Unit: Thermochemistry Chapter 16 in text Heat, Temperature, Energy Conversions

After today you will be able to… Explain what thermochemistry is List different types of energy Explain what happens to particle motion when heat energy is added Convert between units of calories (cal), Joules (J) , and kilo-Joules (kJ)

Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state. Energy: The ability to do work or supply heat.

Different Forms of Energy Kinetic Potential Heat Solar Nuclear Wind Sound Magnetic

Heat: (q) a form of energy Flows from warmer objects to cooler objects If two objects come into contact with one another, heat will move from the warmer object to the cooler object until they are the same temperature. -At this point, equilibrium is reached.

When you add heat energy, molecules move faster; temperature increases. Temperature: a measure of heat, increase heat, increase temperature Chemical Potential Energy: stored energy in chemical bonds Depends on the type of atoms and arrangement

In a chemical reaction, either: PE is converted to heat (heat is given off – feels hot, exothermic) Heat is absorbed and converted to PE (feels cool, endothermic)

Law of Conservation of Energy: Energy can neither be created nor destroyed, only converted from one form to another. Units of Energy: Joule (J) calorie (cal): chemistry calorie! Calorie (Cal): food calorie - 1 food Cal = 1000 chemistry cal - 1 cal = 4.184J

Conversions between energy units: Example: How many calories are in 287J? K: 287 J U: ? cal 287 J 1 cal x = 68.6 cal 1 4.184 J

Conversions between energy units: Example: How many kJ are equal to 1478cal? K: 1478cal U: ? kJ 1478 cal 4.184 J 1 kJ x x = 6.184 kJ 1 1 cal 1000 J

Homework: Read chap 16 (pg 501-514) Questions? Complete WS 1 Homework: Read chap 16 (pg 501-514)

Unit: Thermochemistry Calorimetry and q=mcΔT

After today you will be able to… Describe what a calorimeter is and its function Explain what temperature change is dependant on Explain specific heat (c) and use the specific heat equation: q=mcΔT

Calorimetry: Is the accurate and precise measurement of heat change for chemical and physical processes. A calorimeter is a device used to measure the amount of heat absorbed or released during these processes. Example: Styrofoam cups! 

“Coffee Cup”Calorimeter: Constant-Pressure Calorimeter The thermometer records temperature change as the chemicals react in the water. The temperature change is then converted into units of energy. Most helpful for measuring the heat lost/gained in a chemical reaction.

Bomb Calorimeter: Constant-Volume Calorimeter A food sample is lit on fire. It burns until it is completely gone. Heat from the sample is released and heats up the water. The temperature change is then converted into units of energy. Most helpful in measuring the calorie content in food.

Temperature change is dependant on: Amount of heat added Mass of the substances Composition of the substance (specific heat) Example: Boiling water – metal pot gets hotter faster than water.

Specific heat: (c) amount of heat required to raise the temperature of 1 gram of a substance by 1˚C http://ga.water.usgs.gov/edu/heat-capacity.html

Units: Specific heat of H2O: cal J or g˚C g˚C 4.18 J/g˚C or 1.00 cal/g˚C

q=mcΔT q= m= c= ΔT= heat (J or cal) mass (grams) specific heat g˚C cal g˚C J specific heat or change in temperature (˚C) (Tfinal-Tinitial)

q= (525g) (0.21cal/g˚C) (34.8˚C) q= 3800cal Example: How many calories of heat are required to raise the temperature of 525g of aluminum from 13.0˚C to 47.8˚C? (cAl=0.21cal/g˚C) q= ? cal m= 525g c= 0.21cal/g˚C ΔT= 47.8-13.0= 34.8˚C q= (525g) (0.21cal/g˚C) (34.8˚C) q= 3800cal

5100cal= (m) (1.00cal/g˚C) (71.6˚C) m=71g Example: What mass of water would have its temperature raised from 22.5˚C to 94.1˚C with the addition of 5.1kcal of heat? 1000 cal q= 5.1 kcal x = 5100 cal 1 kcal m= ? g c= 1.00 cal/g˚C ΔT= 94.1-22.5= 71.6˚C 5100cal= (m) (1.00cal/g˚C) (71.6˚C) m=71g

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Unit: Thermochemistry Thermochemical Equations

After today you will be able to… Define what a thermochemical equation is and use its information to calculate the energy used or produced in a chemical reaction Explain what the ΔH of a reaction is and how it can be found

Thermochemical equation: A chemical equation which includes the energy used or produced in a chemical reaction. (Every reaction has an energy change associated!)

Enthalpy: (H) Heat content of a substance at constant pressure. “Heat of a reaction” - heat that is absorbed or released during a chemical reaction ΔH= change in enthalpy

Exothermic Reactions can be shown two ways: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔH= -890.4kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 890.4kJ Negative sign means the reaction is exothermic! Energy is a product in an exothermic reaction!

