The First Law

Classical Thermodynamics

Thermodynamics has at its foundation three fundamental laws named, not very imaginatively: the First Law, the

Second Law, and the Third Law. There are no known exceptions to these laws.

This week will be devoted to examination of the First Law.

This law, remarkably powerful given its simplicity, allows us to address questions such as whether a gas will cool upon expansion, or to calculate the energy changes for chemical reactions.

A colloquial, but perfectly acceptable, statement of the First Law is: Energy can neither be created nor destroyed, but it may be distributed in different ways. Or, more succinctly:

Energy is conserved.

First Law History

Considerable philosophical and scientific effort has gone into addressing the concept of “heat”

Heat is an invisible fluid, called “caloric”, which flows from warmer bodies to cooler ones. — Antoine Lavoisier

First Law History

Heat is a Form of Motion: An Experiment in Boring Cannon Benjamin Thompson (Count Rumford), Philosophical

Transactions (vol. 88), 1798

Careful analysis of weight and temperature changes for various processes led to suggestion of work/heat

equivalence

First Law History

James Prescott Joule

First to quantitatively establish the equivalence of heat and work

many other contributions, including pioneering refrigeration via gas expansion

Some Definitions

There are two ways that energy can be transferred between a system and its surroundings, work (w) and heat (q).

The system:

The part of the world under investigation

The surroundings:

everything else

Work, w: transfer of energy as a result of unbalanced forces

+w –w Convention:

positive, +w, energy of system increases,

“work is done on the system” (opposite case, –w, “the

system does work”)

Heat, q: transfer of energy resulting from a temperature difference (cf. zeroth law of thermodynamics)

Convention:

positive, +q, heat is

input to the system +q

–q

Gas Expansion and Work

Consider the work done by the gas on the surroundings in an expansion as a result of the difference in pressure exerted by and on the gas (unbalanced forces):

Work is required to raise a mass, M, a distance, h, against gravity, g

Mgh w = −

A Ah w = − Mg

pressure area

force

= ^{area⋅height = volume}

V P

w = −

_ext

Δ

negative positive w

V →

Δ

Pext

P_f = A

P Mg P_i > _ext =

remove the pins

P_i ,V_i

M

M

P_f ,V_f h

i = initial f = final

Self assessment insert here

•  If the internal pressure of 1 L of an ideal gas is twice the external pressure, by how much will the volume of the gas expand as it does work on the surroundings if the

temperature is held constant?

•  Answers a) the gas will not expand, b) 0.5 L, c) 1 L, d) 2 L

•  correct answer is c

Gas Compression and Work

Consider the work done on the gas by the surroundings in a compression as a result of the difference in pressure exerted by and on the gas (unbalanced forces):

A P Mg

P_i < _ext =

A Ah Mgh Mg

w = − = − V P

w = −

_ext

Δ

Same result as for expansion, but ΔV < 0 so w is positive

positive negative w

V →

Δ

Pext

P_f =

Work is done by a mass, M, falling a distance, h, against gravity, g

remove the pins

P_i ,V_i

M

M

P_f ,V_f h

i = initial f = final

Variations in Pressure

remove the pins

P_i ,V_i

M

M

P_f ,V_f

In this experiment P_ext remains constant during the expansion and

w = − P

_ext

Δ V

For constant P_ext, we recover

€

w =−P_ext dV

V_i V_f

∫

^{= −Pext}

(

^{Vf −Vi}

)

^{= −PextΔV}

If P_ext is not constant during the expansion, the work must be computed as the integral over the path from P_i,V_i to P_f,V_f , and one must know how P_ext varies with V,

€

w = − P

_ext

dV

V_i V_f

∫

A fully general

expression

i = initial f = final

Work Is Area Under a P

_ext

V Curve

Noting the relationship between a one-dimensional integral and geometric area, work (w) is the area under the P_ext vs. V curve,

€

w = − P

_ext

dV

V_i V_f

∫

Consider an isothermal compression at constant pressure P_ext

The work is equal to the shaded areas and depends on P_ext.

