Energy and Chemical Change: Basic Concepts

A. Energy is the ability to do work or produce heat. It exists in 2 forms:

I. The Nature of Energy

Energy and Chemical Change: Basic ConceptsEnergy and Chemical Change: Basic Concepts

1. Potential energyis energy due to the composition or position of an object.

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2. Kinetic energy is energy of motion.

The Nature of Energy

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Explain the potentialand kinetic energyof this system.

B. Chemical systems contain both kinetic energy and potential energy.

The Nature of EnergyEnergy and Chemical Change: Basic ConceptsEnergy and Chemical Change: Basic Concepts

1. Kinetic Energy is affected by temperature.

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2. The potential energy depends upon its composition:

a. the type of atomsin the substance

b. the number and type of chemical bondsjoining the atoms,

c. the particular way the atoms are arranged.

C. Law of conservation of energy


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The law of conservation of energystates that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

A. The energy stored in a substance because of its composition is called chemical potential energy.

II. Chemical potential energy


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B. Heat, which is represented by the symbol q, is energy that is in the process of flowing from a warmer object to a cooler object.

C. Measuring heat


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1. In the metric system of units, the amount of heat required to raise the temperature of one gram of pure water by one degree Celsius (1°C) is defined as a calorie (cal).

2. The SI unit of heat and energy is the joule(J). One joule is the equivalent of 0.2390 calories, or one calorie equals 4.184 joules.

Relationships Among Energy Units


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D. Specific Heat


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1. The specific heatof any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius.

Specific Heat Capacities Table

SubstanceJ/kg/

oCcal/g/oC

Water (0oC -100oC) 4186 1.000

Methyl Alcohol 2549 0.609

Ice (-10oC - 0oC) 2093 0.500

Steam (100 oC) 2009 0.480

Soil (typical) 1046 0.250

Air (50 oC) 1046 0.250

Aluminum 900 0.215

Glass (typical) 837 0.200

Iron/Steel 452 0.108

Copper 387 0.0924

Silver 236 0.0564

Mercury 138 0.0330

Gold 130 0.0310

Lead 128 0.0305

1. The heat absorbed or released by a substance during a change in temperature depends on:

a. the specific heat of the substance

b. the mass of the substance

c. the amount of the temperature change.

E. Calculating heat evolved and absorbed


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2. You can express these relationships in an equation.

3. In the equation, q = the heat absorbed or released, c = the specific heat of the substance, m = the mass of the sample in grams, and ∆Tis the change in temperature in °C (Tfinal –Tinitial).

Calculating heat evolved and absorbed


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1. In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling.

F. Calculating Specific Heat


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2. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat.

What is the specific heat of iron?

Calculating Specific Heat


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• Solve the equation using the known values.

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F. Measuring Heat


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1. A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

Determining specific heat


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2. Suppose you put 125 g of water into a foam-cup calorimeter and find that its initial temperature is 25.6°C. A 50.0-g sample of an unknown metal is heated to a temperature of 115.0°C and put into the water.



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• The flow of heat stops only when the temperature of the metal and the water are equal. The final temperature is 29.3 C. Solve for the heat capacity of the unknown metal.



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4. Assuming no heat is lost to the surroundings,

heat gained (water)= the heat lost (metal).

First, calculate the heat gained by the water.



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• The heat gained by the water equals the heat lost by the metal, qmetal, so you can write this equation.

Now calculate for cmetal. -

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• Substitute the known values of m and ∆T(50.0 g and 85.7 °C) into the equation and solve.

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• The unknown metal has a specific heat of 0.44 J/(g·°C).

Warm Up: Using Data from Calorimetry


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• A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C.

• What is the specific heat of the metal? • Known

• mass of metal = 4.68 g metal • quantity of heat absorbed, q = 256 J • ∆T = 182°C

• Unknown • specific heat, c = ? J/(g·°C)

Using Data from Calorimetry


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• The calculated specific heat is the same as that of strontium.

III. Energy and the Change of StateSome astonishing demos!

A. During a change of state, energy is absorbed or released, but no temperature change is observed.

B. Calculating energy needed to change the state of matter.1. Freezing – energy needed to freeze (or melt) a substance Q = moles x Hfusion

Hfusion = 6.01 kJ/mol water

2. Boiling – Energy needed to vaporize (or condense) a substance Q = moles x Hvaporization

Hvaporization= 40.7 kJ/mol water

C. Calculate the heat released when 10.0 grams of water freeze.

moles water =10.0 g/18.0 g/mol = 0.556 mol

Q = 0.55mol x -6.01 kJ/mol = -3.34 kJ

`D. Calculate the heat needed to melt 20.0 g

of ice at 0.0 ° C and then warm it up to 25.0°C.

First – calculate heat needed to melt the ice:Qfusion = moles x Hfusion

Next – calculate the heat needed to warm up the water.

Qwarming= mass x c x ∆T Finally – add the Q’s together! 8.76 kJ

IV. Chemical Energy and the Universe


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A. Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes.

B. Enthalpyis the heat content of a system (at constant P).

Enthalpy and enthalpy changes


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C. ∆Hrxn is the difference between the enthalpy of the products of the reaction and the enthalpy of the reactants.

1.