Endothermic Reactions can be shown two ways: C(s) + 2S(s)  CS2(l) ΔH= +89.3kJ C(s) + 2S(s) + 89.3kJ  CS2(l) Positive sign means the reaction is endothermic! Energy is a reactant in an endothermic reaction!

NEW conversion factor: The heat of a reaction (ΔH) is the amount of energy used or produced when the coefficient equals the number of moles in a reaction. Coefficient of any substance = ΔH NEW conversion factor:

Examples K: 1.27 mol S U: ? kJ + 89.3 kJ +56.7kJ 2 mol S How much energy is needed when 1.27 moles of sulfur reacts with excess carbon? C(s) + 2S(s)  CS2(l) ΔH= +89.3kJ 1.27 mol S x __________ = K: 1.27 mol S U: ? kJ + 89.3 kJ +56.7kJ 2 mol S

Examples K: -347.1kJ U: ? mol H2O 2 mol H2O -890.4 kJ 0.7796 mol H2O How many moles of water would be produced if this reaction gave off 347.1kJ of energy? CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 890.4kJ -347.1 kJ x __________ = K: -347.1kJ U: ? mol H2O 2 mol H2O -890.4 kJ 0.7796 mol H2O

C2H5OH(l) + 3O2(g)  2CO2(l) + 3H2O(g) + 1235kJ How much energy is produced when 75.0g of oxygen reacts with excess ethanol (C2H5OH)? C2H5OH(l) + 3O2(g)  2CO2(l) + 3H2O(g) + 1235kJ 75.0gO2 x _________ x _________= K: 75.0 gO2 U: ? kJ 1 mol O2 -1235kJ -965kJ 32.00gO2 3 mol O2

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Unit: Thermochemistry Limiting Reactants and Thermochemical Equations

After today you will be able to… Use thermochemical equations to calculate limiting reactant

Recall, a limiting reactant is the reactant which yields the smaller amount of product. Energy in a chemical reaction can also help to determine which reactant limits the amount of product that can be made.

Example: How much energy is used when 14 Example: How much energy is used when 14.3 moles of hydrosulfuric acid reacts with 17.1 moles of oxygen gas? 2H2S + 3O2 + 175 KJ  2SO2 + 2H2O 14.3 mol H2S x _________ = 17.1 mol O2 x ________ = K: 14.3 mol H2S U: ? kJ + 175 kJ +1250 kJ 2 mol H2S K: 17.1 mol O2 U: ? kJ Used: + 175 kJ +998 kJ 3 mol O2

Example: How much energy is produced when 13. 11g of tin reacts with 2 Example: How much energy is produced when 13.11g of tin reacts with 2.715g of nitrogen? __ Sn + __ N2  __ Sn3N4 + 632 KJ 13.11 gSn x _________ x ________ = 2.715 gN2 x ________ x ________ = 3 2 1 K: 13.11 gSn U: ? kJ Produced: 1 mol Sn -632 kJ -23.27 kJ 118.71 gSn 3 mol Sn K: 2.715 gN2 U: ? kJ 1 mol N2 -632 kJ -30.62 kJ 28.02 gN2 2 mol N2

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Unit: Thermochemistry Chapter 16 Heating Curves

After today you will be able to… Explain what occurs on each point of a heating curve Calculate the total energy required for water to go through phase changes

Heating Curves A heating curve is a graph of temperature versus time. It describes the enthalpy changes that take place during phase changes.

Heating Curves When a solid substance is heated, its temperature will increase until it reaches its melting point (m.p.). Temperature will then stay constant during the melting process.

Heating Curves When a substance is completely melted, its temperature will again increase until it has reached its boiling point (b.p.). Temperature stays constant during boiling. Once completely vaporized, the temperature will again increase.

Heating curve problems involve TWO types of calculations: q=mcΔT, use for temperature changes Dimensional analysis using the ΔH, where there is a phase change (flat line), mass will be the “known.”

The type of phase change will determine the ΔH to be used: Molar heat of fusion (ΔHfus): the amount of energy required to melt a substance (solid liquid) Molar heat of vaporization (ΔHvap): the amount of energy required to vaporize (boil) a substance (liquid gas)

Example: What is the total amount of heat in Example: What is the total amount of heat in Joules (J) that must be added to 165g of ice (solid H2O) at -5.00°C to change it into steam at 103°C? Temp (˚C) Time 100˚C 0˚C -5.00˚C GAS q=mcΔT BOILING (ΔHvap) LIQUID q=mcΔT ΔHfus= +6.01kJ/mol ΔHvap= +40.7kJ/mol cice= 2.01J/g°C cwater= you know this already! csteam= 2.02 J/g°C MELTING (ΔHfus) ICE q=mcΔT