Curve for an ideal gas, at constant T

P

ext

P

_f

= P

_f

< P

_ext

Pf P_f

V P

w=− _extΔ w=−P_extΔV

V V

P nRT 1

∝

=

Reversible Isothermal Compression

Work depends on the path taken from V₁ to V₂. For a compression, the

minimum work is done along the reversible path. In infinitesimally small steps, P_ext is made infinitesimally larger than P_gas. Thus, at every step P_ext is equal to the equilibrium gas pressure P_gas,

€

w_rev = − P_gasdV

V₁ V₂

∫

€

ideal gas :P_gas = nRT V

⎛

⎝ ⎜ ⎞

⎠ ⎟

€

w_rev = − nRT

V dV

V₁ V₂

∫

^{= −nRT} V₁ ^dV_V V₂

∫

1 2

rev ln

V nRT V

w = −

€

P_ext = P_f

Compression implies V₂ < V₁ so work is positive, as it should be

Constant P_ext path with minimum w

Pf P_i reversible

path

Reversible Isothermal Expansion

Constant P_ext path with maximum w

For an expansion, the maximum work is done on the

surroundings along the reversible path. It is the same work (with opposite sign) as that required for compression when traveling in the opposite direction. Thus, it is indeed a reversible path.

reversible path Pf

Pi

€

P_ext = P_f €

w_rev = − nRT

V dV

V₁ V₂

∫

^{= −nRT} V₁ ^dV_V V₂

∫

1

rev ln 2

V nRT V

w = −

Expansion implies V₂ > V₁ so work is negative

Reversibility As a Limit

Consider three ways to isothermally (so PV = constant) expand an ideal gas from 0.5 dm³ and 4 bar to 1.0 dm³ and 2 bar.

w = –100 J w = –117 J w = –139 J

One step Two steps Reversible

State Functions

A state function is a property that depends only upon the state of the system. That is, it is independent of how the system was brought to that state (independent of the path).

Energy is a state function:

€

dU = U₂

1

∫

2 ^{−U1 = ΔU} Energy is independent of the path from 1 to 2; it depends only on the initial (1) and final (2) states.

A key property of a state function is that its differential can be integrated in a normal, path independent way.

Work and heat are not state functions.

€

w = − P

_ext

dV

1

∫

2 We’ve already seen that work depends

on the path from state 1 to state 2 (different P_ext lead to different w)

Different Differentials

Since work depends upon how a process is carried out, work is not a state function, work is a path function, so we write,

€

δ^w

1

∫

2 ^{= w}

(

^{not Δw or w2 − w1}

)

An inexact differential

Cannot be integrated in the normal way

Energy is a state function, and dU is an exact differential.

€

δ^w

1

∫

2 ^{= w}

(

^{not Δw or w2 − w1}

)

Path functions:

State functions:

€

δ^q

1

∫

2 ^{= q}

(

^{not Δq or q2 − q1}

)

€

dU =U₂

1

∫

2 ^{−U1 = ΔU}

The First Law of Thermodynamics

Energy is conserved

Even though δ^{q and} δw are path functions (inexact differentials), their sum is a state function (an exact differential).

w q

dU = δ + δ w q

U = + Δ

(differential form)

(integral form)

All Reversible Roads Lead To…

1 2 2

1 1

1

, V , T P , V , T

P →

^{Path A}: reversible

isothermal expansion

Path B+C: reversible adiabatic expansion followed by heating at constant volume.

Path D+E: reversible constant-pressure

expansion followed by cooling at constant

volume.

ΔU must be the same for all paths, but q and w?

All Reversible Roads Lead To…

1 2 2

1 1

1

, V , T P , V , T

P !

^{Path A}: reversible

isothermal expansion

Path B+C: reversible adiabatic expansion followed by heating at constant volume.

Path D+E: reversible constant-pressure

expansion followed by cooling at constant

volume.

ΔU must be the same for all paths, but q and w?

Path A

Reversible isothermal expansion

1 2 2 1

1

1,V ,T P ,V ,T

P !

Since the process is reversible,

!

"#q_rev,A = #w_rev,A = "P_gasdV = " RT₁

V dV And we have,

!

"q_rev,A = w_rev,A = "RT₁ dV

V = "RT₁lnV₂

V₁

V₁ V₂

#

Since the energy of an ideal gas depends only on T

!

" U

_A

= 0

!

"q_rev,A = w_rev,A

w q U = +

!

Note that heat transfer in is

required to maintain temperature

Path B

!

