Chemistry Thermodynamics HomeWork

General Summary of Thermodynamics

FIRST LAW OF THERMODYNAMICS: The total amount of energy (and

mass) in the universe is constant.

That is, in any process energy can

be changed from one form to

another; but, it can never be

created nor destroyed.

“ You can’t get something for

nothing”

SECOND LAW OF THERMODYNAMICS: In any spontaneous process the

entropy of the universe increases:

Suniverse = Ssystem + Ssurroundings or

Suniverse -Ssurroundings = Ssystem

(Variant)In trying to do work, you

always lose energy to the

surroundings.

“You can’t even break even!”

THIRD LAW OF THERMODYNAMICS: Any pure crystalline substance at

a temperature of absolute zero

(0.0K) has an entropy of zero

(S = 0.0 J/K-mol).

Terminology:

Energy = capacity to do work

System = portion of the universe we are considering

Open system = energy and matter can transfer

Closed system = energy transfers

Isolated system = no transfers

Surroundings = everything else besides the system

Isothermal = system at constant temperature

Heat capacity = amt. of heat required to raise the temperature of a

certain amt. of material by 1C or 1K.

Calorie = amt. of heat required to raise the temperature of 1g of water

by 1ºC.

Signs:

H >0 or (+) heat absorbed (endo)

H <0 or (-) heat released (exo)

S >0 or (+) entropy increasing (becoming disordered)

S <0 or (-) entropy decreasing (becoming ordered)

G >0 or (+) nonspontaneous Kc <1

G = 0, Kc = 1

G < 0 or (-) spontaneous Kc >1

Rules about Entropy: (Entropy increases) 1. w/ increasing temperature* 2. as one goes from s -> l -> aq ->g* 3. if a solid or liquid is dissolved in a solvent* 4. number of particles increases* 5. mass of the molecule increases 6. Entropy is higher for weakly bonded materials than for strong

covalent materials

7. As complexity of a molecule increases.

FORMULA’S:

G = H- TS (T in K = 273 + C) T = H/S (assume G = 0 or when Kc = 1, like fusion/vaporization)

Hrxn =  (#mol.) *(Hf(products)) -  (#mol.) *(Hf(reactants)) = kJ

Srxn =  (#mol.) *(S(products)) -  (#mol.) *(S(reactants) ) = J/K (Watch Out -> J to kJ)

Grxn =  (#mol.) *(Gf(products)) -  (#mol.) *(Gf(reactants)) = kJ

G = -RTlnKc or Kc = e -(G/RT)

Remember: e x is 2

nd function natural log (ln) on calculator and work inside-out

H S G

+ + (+/-) (Spont. Only at High Temp. when TS > H)

- + - (Spontaneous at ALL Temperatures) - - (+/-) (Spont. Only at Low Temp. when TS < H)

+ - + (Non-spontaneous at ALL Temperatures)

G = H- TS when G is – gives a spontaneous reaction

Enthalpy (H): The Energy of motion or transition. q is the heat measured from a reaction. If rxn is at constant pressure q = H

q is measured experimentally by calorimetry where qsystem = (-)qsurroundings

At constant Pressure: Heat (q) = specific heat x mass x T , here you maybe calc qsys, if qsystem = positive number (absorbing heat) it comes from the surroundings where qsystem = (-)qsurroundings

Hrxn per mole of reactant (kJ/mol) = qrxn / mol of reactant

(H°f of elements in stable/natural/elemental form = 0 kJ/mol, units of Hf = kJ/mole)

Hess' Law: The addition of several reactions to obtain a "desired" overall reaction. If a reaction

is reversed, you must "flip" the sign on H. If a reaction is multiplied by a coefficient, then also

multiple H by the coefficient. Cancel out intermediates that appear on both sides of the equation and do not appear in the "desired" equation.

A + B  2C H1 A + B  2C H1

C  D H2 2 x [C  D] 2 x H2

A + B  2 D Hrxn = ? A + B  2 D Hrxn = H1 + (2 x H2)

Enthalpy is a stoichiometric quantity: The amount of heat is proportional to the number of moles of reactants/products.

Entropy: The measure of disorder of the system.

Entropy can NOT be experimentally measured directly.

Ways to predict sign of S rxn without calculations: 1

st , look for changes in physical states (s, l, aq, g) from reactants to products.

2 nd

, look for changes in the number of moles of reactants to moles of products.

Standard States Conditions: Solutions: 1 M; Partial Pressures: 1 atm; Temperatures are generally at 25°C (298 K)

Gibbs' Free Energy: Amount of Energy left over from H and TS to do work & entropy is the “price” one must pay to do work.

Enthalpy, Entropy, and Gibb's Free Energy are all state functions. State functions depend only on the final and initial states of a process.

So also calculate Hrxn, S°rxn, & G°rxn as final state (products) - initial state (reactants).

Consider non-equilibrium conditions:

One can calculate Gibb's Free Energy for non-equilibrium conditions using Q from Equilibium

Chapter: Grxn = G°rxn + RT ln(Q)

All spontaneous reactions (Grxn = - ) move toward equilibrium. As the reaction proceeds, G becomes less positive until it reaches zero-equilibrium.

At equilibrium G=0 and G°rxn = - RT ln(K) (where R = 8.314 J/(Kmol) and K = equil.

constant (WATCH OUT for units: R has units Joules, but G°rxn must be in kJ’s,

where 1000 J = 1 kJ.