Week 8.1 - Thermodynamic and Equilibria

November 8, 2018 | Author: MXR-3 | Category: Heat, Enthalpy, Calorimetry, Thermodynamics, Chemistry
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Chapter 4 Thermodynamic and Equilibria

First Law of Thermodynamics

Prepared by: Second Law of Thermodynamics

Mrs Faraziehan Senusi

Physical Transformation Transformation of Pure Substances Simple Mixtures Chemical Equilibrium

Reference: Chemistry: the Molecular Nature of Matter and Change,


Thermochemistry dynamics dynamics ~  energy changes that accompany • Thermo   physical and chemical processes. Usually these energy changes involve heat.

• Thermochemistry ~  is the study of heat change in chemical reactions. 

concerned with how we observe, measure, and   predict energy changes for both physical changes and  chemical reactions

use energy changes to tell whether or not a given  process can occur under specified conditions to give  predominantly  predominantly products (or reactants) and how to make a process more (or less) favorable.

Basic concepts • System :  specific part of the universe that is of interest to us.  The substances involved in the chemical and

 physical changes that we are studying

• Surroundings: the rest of the universe

 boundary surrounding


Open system: can exchange mass and energy (heat) with surrounding

system where mass and energy can cross the boundary



system energy



allows transfer of heat but not mass

consists of a fixed amount of  mass and no mass can cross its  boundary

Energy in the form of heat or  work can cross the boundary

Isolated system: does not allow transfer either mass or  energy

A system where no mass, heat



system energy

matter  system energy


THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics (the conservation of energy principle) provides a sound basis for studying the relationships among the various forms of energy and energy interactions. •

can be n ei the th er  The first law states that en er gy can cre cr eated ated nor des destr oye oyed; i t can onl y ch ch ange f orms or ms..

When a rock falls, the decrease in potential energy is equals to the increase in kinetic energy. The increase in the energy of a potato in an oven is equal to the amount of  heat transferred to it.


Each particle in a system has potential po tential and kinetic energy; the sum of these energies for all particles in a system is the internal energy, E  energy, E . In a chemical reaction: when reactants are converted to  products, E   products, E changes changes (  E ). ).

 E  =  E final -  E initial =  E products -  E reactants

Energy diagrams for the transfer of internal energy (E ) between a system and its surroundings

Heat and Work  •

Energy transfer outward from the system or inward from the surroundings can appear in two forms, heat and work . work .

 E =  E = q + w •

where q = heat and w = work 

Heat (or thermal energy, symbol q) is the energy transferred  between a system and its surroundings as a result of a difference in their temperatures only. All other forms of energy transfer (mechanical, electrical, and so on) involve some type of work (w), the energy transferred when an object is moved by a force.

Sign Conventions for q, w  and E  •

The numerical values of  q and w can be either  positive either  positive or  negative,, depending on the change the system negative the system undergoes. Energy coming into the system is  positive  positive;; energy  going out   from  from the system is negative negative..

depends on magnitudes of  of q  q  and w 

depends on magnitudes of  of q  q  and w 

For q: (+) means system gains heat, (-) means system loses heat.

Law of Conservation of Energy (First Law of Thermodynamics) Thermodynamics) The energy of the system plus the energy of the surroundings remains constant: energy is conserved.

DE universe


DE system


DE surroundings

Units of Energy  joule (J)

1 J = 1 kg m2/s2

calorie (cal)

1 cal = 4.184 J

British Thermal Unit

1 Btu = 1055 J

= 0

Thermodynamic state of a system •

The properties of a system — such such as P, V, T — are are called state functions The value of a state function depends only on the state of the system and not on the way in which the system came to be in that state. A change in a state function describes a difference  between the two states. It is independent of the  process or pathway by which the change.

The most important use of state functions in thermodynamics is to describe changes changes.. ΔX

= ΔXfinal – ΔXinitial

When X increases, the final value is greater than the initial value, so ΔX is positive; a decrease in X makes ΔX a negative value.

Calorimetry • We can determine the energy change associated with a chemical or physical process by using an experimental technique called calorimetry calorimetry.. • This technique is based on observing the temperature change when a system absorbs or releases energy in the form of heat. • The experiment is carried out in a device called a calorimeter , in which the temperature change of a known amount of substance (often water) of known specific heat is measured. measured. • The temperature change is caused by the absorption or  release of heat by the chemical or physical process under  study.

Specific heat, c : amount required to raise temperature of one gram of the substance by one degree Celsius (J/g. oC) Heat capacity or calorimeter constant, C : amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius (J/ oC) C = mc (where m is the mass)

• If specific heat and amount of substance is known, change in sample temperature ∆T will tell us the amount of heat, q, q, that has been released/ absorbed in particular process. q = mc∆T

• q is positive (endothermic) and negative for exothermic  process

• Calorimeter can be used to measure the amount of  heat absorbed or released when a reaction takes place in aqueous solution.