P

₁

,V

₁

,T

₁

" # " P

₃

,V

₂

,T

₂

Reversible adiabatic expansion Adiabatic means no energy is transferred as heat, i.e., q = 0, and

therefore ΔU = w and dU = δw For an ideal gas, U depends only on T

!

C_V (T) = "U

"T

#

$ % &

' (

V

ideal gas

) ) ) * C_V (T) = dU

dT ) * ) dU = C_V (T)dT

!

w_rev,B = "U_B = dU

T₁ T₂

#

⁼

#

_T^T₁^{2 CV (T)dT}

(from T₁ to T₂)

Path C

Reversibly heat at constant volume

1 2 2

C 2

2

3

, V , T P , V , T

P ! !"

ΔV = 0, so

w

_rev,C

= ! PdV = 0

!

q

_rev,C

= " U

_C

= C

_V

(T ) dT

T₂ T₁

#

(from T₂ to T₁)

That leaves only heat, i.e.,

! U

_C

= q

_rev,C

+ w

_rev,C

= q

_rev,C

+ 0

Paths B + C

For the sum of B + C

!

q_rev,B+C = q_rev,B + q_rev,C = 0 + C_V (T)dT

T₂ T₁

"

!

w_rev,B+C = w_rev,B + w_rev,C = C_V (T)dT

T₁ T₂

"

^{+ 0}

For the energy,

!

"U_B+C = "U_B + "U_C = C_V (T)dT

T₁ T₂

#

⁺ T₂ ^{CV (T)dT} T₁

#

^{= 0}

ΔU=0, the same as for path A (as must be true for a state function), but wrev,A ! wrev,B+C _, q_rev,A ! q_rev,B+C

Paths D + E

For the sum of D + E

w_rev,D+E = w_rev,D + w_rev,E = !P₁

(

V₂ !V₁

)

^{+ 0}

q_rev,D+E = P₁

(

V₂ !V₁

)

! So,

"U_D+E = C_V (T)dT

T₁ T₃

#

⁺

#

_T^T₃^{1CV (T)dT = 0}

Another illustration that it’s usually easiest to get q by

difference from more easily computed ΔU and w

Comparison of Paths

ΔU=0 for all paths (state function), but q_rev and w_rev differ

!

q_rev,B+C = C_V (T)dT

T₂ T₁

"

!

w_rev,B+C = C_V (T)dT

T₁ T₂

"

!

w_rev,A = "RT₁lnV₂ V₁

1 1 2

A

rev, ln

V RT V

q =

(

2 1

)

1 E

D

rev, P V V

q ₊ = !

(

2 1

)

1 E

rev, P V V

w _D+ = ! !

!

P₁,V₁,T₁ " P₂,V₂,T₁

Comparison of Paths

ΔU=0 for all paths (state function), but q_rev and w_rev differ

€

q_rev,B+C = C_V (T)dT

T₂ T₁

∫

€

w_rev,B+C = C_V (T)dT

T₁ T₂

∫

€

w_rev,A = −RT₁lnV₂ V₁

1 1 2

A

rev, ln

V RT V

q =

(

2 1

)

1 E

D

rev, P V V

q ₊ = −

(

2 1

)

1 E

rev, P V V

w _D+ = − −

€

P₁,V₁,T₁ → P₂,V₂,T₁

Quantitative Comparison of Paths

€

P₁,V₁,T₁ → P₂,V₂,T₁

For example if P₁ = 4.0 bar, V₁ = 0.5 dm³, P₂ = 2.0 bar, V₂ = 1.0 dm³, and we have 0.1 moles of ideal monatomic gas:

€

q

_rev,D+E

= 200 J

€

w_rev,D+E = −200 J

€

Δ U

_D+E

= 0

€

w_rev,A = −139 J

€

q_rev,A =139 J

= 0 ΔU_A

€

q

_rev,B+C

= 111 J

€

w

_rev,B+C

= − 111 J

€

ΔU_B+C = 0

Adiabatic Expansion Cools a Gas

Adiabatic, so q = 0 and

€

dU = δw = dw

(note that if either δ^{q = 0 or} δw = 0 then the remaining differential becomes exact)

For an ideal gas reversible expansion:

dT T

C dU

dw = = _V ( )

