everything ACCEPT how fast, nothing about kinetics
first law of thermodynamics
conservation of energy-energy cant be created or destroyed
second law of thermodynamics
for a spontaneous process, entroby (Δs) of the universe increases
for a spontaneous process, ΔSuniverse increases, greater than 0
for a spontaneous process, Δsystem +Δsurrounding is positive-one can be negative while the other positive, but net has to be positive
third law of thermodynamics
a perfect crystal has NO entropy at 0 kelvin. absolute 0 is k=0 or c= –273, cant get lower than this!
gas in a piston
have cylinder with gas and put piston in that seals it, piston can move down or up.
locked piston in place-if you transfer heat, temperature goes up, so pressure goes up
if you take heat away, you decrease the pressure
what happens to expanding gases?
expanding gasses cool down. canned air(to clean computer) gets very cold. gas thats condensed under high pressure, is released and gets cool. if you take expanded gas and fit it in the can, it heats up
equation of first law of thermodynamics
ΔE is the change in internal energy
q is the change in heat
w is the change in work
q and w are NOT state functions, ΔE is a state function--it only depends on initial and final state
when gas expands, it loses energy on surrounding so its w is negative, cuz gas did work on surrounding
when you compress a gas, surrounding is doing work giving gas more energy, so work is positive
change in pressure is 0
no change in volume. if Δv=0, no work is done cuz w= –PΔV, so ΔE=q
ΔT=0. if two gases are at same temp, they have same kinetic energy--internal energy is stored as kinetic energy. if two gases are at same temp than 2 gases have same internal energy
ΔE=0, so q= -w heat change, in or out is doing work.
heat stayes constant-insulate q=0
temperature can change-have gas in container in super insulated so heat doesnt change, but gas expands so the gas cooled down so temp is lower, but heat stayed the same, so internal energy is lower.
single biggest indicator of Δs is by focusing in on gases
entropy will be increased cuz there was an increase in the moles of GAS, more gas, more disorder. ΔS greater than 0
for Δs to be 0, same moles of gas should be on both sides.
phase changes and entropy
when you go from solid to liquid to gas, entropy increases
when you go from gas to ligquid to solid entropy decreases
ΔH-change in heat-state function , q is heat but NOT state function
when + endothermic, needs to absorb heat to happen
when - exothermic-gives off heat
what are the three ways to calculate ΔH of a reaction?
1. bond enthalpies
2. enthalpies of formation
3. hess's law
break bond, endothermic(reactant) make bond, exothermic(products), cuz electrons go to lower energy
enthalpy of formation
ΔΗ°rxn=ΣnΔH°f1prod – ΣnΔH°f1react
ΔH of formation-amount of heat it takes to form exactly one mol of a single product from its individual elements in their standard states
elements in elemental form=0
formation reactions form 1 mol from its individual elements in their standard states
diatomic elements-Never Have Fear Of Ice Cold Beer=O2, N2 etc.
gibbs free energy
ΔG-energy thats free, or available to do work
have system that has 100kj of energy thats free to do work.next day and it only has 50kj, lost free energy and so it must have done work and something spontaneous happens. to gain G or free energy, someone needs to do work on them and need outside influence
thins are only spontaneous in ONE direction, spontaneous is something that happened without outside influence
ΔG= – spontaneous
ΔG=0 at equilibrium
ΔG°rxn=ΣnΔG°f1prod – ΣnΔG°f1react
if your conditions arent standard conditions, what ΔG?
Q is products/reactants reaction quotient
if reactions=products and Q=1, ln(1)=0.
get ΔG° from book.
if reaction reaches equilibrium, ΔG=0
ΔG°less than 0 spontaneous under standard conditions, so at equilibrium, more product wil be created, so Keq should be greater than1
ΔG° greater than 0 its spontaneous in the opposite direction and end up with more reactants than productsduring equilibrium and Keq is less than 1
for a reaction at equilibrium, ΔG=0, and Keq=Q
what equation relates ΔG to ΔS to ΔH
ΔG=ΔH-TΔS "get higher test scores"
for spontaneous ΔG is negative
to get spontaneous, want ΔH to be negative.
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