Example 1 We add 3.358 kJ of heat to a calorimeter that contains 50.00 g of  water . The temperature of the water and the calorimeter, originally at 22.34°C, increases to 36.74°C. 36.74°C . Calculate the heat capacity of the calorimeter in J/°C. The specific heat of water is water is 4.184 J/g.°C. J/g.°C. 

Calculate the amount of heat gained by the water in the calorimeter.

The rest of the heat must have been gained by the calorimeter.

Determine the heat capacity of the calorimeter.

Example 1

amount of heat gained by the water 

amount of heat gained by the calorimeter 

Example 2 A 50.0 mL sample of 0.400 M copper(II) sulfate solution at 23.35°C is mixed with 50.0 mL of 0.600M sodium hydroxide solution, solution , also at 23.35°C,, in the coffee-cup calorimeter. Heat capacity of calorimeter  is 23.35°C 24.0 J/°C. After the reaction occurs, the temperature of the resulting mixture is measured to be 25.23°C 25.23°C.. The density of the final solution is 1.02 g/mL. g/mL. Calculate the amount of heat evolved. evolved. Assume that the specific heat of the solution is the same as that of pure water, 4.184 J/g.°C.

The amount of heat released by the reaction is absorbed by the calorimeter and by the solution.

To find the amount of heat absorbed by the solution, we must know the mass of solution; to find that, we assume that the volume of the reaction mixture is the sum of volumes of the original solutions.

Example 2

The practice of calorimetry • Two common types are :  constant-pressure calorimeters

A "coffee-cup" calorimeter is often used to measure the heat transferred (q p) in processes open to the atmosphere.  constant-volume calorimeters

One type of constantvolume apparatus is the  bomb calorimeter, designed to measure very  precisely the heat released in a combustion reaction.

Constant-pressure Calorimetry One common use is to find the specific heat capacity of a solid that does not react with or  dissolve in water . The solid (system) is weighed, heated to some known temperature, and added to a sample of  water (surroundings) of known temperature and mass in the calorimeter. With stirring, the final water temperature, which is also the final temperature of the solid, is measured. The heat lost by the system (-q sys, or -qsolid) is equal in magnitude but opposite in sign to the heat gained by the surroundings (+qsurn or +qH2O):

- qsolid = q H2O Or, Or, - (c solid x mass solid x ΔT solid ) = c H2O x mass H2O x ΔTH2O

Example 3 Determining the Specific Heat Capacity of a Solid PROBLEM: A 25.64 g sample of a solid was heated in a test tube to 100.00 oC in boiling water and carefully added to a coffee-cup calorimeter containing 50.00 g of  water . The water temperature increased from 25.10 oC to 28.49 oC. What is the specific heat capacity of the solid? solid? (Assume all the heat is gained by the water) PLAN:

It is helpful to use use a table to summarize summarize the data given. Then work the problem problem realizing that heat lost by the system must be equal to that gained by the surroundings. mass (g)

c (J/g.K)













solid H2 O SOLUTION:

25.64 g x csolid =

c x

- 50.00 g x

-71.51 K 4.184 J/g.K  x

25.64 g x

-71.51 K 

4.184 J/g .K  x 3.39 K 

= - 50.00 g x 3.39 K 



0.387 J/g.K 

Constant-volumee Calorimetry Constant-volum •

Figure 6.8 depicts the preweighed combustible sample in a metal-walled chamber (the bomb), which is filled with oxygen gas and immersed in an insulated water bath fitted with motorized stirrer and thermometer. A heating coil connected to an electrical source ignites the sample, and the heat evolved raises the temperature of the bomb, water, and other  calorimeter parts. Because we know the mass of the sample and the heat capacity of the entire calorimeter, we can use the measured ΔT to calculate the heat released.

Example 4 Calculating the Heat of Combustion PROBLEM:


A manufacturer claims that its new diet dessert has ―fewer  than 10 Calories (10 kcal) per  serving‖. serving‖. To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2 (the heat capacity of the calorimeter = 8.151 kJ/K). The temperature increases by 4.937 oC. Is the manufacturer’s claim correct? - q  sample = q  calorimeter q  calorimeter

= heat capacity x DT = 8.151 kJ/K x 4.937 4.937 K  = 40.24 kJ

40.24 kJ x


= 9.62 kcal < 10 Calories = 10 kcal

4.184 kJ The manufacturer’s claim is correct.