€

dw = −PdV = − nRTdV and V

Putting them together,

€

C_V(T)dT = − nRT

V dV ⎯ → ⎯ C _V (T) T

T₁ T₂

∫

^{dT = −R} V₁ ^dV_V V₂

∫

^=−RlnV_V²

1

For a monatomic ideal gas,

2 C_V = 3R

€

3R 2

dT T

T₁ T₂

∫

^{= 3R}₂ ^lnT_T²

1

= −RlnV₂

V₁ ⎯ → ⎯ T₂ T₁

⎛

⎝ ⎜

⎞

⎠ ⎟

3 / 2

= V₁ V₂

The gas cools as it expands

Adiabatic vs Isothermal Ideal Gas Law

Cf. an adiabatic process (ideal monatomic gas):

€

T

₂

T

₁

⎛

⎝ ⎜ ⎞

⎠ ⎟

3 / 2

= V

₁

V

₂

PV=nRT

⎯ ⎯ ⎯ ⎯ → P

₂

V

₂

P

₁

V

₁

⎛

⎝ ⎜ ⎞

⎠ ⎟

3 / 2

= V

₁

V

₂

3 / 5 2 2

3 / 5 1

1

V P V

P =

Boyle’s law for an isothermal process:

2 2

1

1 V P V

P =

less compression; with nowhere to dump heat, temperature rises

Statistical Mechanics of w and q

Recall from stat mech: ( ) ( ) ( ) ^{( )}

( ^!)

!

!

!

, , ,

, with

, ,

,

,

V N Q V e

N p V

N E V

N p U

V N E j

j j

j

" j

=

=

#

Differentiate:

!

dU = p_jdE _j

j

"

^{+ E jdpj}

j

"

^{= pj$ #E}_#V^j

% & ' ( )

N

dV

j

"

^{+ E jdpj}

j

"

(^{N V})

E E_j = _j ,

Compare this to

dU = ! w

_rev

+ ! q

_rev

!^w_rev = p_j

(

N,V,!

)

^!Ej

!V

"

#$ %

&

'

N

dV

j

(

Infinitesimal change in the energy

levels, probability remains the same.

!^q_rev = E_j

(

N,V

)

^dpj

(

^N,V,!

)

j

!

Infinitesimal change in the probability of

levels, energy levels remain the same.

Statistical Mechanics of w and q

!

"w_rev = p_j(N,V,#) ^$Ej

$V

%

&

' (

) *

N

dV

j

+

!

"^q_rev = E_j(N,V)^dpj(^N,V,#)

j

$

Infinitesimal change in the energy levels, probability remains the same.

Infinitesimal change in the probability of levels, energy levels remain the same.

energy energy Increase E Decrease E

energy Increase E

Decrease E ^en

ergy

WORK

HEAT

Pressure

!

"w_rev = p_j

(

N,V,#

)

^{$E j}

$V

%

&

' (

) *

N

dV

j

+

Given,

And the definition of pressure-volume work,

!

" w

_rev

= # P dV

We can determine the pressure as,

!

P = " p_j

(

N,V,#

)

^{$E j}

$V

%

&

' (

) *

N

= " $E _j

$V

%

&

' (

) *

j N

+

Cf. Video 3.4 where we used this relationship without derivation to prove that the partition function for the monatomic ideal gas is consistent with the ideal gas equation of state

Constant Pressure Conditions

For a reversible process where the work is restricted to pressure-volume work:

€

Δ U = q + w = q − PdV

V₁ V₂

∫

q

V

U =

At constant volume V₁=V₂:

Δ

constant volume

ΔU is the heat at constant volume (can be measured using bomb calorimetry)

However, in chemistry it is often more convenient to work at constant pressure, for which the heat is,

V P

U dV

P U

q

^V

P

= Δ +

ext

∫

V1²

= Δ + Δ

constant pressure (not a function of V)

So q_P is not equal to ΔU.

