Loading in 2 Seconds...
Loading in 2 Seconds...
THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins. Concerned with the study of transformation of energy: Heat work. CONSERVATION OF ENERGY – states that:. Energy can neither be created nor destroyed in chemical reactions. It can only be converted from one form to the other.
Concerned with the study of transformation of energy:
↔matter↔energy ↔ energy not mattermatter× Energy ×
HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l) - heat given off.
If heat released to surroundings – exothermic.
If heat absorbed by surroundings – endothermic.
CH3C(CH3)2CH2CH(CH3)CH3 + 25/2 O2→ 8CO2(g) +H2O(l)
5401 kJ of heat is released (exothermic)
Where does heat come from?
From internal energy, U of gasoline. Can represent chemical reaction:
Uinitial = Ufinal + energy that leaves system (exothermic)
Ui = Uf – energy that enters system (endothermic)
(KE) k T (thermal energy)where k = Boltzmann constant
Now,U = KE + PE
w = force × distance moved in direction of force
i.e. w = mg × h = kg × m s-2 × m = kg m2 s-2
(m) (g) (h)
1 kg m2 s-2 = 1 Joule
- Consider work – work against an opposing force, eg: external pressure, pex. Consider a piston
w = h × (pex × A) = pex × hA
Work done on system = pex × ∆V
∆V – change in volume (Vf – Vi)
Since U is decreased
C3H8(g) + 5 O2(g)→ 3CO2(g) + 4H2O(l)at 298 K 1 atm
(1 atm = 101325 Pa), -2220 kJ = q
What is the work done by the system?
For an ideal gas;
pV = nRT (p = pex)
n – no. of moles
R – gas constant
T = temperature
V – volume
p = pressure
6 moles of gas:
Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3
3 moles of gas:
Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3
work done = -pex × (Vf – Vi)
= -101325 (0.0734 – 0.1467) = +7432 J
= (nf – ni) RT
Work done = -∆ngasRT
i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J
Can also calculate ∆U
∆U = q +w q = - 2220 kJ
w = 7432.7 J = 7.43 kJ
∆U = - 2220 + 7.43 = - 2212.6 kJ
qp ∆U why?
Only equal if no work is done i.e.∆V = 0
i.e. qv = ∆U
System at equilibrium when pex = pint (mechanical equilibrium)
Change either pressure to get reversible work i.e.
pex > pint or pint > pex at constant temperature by an infinitesimal change in either parameter
For ideal gas, pV = nRT
p = nRT/ V
work = p dV = nRT dV/ V
= nRT ln (Vf/Vi) because dx/x = ln x
i.e. V 0 (expansion work).
Definition: H = qp i.e. heat supplied to the system at constant pressure.
i.e H = U + pV (p = pex)
Hf = Uf + pVf
Hf – Hi = Uf – Ui +p(Vf – Vi)
H = U + p V
H = (- pex V + q) +pV (pex= p)
H = U + pi V U + p V
H = U + Vi p
eg: U, H, T and p (IUPAC convention).
What about temperature?
C3H8(g) + 5 O2(g)→ 3CO2(g) + 4H2O(l)
at 1 bar pressure, qp = - 2220 kJmol-1.
H = - 2220 kJ mol-1 at 298 K
represents the standard state.
H2O(l)→ H2O(g), Hvap = +44.0 kJmol-1
Calculate U for the following reaction:
CH4(l) + 2 O2(g)→ CO2(g) + 2H2O(l), H = - 881.1kJmol-1
= U + pi V + Vi p + p V
NB: p = 1 bar, i.e. p = 0
H =U + pi V
Since -pi V = - nRT,
U = H - nRT
U = - 881.1 – ((1 – 2)(8.314) 298)/ 1000
= - 881.1 – (-1)(8.314)(0.298) = - 2182.99 kJ mol-1
At 298 K Carbon = Cgraphite
Hydrogen = H2(g)
Mercury = Hg(l)
Oxygen = O2(g)
Nitrogen = N2(g)
eg. Can compare thermodynamic stability of substances in
their standard state.
Hf of C3H8(g) = - 103.9 kJ mol-1
Hf of O2(g) = 0 (reference state)
Hf of CO2(g) = - 393.5 kJ mol-1
Hf of H2O(l) = - 285.8 kJ mol-1
Hrxn = n H (products)- n H(reactants)
= - 1180.5 -1143.2 = - 2323.7 kJ mol-1
Hf(reactants) = - 103.9 + 5 0 = - 103.9 kJ mol-1
Hrxn = - 2323.7 – (- 103.9) = - 2219.8 kJ mol-1 = - 2220 kJ mol-1
Suppose do reaction at 400 K, need to know
Hf at 298 K for comparison with literature value. How?
pV= nRT = constant
= - nRT ln (Vf/Vi)
For an ideal gas, u = 0
Because: U KE + PE
k T + PE (stored in bonds)
Ideal gas has no interaction between molecules (no bonds broken or formed)
Also H = 0 since (pV) = 0 ie no work done
This applies only for an ideal gas and NOT a chemical reaction.
Vi = nRT/pi = 1 x 8.314 x (298)/202650
= 1.223 x 10-2 m3
Vf = 1 x 8.314 x 298/101325 = 2.445 x 10-2 m3
therefore, w = -pex (Vf- Vi)
= -101325(2.445-1.223) x 10-2 = -1239 J
U = q + w; for a perfect gas U = 0
therefore q = -w and
q = -(-1239) = +1239 J
Srxn = 2S(H2Ol) - (2 S(H2g ) + S(O2g) )
S total =( - 327 JK-1mol-1) + 1.92 x 103 = 1.59 x 103 JK-1 mol-1
Consider the reaction:
N2(g) + 3H2(g)= 2NH3 (g)
Kp =( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3
K = [( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3] eqlb
Kp = K (p)2 in this case. ( Rem: (p)2 / (p)4 = p -2