17.0a thermochemistry -...

January 13 1 ThermoChemistry

17.0a ThermoChemistry

Miramar College Chemistry Chemistry 201

Dr. Fred Omega Garces

Nature of Energy Enthalpy and its Sign convention


The Nature of Energy

Matter and Energy in the Universe is constant. Law of Conservation of Matter Law of Conservation of Energy

(1st Law of Thermodynamics).

HomogeneousMixture

PureSubstances

Homogeneous HeterogeneousMixture

Matter Energy

Universe

Compounds Elements


Potential Energy ( P.E. = mgh) Energy of position.

Kinetic Energy (K.E. = 1/2 mv2) Energy of motion.

Classification of Energy

Study of energy and its changes subject of thermodynamics.


Energy- State Function

State function - a property of a system that depends only in its present state. i.e., Measuring the distance between AB is impossible since building in between A and B. But the distance can be indirectly measured by measuring segments 1, 2 and 3. The distance between A and B is path independent (State function).

A B

1

2

3

 Analogy: Measuring distance with big building blocking


Enthalpy- Heat content for system at constant pressure

H-Enthalpy; an extensive property (size dependent)

CH4 (g) + O 2(g) → CO2 (g) + H2O (l) + Heat ΔH = - 890 KJ Stoichiometry coefficient changes ΔH

2 [CH4 (g) + O 2(g) → CO2 (g) + H2O (l) + Heat] ΔH =(2) x - 890 KJ Direction of reaction determines sign of ΔH

CO2 (g) + H2O (l) + Heat → CH4 (g) + O 2(g) ΔH = 890 KJ


The Universe according to ThermoChemistry

The Universe System - part of interest Surrounding - everything else.

System

Surrounding

Universe


Nomenclature; The Universe, Surrounding and the system

Universe = Matter + Energy

Universe

System!!Matter of interest!

Surrounding!!(Rest of World)!

System!!CH4 + O2 → CO2 + H2O!

Surrounding!!(Gets hotter)!


Enthalpy: Heat as a Reactant

Example of Heat as a reactant (Endothermic)

1. NH4NO3 in water; NH4NO3 absorbs energy from surrounding to dissolve.

2. Ice melting, water evaporating, solid

undergoing sublimation.. Gas

Solid

Liquid


Heat Reactant Effects to the Surroundings

Sensation of cold, results from heat being absorb from the surrounding into the system.

1. Because the surrounding cools down, heat must be absorb from the surrounding..

2. Heat is a reactant: Endothermic ( + ΔH)

Universe

System!

Surrounding!

Heat


Enthalpy: Heat as a Product

Example of Heat as a product (Exothermic)

1. KCl + P4 → PCl3 + Heat. 2.Liquid ⎯→ Freezing

3. NaOH ⎯(Η2Ο)→ Na+ + OH-

Gas

Solid

Liquid System!

!Heat!


Heat Reactant Effects to the Surroundings

Sensation of hot, results from heat being release from the system to the surrounding.

1. Because the surrounding warms up, heat must be emitted to the surrounding. 2. Heat is a product: Exothermic ( - ΔH)

Universe

System!

Surrounding!

Heat


Energy: Units

Energy : Joule or calorie 1 cal = 4.184 J. 1000 cal = 1 Cal = 4.184 kJ = 4184 J 9 Cal /gram fat : 3 Cal / gram proteins and Carb. 3500 Cal or 3500 Kcal = 1 lb fat.


Thermodynamics Sign convention, (+ or - )

Heat product (-)Exothermic

Reactant → Product + q Heat reactant

(+)Endothermic Reactant + q → Product

System

En

erg

y

Energy releasedownhill

Surrounding

System

En

erg

y

SurroundingEnergy absorbuphill


Thermodynamics Exothermic (-), TNT Explosion

Heat product (-)Exothermic

Energy release from a reactant (TNT) in a chemical reaction (i.e., exothermic) originates from breaking bonds in the reactants as the atoms rearrange to form more stable products.

Reactant → Product + q


Calculating the Energy Exchange

Example: Reger 4.54 N2 (g) + O2 (g) g 2NO (g) ΔH = + 180 KJ Q, Heat product or reactant?

Heat absorb or release by system ? Exothermic or Endothermic? Stoichiometry relationship : 1 mol N2 = 1 mol O2 = 2 mol H2O = + 180 KJ.

€

ΔH = 2.20 g∗ 1 mol N228.0 g

$

% &

'

( ) ∗

+ 180 KJ1 mol N2

$

% &

'

( ) = 14.1 KJ

H + N2 (g) + O2 (g) --> 2NO (gl)Δ2.20 g

mol N2


Sign Convention ... monetary analogy

$ign convention and the flow of energy/heat between system and surrounding$.

System point of View Exothermic(-) ____________ Surrounding View Endothermic (+)

System point of View Endothermic (+) ___________ Surrounding View Exothermic (-)

Cash Flow Analogy

Wages (+) cash flow (pay in) Debt (-) cash flow (pay out)

System

Surrounding

Heat System

Surrounding

Heat

World

$ (+) (-)

Wages Debt $


17.0b Calorimetry Measuring Heats of Reaction

Dr. Fred Omega Garces Chemistry 201 Miramar College


Measuring Heats of Reaction Calorimetry

Experimental technique to measure the amount of heat absorb or release by a reaction.

Assumption: Heat is not absorb by calorimeter. System (reaction) -heats-> Surrounding (water) [Temp. increase] System (reaction) -cools-> Surrounding (water) [Temp. decrease]

Reaction Mixture(System)

Surrounding(water)

InsulatedCalorimeter

Thermometer


Water is the surrounding; measure heat exchange by monitoring the surrounding with a thermometer.

