pV = nRT
∆ ( pV ) = ∆n ( RT )
N2 + O2 → 2NO
∆n = 0
∆ ( pV ) = 0
at const T, ∆ H= ∆ E
N2 + 3 H2 → 2 NH3
∆n = -2
∆(pV) = -2RT
∆nRT = -2 ( 8.314 ) ( 298 ) = -4955.14 joules.
∆H = ∆E + ∆nRT for reactions of ideal gases
∆H = 2 ∆Hfo ( NH3 ) = 2 ( -46,110 ) = -92,220 joules
∆H = -92,220 joules done at const T=298 K
( i.e. initial and final state T’s are same )
∆E = ∆H - ∆nRT = -92,220 -(-4955) = -87,265 joules
Example : Convert 1 mole of liquid H2O at 100o C into
1 mole of H2O vapor at 100oC and p = 1 atm.
H2O ( l ) → H2O ( vap )
Find qp = +44,013.6 joules/mole
( heat absorbed )
∆H = qp for process done at const p
∴ ∆H = 44,013.6 joules/mole
What is ∆E?
∆H = ∆E + ∆ ( pV ) = ∆E + p ∆V (const p)
∆E = ∆H - p ∆V
1 mole H2O = 18 gm; volume 1 gm of H2O at 100oC is
1.04 mL/gm
vi = 18 ( 1.04 ) = 18.72 mL
vf = nRT / p
assume ideal gas
vf =
(1 mole)(82 mL-atm/mole-deg)(373 deg)/1 atm
vf = ( 82 )( 373 ) mL = 30,586 mL
∆v = 30,586 - 19 = 30,567 mL = 30.57 liters
.082 L-atm = 8.314 joules → 101.4 joules/L-atm
p ∆V = 30.57 L-atm/mole = 30.57 ×101.4 joules/ L-atm =
3099.5 joules/mole
∆E = ∆H - p ∆V
∆E = 44013.6 joules/mole - 3099.5 joules/mole
∆E = 40914.1 joules/mole ∆H = 44013.6 joules/mole
General Definitions of heat capacities (for 1 mole of gas):
dQ 

(sub P means take derivative at constant P)
cp =
 dT  P
dQ

(sub V means take derivative at constant V)
cV =
 dT  V
Substitute dQ = dE + pdV= dE - dw (1st Law)
RT
3
V=
Remember E = RT
(1 mole gas)
P
2
 d 3 RT 
  d RT 
2