Enthalpy • The quantity of heat transferred into or out of a system as it undergoes a chemical or physical change at constant pressure. • Extensive property: property: magnitude depends on amount of substance present • Impossible to determine enthalpy of substance • Measure change in enthalpy, ∆H

• Enthalpy of reaction, ∆H ∆H = H (product)  – H  – H (reactant) •

Exothermic : negative

• Endothermic: positive • EXOTHERMIC PROCESS  –  a process that releases energy in the form of heat into its surroundings. (Ex: combustion reaction) • ENDOTHERMIC PROCESS  –  a process that absorbs energy from its surroundings

Enthalpy diagrams diagrams for exothermic and endothermic processes CH4(g  ) + 2O2(g  )

CO2(g  ) + 2H2O(g  ) + heat heat + H2O(s  )

Heat is released; enthalpy decreases. decreases.

H2O(l )

Heat is absorbed; enthalpy increases. increases.

Some Important Types of Enthalpy Change heat of combustion ( H comb) 1C4H10(l ) + 13/2O2(g  )

4CO2(g  ) + 5H2O(g  )

heat of formation ( H f ) K(s  ) + 1/2Br 2(l )

1KBr(s  )

heat of fusion ( H fus) 1 NaCl(s  )

NaCl(l )

heat of vaporization ( H vap) 1C6H6(l )

C6H6(g  )

Standard quantity of either reactant or product: 1 mol

Thermochemical equations • A balanced chemical equation, together with its value of ΔH • The ΔHrxn value shown refers to the amounts (moles) of substances and their states of matter in that specific equation. Combustion of methane : CH4 ( g   g ) + 2O2 ( g   g )  CO2 ( g   g ) + 2H2O(l  O(l ) ∆H= -890.4kJ

• 1367 kJ of heat is released when one mole of C 2H5OH(l) reacts with three moles of O 2(g) to give two moles of CO 2(g) and three moles of H2O(l). • We can refer to this amount of reaction as one mole of reaction, which we abbreviate ―mol rxn.‖ • We can also write the thermochemical equation as

We always interpret ΔH as the enthalpy change for the reaction as written; as (enthalpy change)/(mole of reaction), where the denominator means ―for the ―for  the number of moles of each substance shown in the balanced equation.‖ equation. ‖

• Guidelines for writing thermochemical equations



 –  Stoichiometric coefficient refer to number of moles of  substance  –  Reverse equation, magnitude of  ∆H same but sign changes  –  Multiply equation by factor of n then ∆H must also change by the same factor   –  Must always specify the physical state

AMOUNT (mol)

Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction

of compound A

AMOUNT (mol) molar ratio from balanced equation

of compound B

HEAT (kJ) H rxn (kJ/mol)

gained or lost

Example PROBLEM:

Using the Heat of Reaction ( H rxn) to Find Fi nd Amounts Amounts

The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by: Al2O3( s)  s)

2Al( s)  s) + 3/2O2( g   g )

D H rxn

= 1676 kJ

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 10 3 kJ of heat is transferred? PLAN:


heat (kJ)

1.000 x 10 3 kJ x

2 mol Al 1676 kJ

1676 kJ = 2 mol Al mol of Al x M (g/mol) g of Al

= 32.20 g Al


26.98 g Al 1 mol Al

Standard heat of reaction Specifying Standard States For a gas, the standard state is 1 atm; ideal gas behavior is assumed. For a substance in aqueous solution, the standard state is 1 M concentration  M concentration (1 mol/liter solution). For a pure substance (element or compound), the standard state is usually the most stable form of the substance at 1 atm and the temperature of interest (usually 25 oC (298 K).

Horxn = standard heat of reaction (enthalpy change determined with all substances in their standard states)

Standard enthalpy of formation

( ∆Hof ) : heat • Standard enthalpy of formation (∆H change that results when one mole of a compound is formed from its elements at a pressure of 1 atm (standard state). • ∆Hof  of any element in its most stable form is zero ∆Hof  (O2) = 0

∆ Hof  (O3) ≠ 0

Direct method of measuring ∆Hof  Formation Equations In a formation equation, 1 mol of a compound forms from its elements. The standard heat of formation (D H of ) is the enthalpy change for the formation equation when all substances are in their standard states. C(graphite) + 2H 2( g   g )

C(graphite) + O2 (g)  CO2 (g)

CH4( g   g )

D H of  =

-74.9 kJ

∆Horxn= - 393.5kJ

An element in its standard state is assigned a D H of  of 0.

