First Law of Thermodynamics

Doba Jackson, Ph.D.Dept of Chemistry & Biochemistry

Huntingdon College

Outline of Chapter 2

• Basic Concepts: Heat, Work, Energy• First Law of Thermodynamics• Expansion work• Measurement of Heat• Enthalpy changes• Adiabatic changes (Cooling)• Thermochemistry• Inexact Differntials

System, Surroundings, Universe

• System- Part of the world of interest

• Surroundings- Region outside the system

• Open system- Allows matter and energy to pass

• Closed system- Cannot allow matter to pass

• Isolated system- Cannot allow matter or energy to pass

Work• Work- the amount of energy transferred by

motion against an opposing force

▪

Heat, and Heat Transfer

Heat can be transferred byConvectionConductionRadiation

Heat- is the transfer of energy from one body to another by thermal contact

Internal Energy: Energy that exist internal to the system

U = q + wsystem

First Law: The internal energy of an isolated system is a constant; Energy cannot be created or destroyed.

ΔU= change in internal energyq = heat (released or absorbed)w = work done on or by the system

Internal energy can have different forms:

Translational Kinetic EnergyRotational Kinetic energyVibrational Kinetic energy

*Energy within the atom

U = U 0total system surroundingsU

Internal Energy of a monoatomic gas

60

2m mU T U RT

50

2m mU T U RT

Linear molecule

Non-Linear molecule

30

2m mU T U RT

Monoatomic gas

RT

RT

RT

RT RT

Expression for WorkWork (W) = force x distance

W = - F Z

Change equation to relate pressure and volume

Force FP = =

Area AF = P A

W = - P A ΔZ

Volume change (ΔV) = A ΔZ

W = - P ΔV

True expression for Work

dW = - P dVexW= -

f

i

V

exVP dV

Types of Expansion Work• Free Expansion- work done by

expansion against zero opposing force or zero pressure

f

i

V

exVw P dV

0

0exP

W

U q

Types of Expansion Work• Expansion against a constant external pressure

(irreversible)

f

i

V

exVw P dV

If external pressure does, not vary, this expression can be integrated

f

1

V

ex Vw = -P dV

ex f iw = -P V -V

exw = -P ΔV

Types of Expansion Work

• Expansion is reversiblef

i

V

exVw P dV

In this case, the external pressure Pex

exactly equals the internal pressure P at each stage of the expansion.

f

1

V

Vw = - P dV

f

i

V

V

nRTw = - dV

V

exP = P

The integral cannot be evaluated becausetemperature is not a constant throughout the integration.

Types of Expansion Work

• Expansion is Isothermal, reversiblef

i

V

exVw P dV

The system is kept at a constant temperature.The external pressure Pex equals the internalpressure P at each stage of the expansion.

f f

i i

V V

V V

nRT 1w = - dV = -nRT dV

V V

exP = P

Moles is always assumed

constant unless otherwise noted.

f

i

Vw = -nRT Ln

VΔT = 0

Problems: Calculate the work needed for a 65 kg person to climb up 4.0 meters on (a) the earth, (b) the moon (g=1.60 m/s2)

Problem 1: A chemical reaction takes place in a container of cross sectional area of 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an constant external pressure of 121 kPa. Calculate the work (in Joules) done by the system

Problem 2: A sample consisting of 2.00 mol of He is expanded at 22ºC from 22.8 dm3 to 31.7 dm3. Calculate q, w and ΔU (in Joules) when the expansion occurs under the following conditions:  (a) Freely (b) Isothermal, and constant external pressure equal to the final pressure of the gas (c) Isothermal, Reversibly 

Reversible, NonspontaneousReversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initialstate leaving all thermodynamic variables for the universe (system + surroundings) unchanged.

A truly reversible change will:

- Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change

IrreversibleReversible

Irreversible, SpontaneousIrreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initialstate will change some thermodynamic variables of the universe.

A truly irreversible change will:

- Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change

IrreversibleReversible

Heat exchange can be measured in a Bomb Calorimeter

U = q + w

First Law: The internal energy of an isolated system is a constant.

