( n ( RT ) ) = pV - - PowerPoint PPT Presentation

SLIDE 1

pV pV = = nRT nRT ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ( ( pV pV ) = ) = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆n ( RT ) n ( RT ) N N2

2 + O

+ O2

2 →

→ → → → → → → 2NO 2NO ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆n = 0 n = 0 ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ( ( pV pV ) = 0 ) = 0 at at const const T, T, ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ H= H= ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ E E N N2

2 + 3 H

+ 3 H2

2 →

→ → → → → → → 2 NH 2 NH3

3

∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆n = n = -

• 2

2 ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆( (pV pV) = ) = -

• 2RT

2RT ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆nRT nRT = = -

• 2 ( 8.314 ) ( 298 ) =

2 ( 8.314 ) ( 298 ) = -

• 4955.14 joules.

4955.14 joules. ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E + E + ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆nRT nRT for reactions of ideal gases for reactions of ideal gases ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H = 2 = 2 ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H Hf

fo

• ( NH

( NH3

3 ) = 2 (

) = 2 ( -

• 46,110 ) =

46,110 ) = -

• 92,220 joules

92,220 joules ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = -

• 92,220 joules done at

92,220 joules done at const const T=298 K T=298 K ( i.e. initial and final state T ( i.e. initial and final state T’ ’s are same ) s are same ) ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E E = = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H -

• ∆

∆ ∆ ∆ ∆ ∆ ∆ ∆nRT nRT = = -

• 92,220

92,220 -

• (

(-

• 4955) =

4955) = -

• 87,265 joules

87,265 joules

SLIDE 2

Example : Convert 1 mole of liquid H Example : Convert 1 mole of liquid H2

2O at 100

O at 100o

• C into

C into 1 mole of H 1 mole of H2

2O vapor at 100

O vapor at 100o

• C and p = 1

C and p = 1 atm atm. . H H2

2O ( l )

O ( l ) → → → → → → → → H H2

2O (

O ( vap vap ) ) Find Find q qp

p = +44,013.6 joules/mole

= +44,013.6 joules/mole ( heat absorbed ) ( heat absorbed ) ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = q qp

p for process done at

for process done at const const p p ∴ ∴ ∴ ∴ ∴ ∴ ∴ ∴ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = 44,013.6 joules/mole 44,013.6 joules/mole What is What is ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E? E? ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E + E + ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ( ( pV pV ) = ) = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E + p E + p ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆V ( V (const const p) p) ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E = E = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H -

• p

p ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆V V 1 mole H 1 mole H2

2O = 18 gm; volume 1 gm of H

O = 18 gm; volume 1 gm of H2

2O at 100

O at 100o

• C is

C is 1.04 1.04 mL mL/gm /gm v vi

i = 18 ( 1.04 ) = 18.72

= 18 ( 1.04 ) = 18.72 mL mL v vf

f =

= nRT nRT / p assume ideal gas / p assume ideal gas

SLIDE 3

v vf

f =

= v vf

f = ( 82 )( 373 )

= ( 82 )( 373 ) mL mL = 30,586 = 30,586 mL mL ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆v = 30,586 v = 30,586 -

• 19 = 30,567

19 = 30,567 mL mL = 30.57 liters = 30.57 liters .082 L .082 L-

• atm

atm = 8.314 joules = 8.314 joules → → → → → → → → 101.4 joules/L 101.4 joules/L-

• atm

atm p p ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆V = 30.57 L V = 30.57 L-

• atm

atm/mole = 30.57 /mole = 30.57 × × × × × × × ×101.4 joules/ L 101.4 joules/ L-

• atm

atm = = 3099.5 joules/mole 3099.5 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E = 44013.6 joules/mole E = 44013.6 joules/mole -

• 3099.5 joules/mole

3099.5 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E = 40914.1 joules/mole E = 40914.1 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = 44013.6 joules/mole H = 44013.6 joules/mole ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆E = E = ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H -

• p

p ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆V V (1 mole)(82 (1 mole)(82 mL mL-

• atm

atm/mole /mole-

• deg

deg)(373 )(373 deg deg)/1 )/1 atm atm

SLIDE 4

General Definitions of heat capacities (for 1 mole of gas): General Definitions of heat capacities (for 1 mole of gas): cp = dQ dT    