Selected Standard Heats of Formation at 25 oC (298 K)

Table 6.5 Formula

H of (kJ/mol)

calcium Ca(s  ) CaO(s  ) CaCO3(s  ) carbon C(graphite) C(diamond) CO(g  ) CO2(g  ) CH4(g  ) CH3OH(l ) HCN(g  ) CS2(l ) chlorine Cl(g  )

0 -635.1 -1206.9

0 1.9 -110.5 -393.5 -74.9 -238.6 135 87.9



H of (kJ/mol)

Cl2(g  )


HCl(g  )


hydrogen H(g  ) H2(g  )

218 0

nitrogen  N2(g  )  NH3(g  )  NO(g  )

0 -45.9 90.3

oxygen O2(g  ) O3(g  ) H2O(g  ) H2O(l )

0 143 -241.8 -285.8


silver  Ag(s  ) AgCl(s  )

H of (kJ/mol)

0 -127.0

sodium  Na(s  )  Na(g  )  NaCl(s  )

0 107.8 -411.1

sulfur  S8(rhombic) 0 S8(monoclinic) 2 SO2(g  ) -296.8 SO3(g  )


Indirect method measuring ∆Hof  • Many compounds cannot be directly synthesized from their elements (proceed too slowly/ side reactions produce other substance than desired compound) • Hess’s law: when reactants are converted to products, the change in enthalpy is the same whether the reaction takes  place in one step or in a series of steps • General rule of applying:  –  Arrange series of chemical equation in a way that when added together all species will cancel except reactant and product that appear in the overall reaction  –  Multiply or reverse equation

Hess’s Law of Heat Summation

The enthalpy change of an overall process is the sum of  the enthalpy changes of its individual steps. steps .

Used to predict the enthalpy change (a) of an overall reaction that cannot be studied directly, and/or (b) of an overall reaction that can be separated into distinct reactions whose enthalpy changes can be measured individually individually..

Example PROBLEM:

Using Hess’s Law to Calculate an Unknown H 

Two gaseous gaseous pollutants that form auto exhaust are CO and NO. An environmental chemist is studying ways to convert them into less harmful gases through the following equation: CO(g  ) + NO(g  )

CO2(g  ) + 1/2N2(g  )

 H = D H =


Given the following information, calculate the unknown D H . Equation A: CO(g  ) + 1/2O2(g  ) Equation B: N2(g  ) + O2(g  ) PLAN:

CO2(g  ) 2NO(g  )



= -283.0 kJ

= +180.6 kJ

Equations A and B have to be manipulated by reversal and/or multiplied by factors in order to sum to the target equation.


Multiply Equation B by 1/2 and reverse it. CO(g  ) + 1/2O2(g  )  NO(g  )

CO(g  ) + NO(g  )

CO2(g  )

1/2N2(g  ) + 1/2O2(g  ) CO2(g  ) + 1/2N2(g  )


= -283.0 kJ


= -90.3 kJ

D H rxn

= -373.3 kJ

• From ∆Hof  values, the standard enthalpy of  reaction ∆Horxn can be calculated aA + bB  cC + dD ∆Horxn = [c∆H [c∆Hof (C) + d∆Hof (D)] –  (D)] – [[a∆Hof (A) + b∆H + b∆Hof (B)] ∆Horxn = ∑n∆Hof (Product) –∑m∆Hof (Reactant)

The general process for determining H orxn from H of  values


Calculating the Standard Heat of Reaction from Standard Heats of Formation

PROBLEM:  Nitric acid, whose worldwide annual production is about 8 billion kg, is used to make many products, including fertilizers, dyes and explosives. The first step in the industrial production process is the oxidation of  ammonia:

4NH3(g  ) + 5O2(g  )

4NO(g  ) + 6H2O(g  )

Calculate D H orxn from D H of values. D H orxn


= S nD H of  (products) -

= [4D H of  NO(g  ) + 6D H of  H2O(g  )] )]


mD H of  (reactants) - [4D H of  NH3(g  ) + 5D H of  O2(g  )] )]

= [(4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol)] [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] D H orxn

= -906 kJ

Bond Energies •

Chemical reactions involve the breaking and making of chemical  bonds.

Energy is always required to break a chemical bond.

The bond energy (B.E.) is the amount of energy necessary to mol e of   bonds in a gaseous covalent substance to form break  one mole

 products in the gaseous state state at constant temperature and pressure. •

The greater the bond energy, the more stable (stronger) the bond is, and the harder it is to break. Thus bond energy is a measure of bond strengths.

A special case of  Hess’s Law involves the use of bond energies to estimate heats of reaction. reaction.

A schematic representation of the relationship between bond energies and Δ H rxn for gas phase reactions.

(a) For a general reaction (exothermic)

(b) For the gas phase reaction : H2 (g) + Br 2 (g)  2HBr (g)

Example 1 Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K for the following reaction:

Example 2 Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K  for the following reaction:

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