Enthalpy: A State Function

€

q

_V

= Δ U

At constant volume, ; q_V is a state function

€

dH = dU + PdV + VdP

^(general)

€

Δ H = Δ U + P Δ V

At constant pressure:

confirming that the more general enthalpy is equal to the heat at constant pressure,

Δ H = q

_P

H has the same role at constant P that U has at constant V

V P

U

q

_P

= Δ + Δ

At constant pressure, is also a state function

PV U

H = +

Define the enthalpy: (general)

Enthalpy vs Internal Energy

Ice (H₂O) melting at 273 K and one atm, q_P = 6.01 kJ•mol^–1, so

€

Δ H = q

_P

= 6.01 kJ • mol

⁻¹

What is ? ΔU

€

ΔU =

(

6.01 kJ• mol⁻¹

)

⁻

(

^{1 atm}

) (

^{0.0180 L •mol−1 −0.0196 L • mol−1}

)

€

ΔU =

(

6.01 kJ • mol⁻¹

)

⁺

(

^{1.60 ×10−3 L • atm• mol−1}

)

0.008314 kJ 0.08206 L • atm

⎛

⎝ ⎜ ⎞

⎠ ⎟ ≈ 6.01 kJ • mol⁻¹

Because ΔV is very small, PΔV is also very small, and so there is neglibgible difference between ΔH and ΔU

€

Δ U ≈ 6.01 kJ • mol

⁻¹

273 K molar volumes are

€

V _s = 0.0196 L • mol⁻¹

€

V _l = 0.0180 L • mol⁻¹

solid:

liquid:

rare and important!

V P H

U = Δ − Δ

Δ (at constant pressure)

Enthalpy vs Internal Energy

Water (H₂O) boiling at 373 K and one atm, q_P = 40.7 kJ•mol^–1, so

€

Δ H = q

_P

= 40.7 kJ • mol

⁻¹

What is ? ΔU

€

ΔU =

(

40.7 kJ •mol⁻¹

)

⁻

(

^{1 atm}

) (

^{30.6 L • mol−1 −0.0180 L • mol−1}

)

€

ΔU =

(

40.7 kJ• mol⁻¹

)

⁻

(

^{30.58 L • atm• mol−1}

)

0.008314 kJ 0.08206 L • atm

⎛

⎝ ⎜ ⎞

⎠ ⎟ = 37.6 kJ• mol⁻¹

The ΔU term relates to the energy to overcome intermolecular forces in the liquid, the ΔH term is larger as it includes the PΔV expansion work on going

from liquid to vapor

€

Δ U = 37.6 kJ • mol

⁻¹

373 K molar volumes are

€

V _g = 30.6 L • mol⁻¹

€

V _l = 0.0180 L • mol⁻¹

gas:

liquid:

V P H

U = Δ − Δ

Δ (at constant pressure)

Heat Capacity is a Path Function

The amount of energy required to raise the temperature of a substance by one degree is different if done at constant V or constant P:

At constant V, the energy added as heat is q_V,

(

ΔU = qV

)

T q T

U T

C U ^V

V

V = Δ

Δ

≈ Δ

⎟⎠

⎜ ⎞

⎝

⎛

∂

= ∂

At constant P, the energy added as heat is q_P,

(

ΔH = qP

)

T q T

H T

C H ^P

P

P = Δ

Δ

≈ Δ

⎟⎠

⎜ ⎞

⎝

⎛

∂

= ∂

Heat Capacities of Ideal Gas

For an ideal gas:

€

H = U + PV

= U + nRT

Differentiating:

€

dH

dT = dU

dT + nR

For an ideal gas, U and H depend only on T, not P or V

So:

€

∂ H

∂ T

⎛

⎝ ⎜ ⎞

⎠ ⎟

P

= ∂ U

∂ T

⎛

⎝ ⎜ ⎞

⎠ ⎟

V

+ nR

Or:

€

C

_P

= C

_V

+ nR

Recall that for a monatomic ideal gas, C_V = (3/2)R, so the difference between C_P and C_V is 67% of C_V

Determining Enthalpy

€

C_P = ∂H

∂T

⎛

⎝ ⎜ ⎞

⎠ ⎟

P

→ dH = C_PdT → H T

( )

₂ ^{− H T}

( )

^{1 =}

∫

_T^T₁^{2 CP}

( )

^{T dT}

The difference in enthalpy at two different temperatures is

determined from integration of C_P over the temperature range:

This is true only if there is no phase transition occurring between T₁ and T₂. At a phase transition, there is no change in the temperature as you add heat (C_P →∞), so one must also add any enthalpy associated with a phase change where needed:

€

H T

( )

^{− H}

( )

^{0 =}

∫

₀^{Tfus CPs}

( )

