO3 dissolves in 60.0 g of H2O in a calorimeter,the temperature drops from 21.0 ºC to 16.9 ºC. Calculate H. Rewrite the thermochemical equation with the heat term as a reactant or product.

NH4NO3(s) → NH4+

(aq) + NO3-(aq) H

= ?

1. qsur=mCpΔT = (64.3g)(4.184J/g°C)(-4.1°C)

qsur=-1100J

qrxn = -qsur = +1100J

2.

3.

4 34 3

4 3

1 mol NH NO4.25 g NH NO = 0.0531 mol

80.0 g NH NO

+q +1100JΔH = = = 21,000 J/mol = +21 kJ/mol

mol .0531 mol

+ 21 kJ

10. Foods as Fuels:A. Carbohydrates typically have highhigh enthalpies;

however, the products of their combustion, CO2 and H2O, have lowlow enthalpies.

B. Therefore, the combustion of carbohydrates and fats, is exothermicexothermic.

C. Sugars and Starches break down to glucoseglucose, which reacts with O2 in a combustion reaction.

D. Nutritional information on food labels can be gathered using a calorimeter.

Carbs = 4 Cal/g Protein = 4 Cal/g Fat = 9 Cal/g