Example: What is the total amount of heat in Joules (J) that must be added to 165g of ice (solid H2O) at -5.00°C to change it into steam at 103°C? ΔHfus= +6.01kJ/mol ΔHvap= +40.7kJ/mol cice= 2.01J/g°C cwater= you know this already! csteam= 2.02 J/g°C Temp (˚C) Time 100˚C 0˚C -5.0˚C Use mass for sig figs! (5) (4) (3) (2) (1) Ice, q=mcΔT q=(165g)(2.01J/g˚C)(5.00˚C) (4) Boiling, ΔHvap DA calc q=+1660J 165gH2O 1 mol H2O 18.02gH2O +40.7 kJ 1 mol H2O 1000 J 1 kJ x x x = +373,000J (2) Melting, ΔHfus, DA calc 165gH2O x 1 mol H2O 18.02gH2O +6.01 kJ 1 mol H2O 1000 J 1 kJ x x = +5.50x104J TOTAL ENERGY (J): +499,660 J (3) Liquid, q=mcΔT (5) Gas, q=mcΔT q=(165g)(4.18J/g˚C)(100˚C) q=(165g)(2.02J/g˚C)(3˚C) q=+6.90x104J q=+1.00x103J

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Unit: Thermochemistry Hess’s Law

After today you will be able to… Add, multiply, divide, or reverse chemical equations Calculate the enthalpy changes for an overall reaction using Hess’s Law

Hess’s Law Hess’s Law: The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process.

Hess’s Law Hess’s Law: If you add two or more chemical equations to get an overall equation, then you can also add the heat changes (ΔHs) to get the overall heat change.

How to add chemical equations: If two identical substances are on opposite sides of the arrow, they will cancel (reduce). If two identical substances are on same side of the arrow, add the coefficients together. Keep substances on the same side of the arrow in the final equation.

Helpful hint: Align arrows underneath each other! Example: Add the following equations. C + O2  CO2 CO2  CO + ½ O2 ½ C + ½O2  CO Helpful hint: Align arrows underneath each other!

Example: Add the following equations Example: Add the following equations. 2 Cu + O2  2 CuO 4 Cu + O2  2 Cu2O 6Cu + 2O2  2CuO + 2Cu2O

How to reverse chemical equations: To reverse an equation, change sides with the products and reactants, and change the sign on the ΔH (ex: - to +)

Example: Reverse this reaction. 2 P + 3 Cl2  2 PCl3 H = -574 kJ Switch sides! Reverse the sign! 2 PCl3  2 P + 3 Cl2 H = +574 kJ

How to multiply/divide a chemical equation: Multiply or divide the coefficients by the same number and also multiply/ divide the H by the same number.

Example: Multiply this equation by 2. Ca + 2 C  CaC2 H = -62.8kJ [ ] [ ] x2 x2 2 Ca + 4 C  2 CaC2 H = -125.6 kJ

These rearrangements can be combined… Example: Reverse and multiply by 3. 2Fe + 3/2 O2  Fe2O3 H = -824.2kJ [ [ ] ] x3 x3 Reverse the sign! Switch sides! 3 Fe2O3  6 Fe + 9/2 O2 H = +2472.6 kJ

It’s time to apply Hess’s law to some examples!

Helpful hint: Pick one substance that appears only in the overall equation!

2 NH3 + 3 Cl2  6 HCl + N2 Example: Find the H for using the following information: ½ H2 + ½ Cl2  HCl H = -92kJ ½ N2 + 3/2 H2  NH3 H = -46kJ

Example: Find the H for 2 NH3 + 3 Cl2  6 HCl + N2 using the following information: ½ H2 + ½ Cl2  HCl H = -92kJ ½ N2 + 3/2 H2  NH3 H = -46kJ (x 6) (Rev, x2)

3 H2 + 3 Cl2  6HCl ΔH= -552kJ 2 NH3  N2 + 3 H2 ΔH= +92kJ Example: Find the H for 2 NH3 + 3 Cl2  6 HCl + N2 using the following information: ½ H2 + ½ Cl2  HCl H = -92kJ ½ N2 + 3/2 H2  NH3 H = -46kJ (x 6) (Rev, x2) 3 H2 + 3 Cl2  6HCl ΔH= -552kJ 2 NH3  N2 + 3 H2 ΔH= +92kJ 3 Cl2 + 2 NH3  6HCl + N2 ΔH= -460kJ

#WBM When 1 mol of methane is burned at constant pressure, 890 kJ of energy is released as heat. If a 3.2 g sample of methane (CH4) is burned at constant pressure, what will be the value of H?

-180kJ

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Entropy Study of randomness and disorder Nature favors the following conditions: G < 0 Low enthalpy and high entropy (there are exceptions) Gibbs energy = available energy G = H - T S