CHEM223-Up to 1st midterm.doc
Course CodeCHEM 223
Chem 223 –Up to 1 st
# moles water delta H = Q
Q consumed = -Q cooled Delta E = 0
Isolated system Delta E = 0
Q(Au) = - Q(H20)
nAu (Cp,m) deltaT Au = -nH20 (Cp,m) deltaT H20
Cp = Cv + TV (B2/K) B= 1/V (dV/dT)p
The above problems are assignment #1 questions.
Quarternization – two types
1. Cardboard box ⇓ Box can be described as an expanding or contracting gas
The particle in a box is not done classically. In order for waves to exist, the boundary limits
from a◊ b disregarding the path of the waves.
The energy levels of the particle that it can take on. Particle in a box can only have discrete
n is the energy level. A is the length of the box.
Energy level spacing goes further and further away in nature.
What happens to energy levels when you make the size of the box double or half it used to
Translational motion not described
2. Harmonic Oscillator ⇓ depend more on Bond Order, and large molecules.
Graph: Energy vs. l = distance between atoms <<< graph has lines equally spaced.
E = (n+1/2)hv ⇓ photon? Since n is not squred, the energy level is going to be evenly
spaced. v = vibrational constant
The labelling of lines on graph is a way of numbering energy levels
v = , k = force constant and m = mass. Reduced mass = greater vibration
Closer and further ⇓ Atom vibrates, the idea of the harmonic oscillator
Lower energy level ⇓ vibrates a little across close distance
Higher energy level ⇓ vibrates across longer distance
The above equations are all about vibrations, not translational motion. They contribute to
heat capacity too.
Translational ⇓ 3/2R, R is the gas constant [J/mol-K]
Rotational ⇓ 0 (monatomic), Linear = R. 3/2R (non-linear) If Cp, +R
Vibrational ⇓ 0 (monatomic), Linear = (3n-5)R, (3n-6)R = non-linear All are Cv
N = # of atoms. Assumes that all vibrational levels are active for linear and non-linear. HT-
limit. Calculate the vibrational contribution to heat capacity? SO2 example on lecture notes.
Heat capacity in SO2(non-linear) Cp = 39.8
Cp – 4R(vibrational part only)/(3n-6)-4R = Vibrational contribution
4R – assuming all levels are on. 4R = Translational 3/2R + Non-linear rotational 3/2R
29th September, 2010.
Work done on the system – LHS
W1 = -P1V1 = integrate P1 dV1 from V1 to 0 = P1V1
P2 on the surroundings - RHS
W2=-P2V2 = integrate P2 dV2 from 0 to V2 = -P1V1
WLHS + W RHS = P1V1 – P2V2
Adiabatic = q = 0
E = q + W = o + w = P1V1 – P2V2
So that E2 – E1 = P1V1 – P2V2
H = E +PV
H1=H2 (Process going on in constant enthalpy)
One can calculate the Joule-Thompson coefficient
Mu = (dT/dP) at constant H ⇓ Enthalpy
At low P to the left of point 4, mu is +ve or T decreases on expansion (gas cools) The
energy required to overcome the attraction interactions between atoms or molecules is
obtained from the kinetic(thermal) energy of the particles, and the temperature falls.
Joule-Thompson coefficient provides information for the change of enthalpy with respect to
For an ideal gas, the E = q + w will depend on the work that’s being done.
External pressure is 0, therefore no work is being done, thus no q (energy change)
At constant pressure, starting out with gas at some pressure because being compressed.
When expanding at constant external pressure, it’s not doing the maximum amount of
work, because compressed as a gas, it’s expanding as a force higher than the opposing
force, therefore it’s not doing its max work.
Opposing pressure = internal pressure. Pressures equalize. PV = 0. Pressure falls, no
At the point when expansion stops, we say that q is reversible and the maximum amount of
work can be done at that particular temperature.
A gas will spontaneously expand, as it does...the driving force of expansion falls as internal
pressure falls. This process is purely energetic.
For example, the E=mgh. All that energy falls and went into heat. However, the driving
force does not change during the falling process. If a counter weight contains a bulk of
sand. As the sand is put away, the opposing weight falls.
A gas as change of pressure. This also applies to concentration. The reason why it’s
expanding is because there’s no energy change. Driving force weakens as expanding takes
place. The bond of breaking energy goes on, it doesn’t change very much. The dropping
driving force does change, because as reaction proceeds, the concentrations/ partial
pressures are falling more and more, the entropy goes higher and higher as they dilute.
Entropy is getting lower and lower, becoming more ordered. Initially it may favour bonding
forming and breaking. As process proceeds, it becomes more and more unfavourable.
We try to do max work, but for it to occur, the opposing pressure has to back off and
accommodate the fact that the internal pressure is falling as expanding occurs.
There’s no useful work as ideal gas expands ◊ w useful = 0
As compression goes, external pressure increases and internal (opposing) pressure
decreases, therefore q is measured reversiblily = qrev(T delta S) q rev = 0 if all is
converted into work. Increasing energy = increasing enthalpy
1st October, 2010.
q = Pex delta V
Expansion V = +ve Wexp = -ve
Q rev = int Pext dV from V to V2 = nRT integrate dV/V from V1 to V2 = nRT ln(V2/V1)
Qrev = T delta S
(P vs V) Isothermal and adiabatic expansion PV=nRT
Adiabatic is the lower curve and Isothermal is the higher one.
As pressure and volume falls, temperature falls. The Isothermal curve is going to have more
Pay attention to A2 Question 3
Delta H = delta E + delta(PV)
Delta H = delta E + P delta V + V delta P
= q + wuseful – Pex delta V /- P delta V + V delta P
During heating or a chemical reaction, P will remain constant if V is allowed to change to
maintain P=Pexternal so that
Delta H = delta E + Pext delta V
Wexp (by the system) = Pext delta V
If reaction is carried out irreversibly, delta H = q irr
½ N2(g) + 3/2 H2(g) ◊ NH3 (g)
Delta H 298K = -45.9 KJ/mol
Delta H 298K = delta E 298K + delta(PV)