∆H = qp



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2. Energy Diagrams

reactants



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3. Example, the highly exothermic combustion of glucose (C6H12O6) occurs in the body as food is metabolized to produce energy for activities.

4. The enthalpy (heat) of combustion(∆Hcomb) of a substance is the enthalpy change for the complete burning of one mole of the substance.



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5. Standard enthalpychanges have the symbol ∆H°

Measured at standard conditions (1 atm pressure and 298 K)

V. Calculating Enthalpy of Reaction


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A. The ∆H°comb for methanol (CH3OH) is –726 kJ/mol. How much heat is released when 82.1 g of methanol is burned?

Calculating Enthalpy of Reaction


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• First, calculate the number of moles of methanol that is burned.

• Now find the enthalpy of reaction for the combustion of 82.1 g (2.56 mol) of methanol.

Basic Assessment QuestionsBasic Assessment Questions

Question 1

A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol?

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Answer

44.3°C

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Question 2

If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?

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Answer

58.5°C

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Question 3

How much heat is evolved when 24.9 g of propanol (C3H7OH) is burned? ∆Hcomb = –2010 kJ/mol

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Answer

833 kJ

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Calculating Enthalpy Change


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B. Hess’s law: If two or more thermochemical equations can be added to produce a final equation for a reaction, then the enthalpy change for the final reaction equals the sum of the enthalpy changes for the individual reactions.

Applying Hess’s Law


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Example: Use thermochemical equations aand b to determine ∆H for the oxidation of ethanol (C2H5OH) to form acetaldehyde (C2H4O) and water.

a.

b.

Applying Hess’s Law


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• Add these two equations, and cancel any terms common to both sides of the combined equation.

Hess’s Law: Heat of Combustion of Mg Lab

• Overall Equation: Mg + ½ O2 � MgO

• Known Equations:– MgO + 2HCl � MgCl2 + H2O– Mg + 2HCl � MgCl2 + H2

– H2 + ½ O2 � H2O

Standard enthalpy (heat) of formation


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• Note: the standard heat of formation of an element in its standard state is zero.

C. Standard enthalpies of formation may be used with Hess’s law to calculate enthalpies of reaction under standard conditions.

Example: Use Heats of formation to calculate the heat of reaction for ethanol reacting with oxygen to form acetaldehyde and water.

Substance Hf°C2H5OH(l) -277.7 kj/mol

C2H4O(g) -166 kJ/mol

H2O(l) -285.82 kJ/mol

[2(-166 kJ/mol) + 2(-285.82 kJ/mol)] – [2(-277.7 kJ/mol)]= -348 kJ

Energy and Chemical Change: Additional ConceptsEnergy and Chemical Change: Additional Concepts

VI. Reaction Spontaneity

A. Entropy (S) is a measure of the disorder or randomness of the particles that make up a system.

1. Spontaneous processes always result in an increase in the entropy of the universe.

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2. The change in the entropy of a system is given by the following equation.


Reaction SpontaneityB. Factors that affect entropy:

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1. Changes of state –increase from solid to liquid to gas; decrease in opposite direction.

2. Dissolving of a gas in a solvent –decrease in entropy3. Change in the number of gaseous particles –increased particles = increase in entropy

4. Dissolving of a solid or liquid to form a solution - Entropy increases.5. Change in temperature -A temperature increase results in increased entropy.


C. Free energyTopic 20

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1. Def. For a reaction or process occurring at constant temperature and pressure, the energy available to do work is the free energy(G).


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3. If ∆Gsystemis negative, the reaction or process is spontaneous

4. If ∆Gsystemis positive, the reaction or process is nonspontaneous.

What conditions of ∆ H and ∆ S support this?-∆ H and +∆ S


Demo: Thermite Reaction!Topic 20

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2Al (s) + Fe2O3(s) �� 2Fe(s) + Al2O3(s) + E

Substance ∆H°f (kJ/mol) S°(J/mol-K)

Fe2O3 -822.16 89.96

Al 0 28.32

Fe 0 113.4

Al 2O3 -1669.8 51.00

Calculate ∆HrxnCalculate ∆SrxnCalculate ∆Grxn


Determining Reaction SpontaneityTopic 20

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• Because ∆Gsystemis negative, the reaction is spontaneous.

Additional Assessment QuestionsAdditional Assessment Questions

Question 1 Topic 20

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Predict the sign of ∆Ssystem for


negative

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Question 2 Topic 20

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positive

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Question 3 Topic 20

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positive

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Calculate ∆Gsystemfor each of the following processes, and state if the process is spontaneous or nonspontaneous.

Question 4 Topic 20

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∆Hsystem = 147 kJ, T = 422 K, ∆Ssystem = – 67 J/K

175 kJ; nonspontaneous

Answer 4a

Question 4a

Additional Assessment QuestionsAdditional Assessment QuestionsTopic 20

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∆Hsystem = –43 kJ, T = 21°C, ∆Ssystem = – 118 J/K

–8 kJ; spontaneous

Answer 4b

Question 4b


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∆Hsystem = 227 kJ, T = 574 K, ∆Ssystem = – 349 J/K

27 kJ; nonspontaneous

Answer 4c

Question 4c


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Energy and Chemical Change: Basic Concepts

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