Constant Volume: no work done

exΔU = q - P ΔV

vΔU = q ΔV = 0

Vq = heat at a constant volume

Calorimetry is the study of heat transfer during a chemical or physical process

vΔU = q

Heat

q = C ΔT

Heat is measured by the change in temperature of the water surrounding thecalorimeter

C = Calorimeter constant

The calorimeter constant is the heat capacity of the system

Heat Capacity (Cv) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant volume.

v

VV

ΔU = q

ΔU dU UC = = =

ΔT dT T

VC T

Heat Capacity at a constant volume

We need partial derivatives because the change holds the volume constant

v

VV

ΔU = q

ΔU U dUC = =

ΔT T dT

VC T

Heat Capacity at a constant volumeWe need partial derivatives becausethe change holds the volume constant

VV

U dU ΔUC =

T dT ΔT

How we can evaluate the Heat capacity equation

If the heat capacity (Cv) does not vary withtemperature during the range of temperaturesused, then the equation can be brokendown to:

VΔU = C ΔT

V Vq = C ΔT

Because at constant volume, change in internalenergy is equal to the heat (qv).

Problem 2.4: A sample consisting of 1.00 mol of a perfect gas, for which Cv,m = (3/2)R, initially at P = 1.00 atm and T = 300 K, is heated reversibly to 400 K at a constant volume. Calculate the final pressure, ΔU, q, and w

We cannot measure internal energy when the volume is not

constant

U = q + w

Typical reactions occur a constant external pressure

ΔU = q - PΔV

Heat exchange can be measured in a Differential Scanning Calorimeter at a

constant pressure

U = q + w

First Law: The internal energy of an isolated system is a constant.

Lets define Enthalpy as:

ΔU = q - PΔV

q = ΔU + PΔV

PΔH = q ΔP = 0

Pq = heat at a constant pressure

H = U + PV

Heat Capacity (Cp) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant pressure.

ΔH = q

ΔH dHC = = =

ΔT dT T

P P

PP

C T

H

Heat Capacity at a constant pressure

We need partial derivatives becausethe change holds the pressure constant

dH ΔHC =

T dT ΔTPP

H

How we can evaluate the Heat capacity equation

If the heat capacity (CP) does not vary withtemperature during the range of temperaturesused, then the equation can be brokendown to:

ΔH = C ΔTP

q = C ΔTP P

Because at constant volume, change ininternal energy is equal to the heat (qv).

Problem 2.20: When 2.25 mg of anthracene, C14H10 (s) was burned in a bomb calorimeter, the temperature rose by 1.35 K. Calculate the calorimeter constant for the system. By how much will the temperature rise if 135 mg of phenol (C6H5OH) is burned in the calorimeter under the same conditions?

(ΔcHΘ(C14H10) = -7061 kJ/mol)

Adiabatic Expansion: work is done but no heat enters the system

• When a gas expands adiabatically, work is done but no heat enters or leaves the system.

• The internal energy falls and the temperature also falls.

q = 0U = q + w

VΔU = w = C ΔT

ad V= U = C ΔTw

Adiabatic changes to an Ideal Gas can be related by the

following equations1c

if i

f

VT T

V

1

if i

f

VP P

V

C = CV/nR

γ = Cp/CV

The derivation of these equations is quite long and follows the next slide

Adiabatic Expansion of an Ideal Gas (Derivation)

VC dT = -P dV

U = q + wFor an adiabatic change: q = 0

addU = w

nRTP =

VV

nRTC dT = - dV

V

V

1 nRC dT = - dV

T V

f f

i i

T V

VT V

1 nRC dT = - dV

T V

f f

i i

T V

V T V

1 1C dT = -nR dV

T V

Assume the heat capacity does notvary during the temperature interval

f fV

i i

T VC Ln = - nR Ln

T V

V f f

i i

C T V Ln = - Ln

nR T V

f i

i f

T Vc Ln = Ln

T VVCc = nR

f i

i f

T V=

T V

c

Problem 2.1- A 3.75 mole sample of an ideal gas with Cv,m = 3R/2 initially at a temperature Ti=298 K, and Pi= 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 725 kg mass on the piston of diameter 25.4 cm. Calculate the work done in this process and the distance the piston travels. Assume the mass of the piston is negligible.

Thermochemistry

The study of energy transferred as heat during the course of chemical reactions.

In thermochemistry, we rely on calorimetry to measurethe internal energy (∆U) enthalpy (∆H).

Changes in matter are:

Physical ChangesSolid Liquid Melting (Fusion)Liquid Gas Boiling (Vaporization)Solid Gas Sublimation

Chemical ChangesFormation reactionsCombustion reactionsOther reactions

The Thermodynamic Standard State

3CO2(g) + 4H2O(g)C3H8(g) + 5O2(g)

3CO2(g) + 4H2O(l)C3H8(g) + 5O2(g) ∆H = -2220 kJ

Thermodynamic Standard State: Most stable form of a substance at 1 bar and at a specified temperature, usually 25 °C (for non-solutions).

