Download 7.7 The Laws of Thermodynamics and How They Relate to the

7.7 The Laws of Thermodynamics and How They Relate to
the Spontaneity of a Reaction
Spontaneous reactions are those which proceed in a forward direction continuously,
once they have been given the right amount of energy (the activation energy). For
example once you light a candle it will continue to burn until all the wax and wick is
gone.
Spontaneous reactions can be both endothermic (cold packs) and exothermic
(combustion of a fuel), and non-spontaneous reactions can be can be endothermic
(making H2 and O2 from water) or exothermic.
To understand why some reactions are spontaneous and other are not we need to
understand the three laws of thermodynamics and how they relate to enthalpy H,
entropy S and Gibbs Free energy G
The First Law – The Law of Conservation of Energy
The Total energy of the universe is constant so that energy can not be created or
destroyed, but it can be converted from one form to another (P.E. to K.E.) or from
one place or system to another (reaction vessel to surroundings)
So if a reaction gains energy, then the energy it gained must have been lost from
somewhere else, so that the total energy of the universe is the same – Cold packs work
by the dissolving of a salt that requires energy and the surrounding water loses energy.
The total energy has remained the same.
Hess’s Law follows the first law of conservation of energy. The sum of the enthalpies of
a series of steps of reactions will equal the enthalpy of the overall reaction. Enthalpy
(H) is related to bond energy. The activation energy is the minimum energy required
break the intermolecular forces (bonds) of the reactant so that it can form the activated
intermediate. The difference between this energy and the bond energy released when
the new bonds are made to form the product gives us the enthalpy of the reaction.
Entropy S
Entropy is a measurement of the disorder of a system. Reactions that give greater
disorder to a system tend to be favoured over those which give order to a system. For
example when an organized crystal of salt is dissolved in water it will lose order and
become more disorganized. For most dissolution processes S of products is greater
than S of reactants so S = Sproducts – Sreactants is a positive value. The greater the
increase in disorder S the more likely the reaction is to be spontaneous.
Second Law of Thermodynamics
All changes will either directly or indirectly increase the entropy disorder of the
universe. Suniverse will be greater than zero.
In terms of statistics if you picked 52 cards up from the floor there is a greater
probability that they will be organized randomly than in order following their suits. To
put them in order (organize them) will require energy. Any reaction that can lead to an
increase in disorder is more likely to be spontaneous.
Increases in entropy occur when
1) a volume of a gas is increased
2) the temperature of a system is increased
3) A change of state occurs. (entropy increases from solid, to liquid to gas)
4) When a molecule decomposed or combusts into smaller molecules (N 2O4  2NO2)
5) When a chemical reaction produces a product with a greater state of entropy (for
example a solid produces a gas) HCl + Zn  ZnCl2 + H2
A system can become more ordered during a reaction provided there is a greater
increase in diorder of the surroundings as it occurs.
Suniverse = Ssystem + Ssurroundings
In cellular respiration there is a great amount of energy released, only a portion of which
is used to create order in building new molecule (like proteins and DNA) This is why
energy food chains always decrease as you move up them.
Gibb’s Free Energy
Gibb’s free energy is the energy available to do work. If energy to do work is less in
products (final free energy) than reactants (initial free energy) then the reaction is
spontaneous. G = Gfinal – Ginitial where G is less than zero. Or more simply
if G is less than zero the reaction is spontaneous.
The Gibbs Free energy of a reaction can be calculated using the:
Gibbs-Helmholtz Equation
G = H - TS
Where G is Gibbs free energy (kJ)
H is the enthalpy change (kJ)
T is temperature in Kelvin (K)
S is the change in entropy (kJ/K)
Note that G is dependent on three factors and each one has an influence on
whether the free energy will be negative and give a spontaneous reaction.
H -the more exothermic a reaction is the more likely it is to be spontaneous
temperature - the greater the temperature the more likely the reaction will occur
spontaneously, provided the entropy is a positive value
S the more disorder the reaction creates the more likely it is to be spontaneous.
General trends for G = H - TS
If H is negative (exothermic) and S is positive (increases disorder) then the reaction
will be spontaneous.
If H is positive (endothermic) and S is negative (increases order) then the reaction
will be not be spontaneous.
If H is negative (exothermic) and S is negative (increases order) or if H is positive
(endothermic) and S is positive (increases disorder) then the reaction is dependent on
the temperature.
Consider a pot of water. To change water from liquid to gas requires energy
(endothermic with H positive) but the gas has greater disorder than the liquid (S
positive). At lower temperatures H  TS making G positive so the reaction will
be......
At higher temperatures H  TS making G negative so the reaction will be......
Knowing the temperature that water boils at tell us when H  TS which is when the
system is at equilibrium. (There will both liquid and gas in equilibrium) For a system at
equilibrium G will be zero. By manipulating the equation the T or temperature of
boiling can be estimated.
T=
At absolute zero TS is zero so G = H This follows the third law of
thermodynamics:
The entropy of a perfectly ordered crystalline substance is zero at absolute zero.
or S = 0 at T = O K
Standard entropy changes can be calculated using a similar equation to the one used to
find enthalpy changes.
S = (n ·Sproducts)  (n · Sreactants)
The entropy is the amount of disorder of the reactants and products. It is alway a
positive value.
The units of the standard entropy values are J/(molK)
Note that if the Free energy of a reaction is positive it is not spontaneous but the
reverse reaction will have a negative free energy and it will be spontaneous.
Question:
Use the equation for Standard entropy changes and standard enthalpy changes to
calculate the Standard Gibbs free energy for the hydrogenation of ethene at SATP
1) Write the chemical equation
2) Calculate the standard entropy
S = (n ·Sproducts)  (n · Sreactants)
3) Calculate the standard enthalpy
H = (n ·Hf products)  (n · Hf reactants)
4) Substitute the values for S and H into the Gibbs-Helmholtz Equation
G = H - TS
2. Use the equation for Standard entropy changes and standard enthalpy changes to
estimate the condensation point of hydrogen peroxide. Remember G = 0 when at bpt
and mpt. S for H2O2(l) = +109.6 J/(molK) , S for H2O2(g) = +233 J/(molK) and
Hf for H2O2(l) = 187.8 kJ/mol and Hf for H2O2(g) = -136.3 kJ/mol
1) Write the equilibrium equation
2) Calculate the standard entropy
S = (n ·Sproducts)  (n · Sreactants)
3) Calculate the standard enthalpy
H = (n ·Hf products)  (n · Hf reactants)
4) Substitute the values for S and H into the Gibbs-Helmholtz Equation making
G equal to zero and solving for T.
G = H - TS
Homework p 511 # 3, 4 and p 512 # 1, 2, 3, 5, 6, 8, 9, 11