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Thermodynamic Review

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Law of Conservation of Energy

• Energy is neither created nor destroyed; it can be transferred or transformed.

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Difference between Heat and Temperature

• Heat → transfer to thermal energy between two systems with different temperatures

• Temperature → measure of agitation of particles in a system

– Greater the agitation = more energy = higher temperature

Pg. 128-130

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How is heat transferred?

• From hot substance to cold substance!

– Why?

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http://www.youtube.com/watch?v=2_SJIFfkdN0

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How is heat transferred?

• Hot substance = higher degree of agitation = more energy

• Cold substance = lower degree of agitation = less energy

• When mixed = reach energy equilibrium where both systems have same degree of agitation

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How much heat is transferred?

• Depends on 3 key things

– Its agitation of particles (i.e. Temp of the substance)

– How much of the substance you have (mass)

– The nature of the substance

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Specific Heat Capacity

• Def: The quantity of energy required to raise the temperature of 1 g of substance by 1°C

• Symbol : c

• Unit: J/g°C

Pg. 134-135

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Change in thermal energy is therefore…

Q=mcΔT

• Q = Heat flow to or from a substance (J)

• m = mass of the substance (g)

• c = specific heat capacity (J/g oC)

• T = Tfinal – Tinitial (oC)

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Exothermic Rxn

• Reaction which is loosing thermal energy

– Heat is released

– T = negative value

– Why?

Pg. 150-154

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Endothermic Rxn

• Reaction which is gaining thermal energy

– Heat is absorbed

– T = positive value

– Why?

Pg. 150-154

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Energy is not created or destroyed …

Heat Lost = Heat Gained

Or

Qexothermic = Qendothermic - +

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Practice

1. In an oil refinery, 200.0 kg of steam is heated from 100.0 oC to 250.0 oC. The specific heat capacity of steam is 2.01 J/(g. oC). Calculate the quantity of heat that flows intothe steam.

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Practice

2. In order to conduct a research experiment, a laboratory requires 150.0 mL of lukewarm water at 35.0 oC. How could they obtain this water by mixing hot and cold water. Assume that 60.0 g of hot water at 70.0 oC is available.

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Practice

3. A 500.0 g piece of metal is heated in a flame to 350.0 oC and then plunged into a beaker containing 0.50 kg of water. The original temperature of the water was 20.0 oC and the final temperature after they have come to thermal equilibrium is 40.0 oC. What is the specific heat of the metal?

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Answers to Practice Questions

1. Q= 6.03 x 107 J or Q= 6.03 x 104 kJ

2. m=90.0g Ti = 11.7°C

3. c= 0.27 J/g°C

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Enthalpy

• Def: Total energy of a system

– But we cannot measure it directly, hence…

• Enthalpy change!

– Def: Energy exchange between a system and its environments during a physical or chemical rxn at constant pressure

Pg. 148

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Molar Heat of Reaction/ Enthalpy Change

Molar heat of reaction is the quantity of heat released or absorbed per mole of product produced or consumed. Formula: H = Q/n Units: J/mol or kJ/mol H represents the change in heat content, therefore a sign must always be included.

A positive H indicates …

A negative H indicates …

an endothermic reaction

an exothermic reaction

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Types of Molar Heats of Reaction

• ΔH fusion/ ΔH solidification

• ΔH vaporization / ΔH condensation

• ΔH solution

• ΔH combustion

• ΔH neutralization

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Practice

1. In an experiment, 25.0 g of urea is dissolved in 200.0 mL of water in a calorimeter. The temperature of the water changes from 21.0oC to 13.6 oC. Calculate the molar heat of solution of urea, NH2CONH2.

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Practice

2. A student pours 35.3 g of liquid paraffin, C25H52, at its melting point into a calorimeter that contains 200.0 mL of water at 22.3 oC. The temperature of the water when the wax just solidifies rises to 29.5 oC. Calculate the molar heat of solidification of paraffin wax.

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Answers to Practice

1. ΔHsolution= 15kJ/mol (n=0.416 mol, Q=6.201

kJ)

2. ΔHsolidification= -60 kJ/mol (n=0.100 mol, Q= -

6.033kJ)

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Hess’s Law

• The change in heat energy in a chemical reaction is the same no matter what the path taken to arrive at the end product.

• Thermochemical reactions can be added together like algebraic equations.

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Hess’s Law

H (kJ/mol)

i.e. NaOH(s) NaOH(aq)

NaOH(aq) + HCl(aq) NaCl(aq) + H20(l)

NaOH(s) + NaOH(aq) + HCl(aq) NaCl(aq) + H20(l) + NaOH(aq)

NaOH(s) + HCl(aq) NaCl(aq) + H20(l)

-35.0

-52.0 +

-87.0

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1. Write the H to the right of the equation.

N2 (g) + 3 H2 (g) 2 NH3 (g) H = - 92.0 kJ

- sign means exothermic

Ways of expressing Molar Heat of Reaction

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2. Specify the H in relation to a particular substance

N2 (g) + 3 H2 (g) 2 NH3 (g)

- 92.0 kJ/mol N2

or - 92.0 kJ/3 mol H2 = - 30.7 kJ/mol H2

or - 92.0 kJ/2 mol NH3 = - 46.0 kJ/mol NH3

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3. Write the amount of heat directly in the chemical equation

N2 (g) + 3 H2 (g) 2 NH3 (g) + 92.0 kJ

Heat is a product in an exothermic reaction

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Changing the direction of a chemical equation …

N2 (g) + 3 H2 (g) 2 NH3 (g) H = - 92.0 kJ

2 NH3 (g) N2 (g) + 3 H2 (g) H = + 92.0 kJ

+ sign means endothermic

changes the sign of the molar heat

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Practice

Some automobiles and buses are equipped to burn propane gas, C3H8, as a fuel. The complete combustion of propane is shown by the following chemical equation: Given the following heats of formation. H2(g) + O2(g) H2O(g) H = 242.0 C(s) + O2(g) CO2(g) H = 393.5 What is the heat of formation of propane? 3 C(s) + 4 H2(g) C3H8(g) H = ?

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