 P 
dE
dV 




+P
=
cP =
+P
=
 dT P
 dT  P
 dT 
 dT 

P

P
PR  dT
3
R+
= (3/2)R + R = (5/2)R
P  dT P
2
Bonus * Bonus * Bonus * Bonus * Bonus * Bonus
Thermochemistry:
Can measure ∆H directly by making a laboratory determination
of qp. Want to set up a table of enthalpy changes for chemical
reactions
Standard Enthalpy Change ≡ ∆H° for a system
∆H° ≡ enthalpy change in a chemical reaction for converting
reactants in their standard state to products in their standard state.
Standard State of a Substance ≡ That form of the substance which
is most stable at a pressure of 1 atm and T = 298 ° K
Examples
Carbon : at 1 atm, 25°C, Stand. St. is graphite (not diamond or coal)
Oxygen : 1 atm, 25° C
Std. St. is O2 gas
Bromine : 1 atm, 25° C
Std. St. is liquid Bromine
For a reaction:
C ( graphite ) + O2 ( g ) → CO2 ( g )
measure qp in lab
(Heat released when
°
∆H298 = - 393.52 kjoules/mole = qp
burn graphite at const p)
This means evolve 393.52 kjoules of heat in converting 1 mole
C graphite, 1 mole O2 gas, into 1 mole CO2 gas at 1 atm and 298° K.
Can also burn CO in lab to produce CO2:
CO ( g ) + 1/2 O2 ( g ) → CO2 ( g )
measure qp in lab.
∆H298° = - 282.98 kjoules/mole (Heat released when burn CO at
const p)
CO ( g ) + 1/2 O2 ( g ) → CO2 ( g ) Can measure qp in lab.
Just burn CO in oxygen.
C ( graphite ) + O2 ( g ) → CO2 ( g )
Can measure qp in lab. Just burn C in oxygen.
C (graphite) + O2 (g)
→ CO (g) + (1/2) O2 (g)
Can’t measure in the lab.
(Always get a little CO2)
QuickTime™ and a
Video decompressor
are needed to see this picture.
Enthalpy's of Formation
Standard Enthalpy of formation is ∆H for a reaction where
a pure compound is formed from its elements with all
substances in their standard states ( 25° C )
C (s) + 1/2 O2 (g) = CO (g)
H2 (g) + 1/2 O2 (g) = H2O(l)
∆H° = ∆Hf° ( CO ) = -110.5 kjoules
∆Hf° ( H2O ) = -285.8 kjoules
H2 (g)+O2(g)+C(s) = HCOOH(l) ∆Hf° ( HCOOH ) = -409.2 kjoules
Enthalpy of formation of elements in their standard
state is defined to be zero:
H2 (g) = H2 (g)
Why are ∆Hf° useful?
∆Hf° ( H2, g ) = 0
Suppose we want to know ∆H° for the reaction:
CH3COOH = CH4 + CO2
∆H° = ?
Consider the path at 25°
1
2
CH3COOH = 2 C (s) + 2 H2 (g) + O2 (g) = CH4 (g) + CO2
1 : ∆H° = -∆Hf° ( CH3COOH )
2 : ∆H° = + ∆Hf° ( CH4 ) + ∆Hf° ( CO2 )
∆H° (total) = - ∆Hf° ( CH3COOH ) + ∆Hf° ( CH4 ) + ∆Hf° ( CO2 )
because ∆H° is independent of path.
In principle can always accomplish a chemical transformation
by following a path which 1st decomposes reactants into
elements then reforms product.
General result:
∆H° = Sum [ ∆Hf° (products) ] - Sum [ ∆Hf° (reactants) ]
∆H° = Σ ∆Hf° (products) - Σ ∆Hf° (reactants)
There are lots of tables of ∆Hf° at 298 K, but not at other
temperatures. Therefore, need T dependence of ∆Hf° .
Temperature Dependence of ∆H
Can use Heat capacities to see how ∆H varies with temperature
Suppose want to know ∆H for the following reaction at two
different temperatures:
H
F
C
F
H
H
C
C
H
trans 1,2 difloroethylene
F
C
F
cis 1,2 difloroethylene
Know ∆H = qp :
Two paths:
1) Trans ( T1 ) → Cis ( T1 )
2) Trans ( T1 ) → Trans ( T2 ) → Cis ( T2 ) → Cis ( T1 )
H
T1
C
∆H1
C
F
C
∆H12(trans)
T2
F
F
∆H21(cis)
∆H2
C
H
H
H
F
C
T1
C
F
H
H
H
H
F
C
F
C
F
Note
Direction!
T2→T1
T2
∆H1 = ∆H12 (Trans) + ∆H2 + ∆H21 (Cis)
Cis = ∆H
H
C
∆H1
C
F
C
∆H12(trans)
F
F
∆H21(cis)
∆H2
C
H
H
F
C
T1
C
F
H
H
H
H
F
H
C
T2
C
F
Rearrange this to get:
∆H2 + ∆H12 ( Trans) = ∆H1 - ∆H21 ( Cis )
F
Note: ∆H21 ( Cis )
= -∆H12 ( Cis )
∆H2 + ∆H12 ( Trans) = ∆H1 + ∆H12 ( Cis )
∆H2 (T2) = ∆H1 (T1) + ∆H12(cis) – ∆H12(trans)
Since ∆H = qp →
∆H12(cis) = qp(cis) and
∆H12(trans) = qp(trans)
qp(cis) is heat needed to
increase temperature of
cis 1,2 from T1→T2
qp(trans) is heat needed to
increase temperature of
trans 1,2 from T1→T2
H
C
∆H1
C
F
H
qp(trans)=∆H12(trans)
H
F
C
F
C
H
C
C
C
F
However, qp(cis) and qp(trans) correspond to physical
changes not chemical changes, so →
dH = dqp
T1
F
F
qp(cis) = ∆H12(cis)
H
H
∆H2
C
H
H
F
dH/dT = dqp/dT = Cp
Assume q depends linearly on T, or Cp = const indep T →
T2
F