^{T ʹ′ dT ʹ′ + ΔfusH +}

∫

_T^T_fus ^CPl

( )

^{T ʹ′ dT ʹ′}

Example:

Solid, from T=0 to T=T_fus

Enthalpy of fusion

( ) ( )

fus s

fus l

fusH = H T − H T

Δ

Liquid, from T=T_fus to T=T

Enthalpy of Benzene

Benzene: T_fus=278.7 K, T_vap=353.2 K

Measuring the heat capacity, temperature by temperature

€

H T

( )

^{− H}

( )

^{0 =}

∫

₀^{Tfus CPs}

( )

^{T ʹ′ dT ʹ′ + ΔfusH +}

∫

_T^T_fus^vapCPl

( )

^{T ʹ′ dT ʹ′ + ΔvapH +}

∫

_T^T_vap^CPg

( )

^{T ʹ′ dT ʹ′}

For T > T_vap,

fusH Δ

vapH Δ

solid

gas

Integrating the heat capacity, adding phase changes

Heat of Reaction

Heat may be absorbed or evolved in a chemical reaction.

When that occurs at constant pressure ΔH = q_P and we define:

€

Δ

_r

H = H

_products

− H

_reactants

endothermic

r > 0 Δ

= H

q_P

Absorbs energy as heat, ‘uphill’. Heat must be supplied to

drive the reaction.

reactants products

r >0 Δ H

(endo=in)

exothermic

r < 0 Δ

= H

q_P

Releases (evolves) energy as heat,

‘downhill’. Heat is

produced. ^products

reactants

enthalpy

r <0 Δ H

(exo=out) (reaction)

Thermochemistry Examples

Exothermic: combustion of methane,

€

CH

₄

( ) g + 2O

²

( ) g ⎯ → ⎯ CO

²

( ) g + 2H

²

O l ( )

€

Δ_rH = −890.36 kJ at 298 K (heat is evolved)

Endothermic: water-gas reaction,

€

C s ( ) + H

²

O g ( ) ⎯ → ⎯ CO g ( ) + H

²

( ) g

€

Δ_rH =131 kJ at 298 K (heat is required to drive the reaction)

(also referred to as a “heat of combustion” when O₂ is a reactant)

Enthalpy Is Additive

ΔH is a state function, which means it is an additive property

€

C s ( ) + 1

2 O

₂

( ) g ⎯ → ⎯ CO g ( )

^ΔrH

( )

^{1 = −110.5 kJ}

€

CO g ( ) + 1

2 O

₂

( ) g ⎯ → ⎯ CO

²

( ) g

€

Δ_rH

( )

2 ^{= −283.0 kJ} Given Δ_rH values for (1) and (2)

(1) (2)

Summation provides Δ_rH for (3)

€

C s ( ) + O

2

( ) g ⎯ → ⎯ CO

²

( ) g

(3)

€

Δ_rH

( )

3 ⁼

(

^{−110.5 kJ}

)

⁺

(

^{−283.0 kJ}

)

^{= −393.5 kJ}

+

Hess’ Law

The additivity of Δ_rH is known as Hess’ Law Another example:

€

2P s

( )

^{+ 3Cl}2

( )

g ^{⎯ → ⎯ 2PCl3}

( )

^l

given (1) (2)

€

2P s

( )

^+5Cl2

( )

g ^{⎯ → ⎯ 2PCl5}

( )

^s

€

Δ_rH( )1 ^{= −640 kJ}

( )

^{2 887 kJ}

r = −

Δ H

calculate (3)

€

PCl₃

( )

l ^{+ Cl}2

( )

g ^{⎯ → ⎯ PCl5}

( )

^s

€

Δ_rH( )3 ^=?

€

2PCl₃

( )

l ^{⎯ → ⎯ 2P s}

( )

^{+ 3Cl}2

( )

g

(–1)

€

Δ_rH( )−1 ^{= 640 kJ}

Reverse (1), add (2), divide sum by 2:

(2)

€

2P s

( )

^+5Cl2

( )

g ^{⎯ → ⎯ 2PCl5}

( )

^{s ΔrH}

( )

^{2 = −887 kJ}

€

2PCl₃

( )

l ^+2Cl2

( )

g ^{⎯ → ⎯ 2PCl5}

( )