Reaction is system; cannot

directly monitor energy change of system, must do it indirectly by monitoring surroundings.

Calorimetry ...and Sign convention

Reaction Mixture(System)

Surrounding(water)

InsulatedCalorimeter

Thermometer


Temperature Relationship • Temp. provides clue on energy exchange between system & surrounding.

A. Temp. increase: Surrounding absorb heat, system release heat. B. Temp. decrease: Surrounding release heat, system absorb heat.

Temp. change of water (surrounding) provides info. on (qsurr) exchange. qabs = m Cs ΔT ( → ΔHsurr = -ΔH sys ) where, m = mass of water

Cs = specific heat (Cs H2O = 4.184 J / g•K)

ΔT = Tf - Ti (remember ΔT is the same in °C and K) qabs Intensive property; similar to density, physical property to cmpd. qabs = ΔHsurr = - ΔHsys


Calorimetry, Sign Convention Example

• A 30-mL solution of a dilute acid is added to 70mL of a base solution in a coffee cup calorimeter. The temperature of the solution increases from 22.3°C to 24.°C. Assuming that the mixture has the same specific heats as water and a mass of 100g, calculate the heat flow in the reaction. Is the process exothermic or endothermic?

A + B g Product 30mL 70mL

Α + Β (system)

Water (Surr)

Ti = 22.3°C

Tf = 24.1°C

Cs = 4.184 J / g ° m = 100 g

Temp increase of solution (water) qsurr (+) gained heat/Energy

qsys (-) lose heat

note: qevolve(sys) = qabs(surr)

qsys= m Cs ΔT

= 100g x 4.184 J/g° x (24.1-22.3)°C

= 711.3 J releasesys

= - 711.3 J

q


Calorimetry, Heats or Reaction Example

• Dissolving 6.00 g of CaCl2 in 300 mL of water causes the temperature of the solution to increase by 3.43°C. Assuming that the specific heat of the solution is 4.18 J/g-K, determine ΔH for the reaction: CaCl2 -H2O→ Ca+2

(aq) + 2 Cl-( aq)

CaCl2 -H2Og Ca+2(aq) + 2 Cl-( aq)

CaCl2 → Ca+2(aq) + 2 Cl-( aq)

(system)

Water (Surr)

m soln= 6.0 g + 300 g = 306 g

ΔT = + 3.43 °C

Cs = 4.184 J / g °

Temp increase of solution (water)

qsurr (+) gained heat/Energy

qsys (-) lose heat note: qevolve(sys) = qabs(surr)

qsurr= mH2O Cs ΔT = 306. g x 4.184 J/g° x (3.43)°C

= 4,391 J abs (surr)

q sys= - 4.39 kJ

q


Energy- State Function

State function - a property of a system that depends only in its present state. i.e., Measuring the distance between AB is impossible since building in between A and B. But the distance can be indirectly measured by measuring segments 1, 2 and 3. The distance between A and B is path independent (State function).

A B

1

2

3

 Analogy: Measuring distance with big building blocking


Enthalpy of a Reaction: Hess’ Law

Take for example the following: C(s) + 1/2 O2 (g) → CO(g) Can't measure ΔH by Calorimetry since this reaction will always produce CO2

Can determine enthalpy for this reaction by finding segments that will produce this reaction.

C (s) + O2 (g) → CO2 (g) Δ H = - 393.5 KJ CO (g) + 1/2 O2 (g) → CO2 (g) ΔH = - 283.0 KJ Combination of these two equation in the following fashion (1) C (s) + O2 (g) → CO2 (g) ΔH = - 393.5 KJ ( rev 2) CO2 (g) → CO (g) + 1/2 O 2 (g) ΔH = + 283.0 KJ net C(s) + 1/2 O 2 (g → CO(g) ΔH = - 110.5 KJ


Energy Level Diagram

ΔH3 = ΔH1 - ΔH2 = - 393.5 - (-283.0) = -110.5 kJ = - ΔH, down hill (exothermic)

Ener

gy

C(s) + O2(g)

ΔΗ1-393.5 Kj

CO(g) + 1/2 O2(g)

CO2(g)

-283.0 KjΔΗ2

ΔΗ3


Hess Law; Example

Consider the reaction: C2H2(g) + 2 Cl2 (g) → C2H2Cl4 (l) ΔH = ? 1) 2 C(s) + H2 (g) → C2H2 (g) ΔH = 227 KJ

2) 2 C(s) + H2 (g) + 2Cl2 (g) → C2H2Cl4 (l) ΔH = 130 KJ

Arrange the above equation to add up to desired reaction:

(rev 1) C2H2 (g) → 2 C(s) + H2 (g) ΔH = - 227 KJ 2) 2C(s) + H2 (g) + 2 Cl2 (g) → C2H2Cl4 (l) ΔH = + 130 KJ

Net: C2H2 (g) + 2 Cl2 (g) → C2H2Cl4 (l) ΔH = - 97 KJ

ΔH3 = -(ΔH1) + ΔH2

= - ( 227 ) + 130 KJ = - 97 kJ down hill (exothermic)


Energy Diagram

2 C(s) + H2 (g) → C2H2 (g) 2 C(s) + H2 (g) + 2 Cl2 (g) → C2H2Cl4 (g)

2 C(s) + H2 (g)

C2H2 (g)

ΔΗ = + 227 2 C(s) + H2 (g) + 2 Cl2(g)

C2H2Cl4 (g)ΔΗ = + 130

C2H2 (g) + 2 Cl2(g)

ΔΗ1 = + 227

2 C(s) + H2 (g) + 2 Cl2(g)

C2H2Cl4 (g)

ΔΗ2 = +130

ΔΗ3

+ =


ThermoChemistry Key Concepts

17.0a thermochemistry -...

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