Part 1. - Chem - Thermodynamics
Part 1. - Chem - Thermodynamics
Part 1. - Chem - Thermodynamics
KE = ½ mv2
Where m is mass (kg), v= velocity (m/s)
4184 J = 4.184 kJ
absorbed by
In endothermic processes, heat is _________
the system.
melting
boiling
sublimation
released by
In exothermic processes, heat is ________
the system.
freezing
condensation
deposition
Water Phase Change
Diagram
• Exothermic process is any process that gives off heat –
The energy will be listed as a product.
ΔH = Hproducts – Hreactants
Practice Question:
How much heat is evolved when 266 g of white
phosphorus (P4) burns in air? ΔHreaction = -3013 kJ
2 H2(g) + O2(g) → 2 H2O(g) ΔH = – 483.6 kJ
q = m c ΔT AND q = m cX
– q )
d (
o ve g
rem l/g
e a t
h
Temp
← l
) →
.
s/l d ( +q
a d de
s t
h ea
HEAT
Constant Pressure Calorimetry
Calorimetry:
The measurement of heat flow
• Commonly called
“COFFEE CUP” calorimetry
• Atmospheric pressure
remains constant during the
reaction .
Practice Problem
A lead (Pb) pellet having a mass of 26.47 g at 89.98°C
was placed in a constant-pressure calorimeter
containing 100.0 mL of water. The water temperature
rose from 22.50°C to 23.17°C.
We know the masses of water and the lead pellet as well as the initial and final
temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost
by the lead pellet to the heat gained by the water. Knowing the specific heat of water,
we can then calculate the specific heat of lead.
Because the heat lost by the lead pellet is equal to the heat
gained by the water,
qPb = −280.3 J.
Combustion reactions
are studied using constant
volume calorimetry.
It requires a BOMB
CALORIMETER.
We assume that no energy escapes into the
surroundings, so that the heat absorbed by the bomb
calorimeter equals the heat given off by the reaction.
Hess’ Law 1840
–477.2 kJ –1754.2 kJ
ΔHorxn = (–1754.2 kJ) – (–477.2) = –1277 kJ
kJ)
So…
for 2 mol
(i.e., 64 g)
X = ΔH = –4910 kJ of CH3OH
Practice problem #1
Chemical
Thermodynamics
Entropy on the Molecular Scale
Chemical
Thermodynamics
Entropy on the Molecular Scale
Chemical
Thermodynamics
Entropy on the Molecular Scale
Implications:
• More Particles
-> more states -> more entropy
• Higher Temp
-> more energy states -> more entropy
Chemical
Thermodynamics
For a constant-temperature process:
Gibbs free energy (G)