P

(sub P means take derivative at constant P) (sub P means take derivative at constant P) cV = dQ dT     V (sub V means take derivative at constant V) (sub V means take derivative at constant V) Substitute Substitute dQ dQ = = dE dE + + pdV pdV= = dE dE -

• dw

dw (1st Law) (1st Law) Remember Remember (1 mole gas) (1 mole gas) E = 3 2RT V = RT P cP = dE dT    

P

+ P dV dT    

P

= d 3 2RT     dT        

P

+ P d RT P     dT        

P

= 3 2 R + PR P dT dT    

P

= (3/2)R + R = (3/2)R + R = (5/2)R = (5/2)R

SLIDE 5

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

SLIDE 6

Thermochemi s t ry Thermochemi s t ry : :

Can measure Can measure ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H directly by making a laboratory determination H directly by making a laboratory determination

• f
• f q

qp

• p. Want to set up a table of enthalpy changes for chemical

. Want to set up a table of enthalpy changes for chemical reactions reactions Standard Enthalpy Change Standard Enthalpy Change ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H°

° ° ° ° ° ° ° for a system

for a system ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H°

° ° ° ° ° ° ° ≡

≡ ≡ ≡ ≡ ≡ ≡ ≡ enthalpy change in a chemical reaction for converting enthalpy change in a chemical reaction for converting reactants in their standard state to products in their standard reactants in their standard state to products in their standard state. state. Standard State of a Substance Standard State of a Substance ≡ ≡ ≡ ≡ ≡ ≡ ≡ ≡ That form of the substance which That form of the substance which is most stable at a pressure of 1 is most stable at a pressure of 1 atm atm and T = 298 and T = 298°

° ° ° ° ° ° ° K

K

SLIDE 7

Exampl e s Exampl e s Carbon : at 1 Carbon : at 1 atm atm, 25 , 25°

° ° ° ° ° ° °C, Stand. St. is graphite (not diamond or coal)

C, Stand. St. is graphite (not diamond or coal) Oxygen : 1 Oxygen : 1 atm atm, 25 , 25°

° ° ° ° ° ° ° C

C

• Std. St. is O
• Std. St. is O2

2 gas

gas Bromine : 1 Bromine : 1 atm atm, 25 , 25°

° ° ° ° ° ° ° C

C

• Std. St. is liquid Bromine
• Std. St. is liquid Bromine

For a reaction: For a reaction: C ( graphite ) + O C ( graphite ) + O2

2 ( g )

( g ) → → → → → → → → CO CO2

2 ( g )

( g ) measure measure q qp

p in lab

in lab ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H298

298° ° ° ° ° ° ° ° =

= -

• 393.52

393.52 kjoules kjoules/mole = /mole = q qp

p

This means evolve This means evolve 393.52 393.52 kjoules kjoules of heat in converting 1 mole

• f heat in converting 1 mole

C graphite, 1 mole O C graphite, 1 mole O2

2 gas, into 1 mole CO

gas, into 1 mole CO2

2 gas at 1

gas at 1 atm atm and 298 and 298°

° ° ° ° ° ° ° K.

K. CO ( g ) + 1/2 O CO ( g ) + 1/2 O2

2 ( g )

( g ) → → → → → → → → CO CO2

2 ( g )

( g ) measure measure q qp

p in lab.

in lab. ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H298

298° ° ° ° ° ° ° ° =

= -

• 282.98

282.98 kjoules kjoules/mole /mole Can also burn CO in lab to produce CO Can also burn CO in lab to produce CO2

2:

: (Heat released when burn CO at (Heat released when burn CO at const const p) p) (Heat released when (Heat released when burn graphite at burn graphite at const const p) p)

SLIDE 8

QuickTime™ and a Video decompressor are needed to see this picture.

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

2 ( g )

( g ) → → → → → → → → CO CO2

2 ( g )

( g ) Can measure Can measure q qp

p in lab.

in lab. Just burn CO in oxygen. Just burn CO in oxygen. C ( graphite ) + O C ( graphite ) + O2

2 ( g )

( g ) → → → → → → → → CO CO2

2 ( g )

( g ) Can measure Can measure q qp

p in lab. Just burn C in oxygen.

in lab. Just burn C in oxygen. C (graphite) + O C (graphite) + O2

2 (g)

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

2 (g)

(g) Can’t measure in the lab. Can’t measure in the lab. (Always get a little CO (Always get a little CO2

2)

)