3CO2(g) + 4H2O(g)C3H8(g) + 5O2(g) ∆H = -2044 kJ

∆H° = -2044 kJ

The standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. The International Union of Pure and Applied Chemistry (IUPAC) recommends using a standard pressure po = 1 bar (100 kilopascals).

Physical and Chemical Transitions

State FunctionsInternal energy and enthalpyare is a state function.

f

iΔU= dU

f

iΔH= dH

Work and Heat are pathfunction. This means we cannever write Δq or Δw

Enthalpies of Physical Change

Hess’s Law

Example of Hess’s Law

Solve for the reaction below

N2H4(g)2H2(g) + N2(g)

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

2NH3(g)3H2(g) + N2(g) ∆H°1 = -92.2 kJ

∆H°3 = ?

2NH3(g)N2H4(g) + H2(g) ∆H°2 = -187.6 kJ

Germain Hess

Example 1: The two reactions below are known methods to produce ammonia by theHaber process. Use the information to solve for the enthalpy change in thereaction below.

Solve for the reaction below

N2H4(g)2H2(g) + N2(g)

2NH3(g)3H2(g) + N2(g) ∆H°1 = -92.2 kJ

∆H°3 =

2NH3(g)N2H4(g) + H2(g) ∆H°2 = -187.6 kJ

Example 1: The two reactions below are known methods to produce ammonia bythe Haber process. Use the information to solve for the enthalpy changein the reaction below.

Steps to solve the problem - Find known reactions that will add up to the desired reaction

- Add up the reactions and add up the ΔH, ΔU, ΔG or other state functions

Θ ΘΔH A B = -ΔH B A

3 2 4 22NH N H (g) + H (g) 2NH3(g)3H2(g) + N2(g)

∆H°2 = 187.6 kJ∆H°1 = -92.2 kJ

2

-92.2 kJ + (187.6 kJ) = 95.4 kJ

Hess’s Law Example

Standard Heats of FormationStandard Heat of Formation (∆fH° ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.

∆H°f = -74.8 kJCH4(g)C(s) + 2H2(g)

Standard states

1 mol of 1 substance

Standard Heats of Formation are used to determine reaction enthalpies (ΔrHº)

cC + dDaA + bB

ReactantsProducts

∆rH° = {c ∆fH°(C) + d ∆fH°(D)} – {a ∆fHº(A) + b ∆fH°(B)}

Reaction enthalpies (ΔrHº) can be determined by the difference of the product enthalpies of formation and the reactant enthalpies of formation.

0 0 0r f f

Products Reactants

Δ H = v Δ H - v Δ H Example:

Standard Heats of Formation

Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.

C6H12O6(s) + 6*O2(g)6*CO2(g) + 6*H2O(l) ∆rH° = ?

ΔrH° = [∆H°f (C6H12O6)] - [6*∆H°f (CO2) + 6*∆H°f (H2O)]

[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]

∆rH° = [(1 mol)(-1273.3 kJ/mol)] -

= 2802.5 kJ

Answer

Products Reactants

Standard State (ΔfHº)= 0

Heats of Combustion (ΔcHº) is a special type of reaction enthalpy

CO2(g) + H2O(l)CH4(g) + O2(g)

= -890.3 kJ

∆cH°= [∆fH°(CO2) + 2 ∆fH°(H2O)] - [∆fH°(CH4)]

[(1 mol)(-74.8 kJ/mol)]

= [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] -

Calculate the standard enthalpy of combustion for methane (CH4(g)).

Answer

ΔfHº (CH4) = -74.8 kJ/mol

ΔfHº (CO2) = -393 kJ/mol

ΔfHº (H2O) = -285 kJ/mol

Combustion reactions are always balanced with 1 as the coefficient of the hydrocarbon

2 2

Temperature dependence of reaction enthalpy (ΔrHº)

• Often standard reaction enthalpies need to be corrected for temperature differences.

o o or p p,m p,m

Products Reactants

Δ C = v C - v C

2

1

T

1 2 pTH T = H T + C dT

2

1

To o or 1 r 2 pT

Δ H T = Δ H T + C dTr

If Cp is independent of temperature during the temperature range, the integral can be evaluated.

The equation doesn’t work if there is a phase transitionthrough the temperature range.

Problem: Balance the reactions and determine ΔrHΘ and ΔrUΘ

23 2g g g gNH NO N H O

NO: 91.3 kJ/molNH3: -45.9 kJ/molH2O: -241.8 kJ/mol