Ch 17 - Thermodynamics Flashcards
entropy
the dispersion(spreading out) of energy
in our universe entropy always
increases
heat tax
an unavoidable cut of energy is taken from ever transaction by nature
- can not be 100% efficient
in general, the most efficient use of energy occurs with
the smallest number of transactions
there is always a loss of energy as
heat
fundamental goal of thermodynamics is to
predict spontaneity
spontaneous process
a process that occurs without ongoing outside intervention(such as the performance of work by some external force)
spontaneity of a process does not depend on
the speed of a reaction
Entropy(S)
a thermodynamic function that increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state
- S = k(lnW) - K = 1.38*10^-23J/K - w = number of possible microstates
the more microstates present the greater entropy
more energetically equivalent ways to arrange the components of the system
the state with the highest entropy also has the greatest
dispersal of energy
second law of thermodynamics
for any spontaneous process, the entropy of the universe increases(deltaSuniverse > 0)
the criterion for spontaneity is the
entropy of the universe
the change in entropy is the entropy of the final state minus the initial
deltaS = Sfinal – Sinitial
solids -> gas
entropy increases as state becomes more fluid
even though the entropy of water decreases during freezing the entropy of the universe
increases
deltaSuniv =
deltaSsys+ deltaSsurr
the entropy of a system can decrease(deltaSsys < 0) as long as the entropy of the surroundings increases by a greater amount
deltaSsurr > -deltaSsys
according to 2nd law of thermodynamics the entropy of the universe must increase for a process to be spontaneous
- deltaSuniv > 0
- overall net entropy of the universe has a net increase
the release of heat energy by the system disperses that energy into the surroundings,
increasing the entropy of the surroundings
water freezing becomes more ordered but is spontaneous
the heat given off to the surroundings increases the entropy to a sufficient degree to overcome the entropy decrease in the water
An exothermic process increases
the entropy of the surroundings
an endothermic process decreases
the entropy of the surroundings
a process that emits heat into the surroundings(Qsys negative) increases
the entropy of the surroundings(+ deltaSsurr)
a process that absorbs heat from the surroundings(Qsys positive) decreases
the entropy of the surroundings(- deltaSsurr)
the magnitude of the change in entropy of the surroundings is
proportional to the magnitude of Qsys
in general the higher the temperature the lower the
magnitude of deltaSsurr(inverse relationship)
-Qsys = Qsurr as long as there is
constant temperature and pressure
as temperature increases in exothermic process the spontaneity tends to
decrease and being exothermic becomes less of a determining factor for spontaneity as temperature increases
at constant pressure Qsys = deltaHsys
deltaSsurr = (-deltaHsys)/T
- constant P,T
Gibbs free energy(G)
- G = H – TS
- H = enthalpy
- T = temp in K
- S = entropy
- deltaG = deltaH – T(deltaS)
the change in Gibbs free energy for a process occurring at constant temperature and pressure is proportional to
the negative deltaSuniv
deltaG is also a criterion for spontaneity
although opposite in sign
Gibbs free energy is also called chemical potential because
it is analogous to mechanical potential
- chemical systems tend toward lower Gibbs free energy(lower chemical potential)
deltaG is proportional to the
negative of deltaSuniv
a decrease in Gibbs free energy(deltaG<0) corresponds to
a spontaneous process
an increase in Gibbs free energy(deltaG>0) corresponds to
a nonspontaneous process
to calculate Gibbs free energy you only need to determine the entropy of the system(deltaSsys) and the change in enthalpy of the system(deltaHsys)
combined you can calculate deltaG = deltaH – T(deltaSsys)
delta H(-), delta S(+)
- Low temp(delta G < 0)
- High Temp(delta G < 0)
- 2N2O(g) -> 2N2(g) + O2(g)
delta H(+), delta S(-)
- Low temp(delta G >0) = Nonspontaneous(delta G > 0)
- High Temp(delta G >0) = Nonspontaneous(delta G > 0)
- 3O2(g) -> 2O3(g)
delta H(-), delta S(-)
- Low temp(delta G < 0)
- High Temp(delta G >0) = Nonspontaneous(delta G > 0)
- H2O(l) -> H2O(s)
delta H(+), delta S(+)
- Low temp(delta G >0) = Nonspontaneous(delta G >0)
- High Temp(delta G < 0)
- H2O(l) -> H2O(g)
when deltaH and deltaS have opposite signs the spontaneity of the reaction
does not depend on temperature
when deltaH and deltaS have the same sign the spontaneity
depends on temperature
the point where delta G changes sign is where the reaction
changes from spontaneous to nonspontaneous
standard state definitions:
Gas
the standard state is the pure gas at a pressure of exactly 1atm
standard state definitions:
Solid and Liquid
the standard state in pure substance force is pressure of 1atm and at the temperature of interest
standard state definitions:
Solutions
standard state is 1M
standard entropy change for a reaction(deltaSrxn)