SLIDE 9

Enthalpy's of Formation Enthalpy's of Formation

Enthalpy of formation of elements in their standard Enthalpy of formation of elements in their standard state is defined to be zero: state is defined to be zero: Standard Enthalpy of formation is ∆H for a reaction where Standard Enthalpy of formation is ∆H for a reaction where a pure compound is formed from its elements with all a pure compound is formed from its elements with all substances in their standard states ( 25 substances in their standard states ( 25°

° ° ° ° ° ° ° C )

C ) C (s) + 1/2 O C (s) + 1/2 O2

2 (g) = CO (g)

(g) = CO (g) ∆H ∆H°

° ° ° ° ° ° ° = ∆

= ∆H Hf

f° ° ° ° ° ° ° ° ( CO ) =

( CO ) = -

• 110.5

110.5 kjoules kjoules H H2

2 (g) + 1/2 O

(g) + 1/2 O2

2 (g) = H

(g) = H2

2O(l)

O(l) ∆ ∆H Hf

f° ° ° ° ° ° ° ° ( H

( H2

2O ) =

O ) = -

• 285.8

285.8 kjoules kjoules H H2

2 (g)+O

(g)+O2

2(g)+C(s) = HCOOH(l) ∆

(g)+C(s) = HCOOH(l) ∆H Hf

f° ° ° ° ° ° ° ° ( HCOOH ) =

( HCOOH ) = -

• 409.2

409.2 kjoules kjoules H H2

2 (g) = H

(g) = H2

2 (g)

(g) ∆ ∆H Hf

f° ° ° ° ° ° ° ° ( H

( H2

2, g ) = 0

, g ) = 0 Why are ∆ Why are ∆H Hf

f° ° ° ° ° ° ° ° useful?

useful?

SLIDE 10

Suppose we want to know ∆H Suppose we want to know ∆H°

° ° ° ° ° ° ° for the reaction:

for the reaction: CH CH3

3COOH = CH

COOH = CH4

4 + CO

+ CO2

2

∆H ∆H°

° ° ° ° ° ° ° = ?

= ? Consider the path at 25 Consider the path at 25°

° ° ° ° ° ° °

CH CH3

3COOH = 2 C (s) + 2 H

COOH = 2 C (s) + 2 H2

2 (g) + O

(g) + O2

2 (g) = CH

(g) = CH4

4 (g) + CO

(g) + CO2

2 1 1 2 2

1 : ∆H 1 : ∆H°

° ° ° ° ° ° ° =

= -

• ∆

∆H Hf

f° ° ° ° ° ° ° ° ( CH

( CH3

3COOH )

COOH ) 2 : ∆H 2 : ∆H°

° ° ° ° ° ° ° = + ∆

= + ∆H Hf

f° ° ° ° ° ° ° ° ( CH

( CH4

4 ) + ∆

) + ∆H Hf

f° ° ° ° ° ° ° ° ( CO

( CO2

2 )

) ∆H ∆H°

° ° ° ° ° ° ° (total) =

(total) = -

• ∆

∆H Hf

f° ° ° ° ° ° ° ° ( CH

( CH3

3COOH ) + ∆

COOH ) + ∆H Hf

f° ° ° ° ° ° ° ° ( CH

( CH4

4 ) + ∆

) + ∆H Hf

f° ° ° ° ° ° ° ° ( CO

( CO2

2 )

) because ∆H because ∆H°

° ° ° ° ° ° ° is independent of path.

is independent of path. In principle can always accomplish a chemical transformation In principle can always accomplish a chemical transformation by following a path which 1st decomposes reactants into by following a path which 1st decomposes reactants into elements then reforms product. elements then reforms product.

SLIDE 11

General result: General result: ∆H ∆H°

° ° ° ° ° ° ° = Sum [ ∆

= Sum [ ∆H Hf

f° ° ° ° ° ° ° ° (products) ]

(products) ] -

• Sum [ ∆

Sum [ ∆H Hf

f° ° ° ° ° ° ° ° (reactants) ]

(reactants) ] ∆H ∆H°

° ° ° ° ° ° ° =

= Σ Σ Σ Σ Σ Σ Σ Σ ∆ ∆H Hf

f° ° ° ° ° ° ° ° (products)

(products) -

• Σ

Σ Σ Σ Σ Σ Σ Σ ∆ ∆H Hf

f° ° ° ° ° ° ° ° (reactants)

(reactants)

There are lots of tables of ∆ There are lots of tables of ∆H Hf

f° ° ° ° ° ° ° ° at 298 K, but not at other

at 298 K, but not at other

• temperatures. Therefore, need T dependence of ∆
• temperatures. Therefore, need T dependence of ∆H

Hf

f° ° ° ° ° ° ° ° .