^s

€

Δ_rH(−1+2) ^{= −247 kJ}

€

PCl₃

( )

l ^{+ Cl}2

( )

g ^{⎯ → ⎯ PCl5}

( )

^{s ΔrH}

( )

^{3 = −124 kJ} (3)

÷2

Standard Enthalpy of Reaction

Δ_rH is extensive; its value depends on the number of moles of the reacting species

To facilitate tabulation, the International Union of Pure and Applied Chemistry created the standard enthalpy of reaction, which is intensive

€

Δ

_r

H °

E.g.,

€

C s ( ) + O

2

( ) g ⎯ → ⎯ CO

²

( ) g

€

Δ_rH° = −393.5 kJ • mol⁻¹

(one mole of C is combusted)

€

2C s

( )

^{+ 2O2}

( )

^{g ⎯ → ⎯ 2CO2}

( )

^g

Then, Δ_rH = 2Δ_rH° = −787 kJ

(intensive)

(extensive)

(at 298 K)

o implies one mole of a specified reagent and all reactants and products in their standard states at a given temperature.

(standard states are chosen by convention, for example, for a gas it is a pressure of one bar)

Standard Molar Enthalpy of Formation

€

Δ

_f

H °

The standard enthalpy of reaction to form one mole of a substance from its constituent elements in their naturally

occurring elemental forms defines the intensive standard molar enthalpy of formation,

indicates all reactants and products in their conventional standard states

€

H

₂

( ) g + 1

2 O

₂

( ) g ⎯ → ⎯ H

²

O l ( )

^{Δf H° = −285.8 kJ ⋅mol−1}

(at 298.15 K)

Standard-state phases at 1 bar and 298.15 K

One mole of H₂O(l) is 285.8 kJ downhill in enthalpy from its constituent elements

Thermochemical Abbreviations

subscript meaning vap

sub fus trs mix ads c f r

vaporization, evaporation sublimation

melting, fusion

transition between phases mixing of fluids or gases adsorption

combustion formation reaction

Elemental Heats of Formation

To assign specific values for Δ_fH°, the values of Δ_fH° for

pure elements in their most stable forms at one bar and the temperature of interest is set to zero.

298 K:

€

Δ_fH° = 0 kJ • mol⁻¹

( )

^g

H₂

( )

^g

O₂

€

Δ_fH° = 0 kJ • mol⁻¹

( )

^g

Cl₂

€

Δ_fH° = 0 kJ • mol⁻¹

( )

^g

Br₂

€

Δ_fH° = 30.907 kJ • mol⁻¹

( )

^g

I₂

€

Δ_fH° = 62.438 kJ • mol⁻¹

(

^diamond

)

C

€

Δ_fH° =1.897 kJ • mol⁻¹

pure elements, but not in their most stable forms at 1 bar and 298 K

Using Δ

_f

H° to Get Δ

_r

H

€

aA + bB ⎯ ⎯ → yY + zZ

One can use Hess’ law, together with the standard enthalpies of formation for each of the reactants and products, to compute a heat of reaction at a temperature of interest, e.g., 298 K:

€

aΔ_fH°

[ ]

A

€

bΔ_fH°

[ ]

B ^{y Δf H°}

[ ]

^Y

€

zΔ_fH°

[ ]

Z

(# moles)

(per mole)

Δ

_r

H = Δ

_f

H °(products) – Δ

_f

H °(reactants)

€

Δ

_r

H = ( y Δ

_f

H ° [ ] Y + z Δ

^f

H ° [ ] Z ) − ( a Δ

^f

H ° [ ] A + b Δ

^f

H ° [ ] B )

Relating Δ

_r

H Values at Different T

To convert Δ_rH from T₁ (e.g., 298 K) to T₂ requires C_P

€

Δ H

₃

=

_T

C

_P

( products )

1

T₂

∫ dT

€

Δ H

₁

= −

_T

C

_P

( reactants )

1

T₂

∫ dT

(note sign and order of limits)

€

Δ_rH T

( )

₂ ^{= ΔrH T}

( )

^{1 +}

∫

_T^T₁²

[

^CP

(

^products

)

^{− CP}

(

^reactants

) ]

^dT

€

Δ

_r

H T ( )

₂

= Δ H

¹

+ Δ H

²

+ Δ H

³

T₂ T₂

T₁ T₁

€

Δ_rH T

( )

₂