the change of entropy for a process in which all reactants and products are in their standard states
- deltaSrxn = Sproducts – Sreactants - standard molar entopies(S) are used in calculating deltaSrxn
for entropy there is
an absolute zero
third law of thermodynamics
the entropy of a perfect crystal at absolute zero(0K) is zero
standard entropy values are listed in
J/mol*K
entropy is an extensive property
depends on the amount of the substance
allotrope
a substance existing in two or more forms within the same state
5 factors on the number of places to put energy and the standard entropy
- state of substance
- molar mass
- particular allotrope
- molecular complexity
- extent of dissolution
the entropy of a gas is
generally higher than a liquid than a solid
the greater the molar mass the
greater the entropy in the same state
the less constrained the allotrope results in
more places to put energy and therefore greater entropy(graphite has more entropy than diamond)
for a given state of matter entropy generally
increases with increasing molecular complexity
the dissolution of a crystalline solid into solution usually results in
an increase in entropy become of greater energy dispersal
to calculate deltaSrxn
subtract the standard entropies if the reactants multiplied by their stoichiometric coefficients from the standard entropies of the products multipled by their stoichiometric coefficients
deltaSrxn =
the sum of nS(products) – sum of nS(reactants)
unlike enthalpies of formation which are 0 for elements in their standard states
entropies are always nonzero at 25C.
standard change in free energy(deltaGrxn)
the criterion for spontaneity at standard conditions
the more negative deltaGrxn the more spontaneous the process is
the further it will go toward products to reach equilibrium
using deltaGrxn = delta Hrxn – T(deltaSrxn)
- deltaHrxn = H,productsmoles – H,reactantsmoles
- deltaSrxn = S,productsmoles – S,reactantsmoles
- T in Kelvin
using tabulated values of free energies of formation
- free energy of formation(deltaGf) – the free energy of formation(deltaGf) is the change in free energy when 1 mol of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is 0.
- deltaGrxn = molesdeltaGf(products) – molesdeltaGf(reactants)
free energy of formation(deltaGf)
the free energy of formation(deltaGf) is the change in free energy when 1 mol of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is 0.
deltaGrxn is a stepwise reaction so you can combine reverse and switch parts as long as you
multiply the deltaGrxn by the appropriate factor
free energy is the amount of energy available
to do work
the deltaGrxn is the theoretical maximum amount of free energy available to do work
in reality the free energy is even less because additional energy is lost to the surroundings as heat
reversible reaction
a reaction that achieve the theoretical limit with respect to free energy
- occurs infinitesimally slowly and the free energy can only be drawn out in infinitesimally small increments that exactly match the amount of energy that the reaction is producing during that increment
irreversible reactions
all real reactions do not achieve the theoretical limit of available free energy
if the change in free energy of a chemical reaction is positive then
deltaGrxn represents the minimum amount of energy required to make the reaction occur
deltaGrxn represents a theoretical limit
making a real nonspontaneous reaction occur always required more energy than the theoretical limit
ordinary life conditions are not
standard conditions
deltaGrxn(degree symbol) only applies at
standard conditions
for nonstandard conditions:
deltaGrxn = deltaGrxn(degree sign) + RT(lnQ)
- Q = reaction quotient(partial pressure in atm)
- T = kelvin
- R = gas constant(8.314 J/mol*K)
under standard conditions Q always equals 1
ln1 = 0 so its ignored
under equilibrium conditions the value of RT(lnQ) is always equal in magnitude but opposite in sign to the value of deltaGrxn(degree sign)
- making deltaGrxn = 0
- nonspontaneous in either direction of the reaction
deltaGrxn<0 is spontaneous in
the forward direction
deltaGrxn>0 is spontaneous in
the reverse direction
the equilibrium constant becomes larger as
the standard free energy change becomes more negative
large negative free energy change = large equilibrium constant =
product strongly favored at equilibrium
large positive free energy change = small equilibrium constant =
reactants strongly favored at equilibrium
deltaGrxn(degree sign) = -RT(lnK)
use Kp for reactions involving gases and use Kc for reactions involving substances dissolved in solution
K< 1, lnK is negative and deltaGrxn(degree sign) is positive
under standard conditions(Q=1) the reaction is spontaneous in the reverse direction
K> 1, lnK is positive and deltaGrxn(degree sign) is negative
under standard conditions(Q=1) the reaction is spontaneous in the forward direction
K= 1, lnK is zero and deltaGrxn(degree sign) is zero
under standard conditions(Q=1) the reaction is in equilibrium
lnK = (-deltaHrxn/R)(1/T)+(deltaSrxn/R)
ln(K2/K1)=(-deltaHrxn/R)(1/T2-1/T1)