.

SLIDE 12

C C F H F H C C F H F H

Temperature Dependence of Temperature Dependence of ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H H

Can use Heat capacities to see how Can use Heat capacities to see how ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H varies with temperature H varies with temperature Suppose want to know Suppose want to know ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H for the following reaction at two H for the following reaction at two different temperatures: different temperatures: trans trans 1,2 1,2 difloroethylene difloroethylene cis cis 1,2 1,2 difloroethylene difloroethylene

SLIDE 13

Know Know ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆H = H = q qp

p :

:

Two paths: Two paths:

1) 1) 2) 2) T T1

1

T T2

2

∆H ∆H1

1

∆H ∆H2

2

∆H ∆H12

12(

(trans trans) ) ∆H ∆H21

21(

(cis cis) )

Trans ( T Trans ( T1

1 )

) → → → → → → → → Trans ( T Trans ( T2

2 )

) → → → → → → → → Cis Cis ( T ( T2

2 )

) → → → → → → → → Cis Cis ( T ( T1

1 )

) Trans ( T Trans ( T1

1 )

) → → → → → → → → Cis Cis ( T ( T1

1 )

) Note Note Direction! Direction! T T2

2→

→ → → → → → →T T1

1

T T2

2

T T1

1

C C F H F H C C F H F H C C F H F H C C F H F H

SLIDE 14

T T1

1

T T2

∆H ∆H1

1

∆H ∆H2

2

∆H ∆H12

12(

(trans trans) ) ∆H ∆H21

21(

(cis cis) )

∆H ∆H1

1

∆H2 + ∆H12 ( Trans) = ∆H1 - ∆H21 ( Cis ) Rearrange this to get: ∆H ∆H12

12 (Trans)

(Trans) ∆H ∆H = + ∆H ∆H21

21 (

(Cis Cis) + ∆H ∆H2

2

=

∆H ∆H2

2 + ∆H

+ ∆H12

12 ( Trans) = ∆H

( Trans) = ∆H1

1 + ∆H

+ ∆H12

12 (

( Cis Cis ) )

Note: ∆H Note: ∆H21

21 (

( Cis Cis ) ) = = -

• ∆H

∆H12

12 (

( Cis Cis ) )

C C F H F H C C F H F H C C F H F H C C F H F H

SLIDE 15

T T1

1

T T2

2

∆H ∆H1

1

∆H ∆H2

2

q qp

p(

(trans trans)=∆H )=∆H12

12(

(trans trans) ) q qp

p(

(cis cis) = ∆H ) = ∆H12

12(

(cis cis) )

∆H ∆H2

2 (T

(T2

2) = ∆H

) = ∆H1

1 (T

(T1

1) + ∆H

) + ∆H12

12(

(cis cis) ) – – ∆H ∆H12

12(

(trans trans) ) Since ∆H = Since ∆H = q qp

p →

→ → → → → → → q qp

p(

(cis cis) ) is heat needed to is heat needed to increase temperature of increase temperature of cis cis 1,2 from 1,2 from T T1

1→

→ → → → → → →T T2

2

Assume q depends linearly on T, or Assume q depends linearly on T, or C Cp

p =

= const indep const indep T T → → → → → → → → dH dH = = dq dqp

p

q qp

p(

(trans trans) ) is heat needed to is heat needed to increase temperature of increase temperature of trans trans 1,2 from 1,2 from T T1

1→

→ → → → → → →T T2

2

However, However, q qp

p(

(cis cis) ) and and q qp

p(

(trans trans) ) correspond to correspond to physical

physical

changes not chemical changes, so changes not chemical changes, so → → → → → → → → dH dH/ /dT dT = = dq dqp

p/

/dT dT = = C Cp

p

∆H ∆H12

12(

(cis cis) = ) = q qp

p(

(cis cis) ) and and ∆H ∆H12

12(

(trans trans) = ) = q qp

p(

(trans trans) )

C C F H F H C C F H F H C C F H F H C C F H F H